Does an increasing sequence of reals converge if the difference of consecutive terms approaches zero?Bounded...
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Does an increasing sequence of reals converge if the difference of consecutive terms approaches zero?
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Does an increasing sequence of reals converge if the difference of consecutive terms approaches zero?
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If $a_n$ is a sequence such that $$a_1 leq a_2 leq a_3 leq ...$$
and has the property that $space$$a_{n+1}-a_n longrightarrow 0$,
Then can we conclude that $a_n$ is convergent?
$$space$$
I know that without the condition that the sequence is increasing, this is not true, as we could consider the sequence given in this answer to a similar question that does not require the sequence to be increasing: $space$ https://math.stackexchange.com/a/1437395/625467
$0, 1, frac12, 0, frac13, frac23, 1, frac34, frac12, frac14, 0, frac15, frac25, frac35, frac45, 1...$
This oscillates between $0$ and $1$, while the difference of consecutive terms approaches $0$ since the difference is always of the form $pmfrac1m$ and $m$ increases the further we go in this sequence.
So how can we use the condition that $a_n$ is increasing to show that $a_n$ must converge? Or is this still not sufficient?
real-analysis sequences-and-series convergence
edited 17 hours ago
José Carlos Santos
162k22130233
asked yesterday
M DM D
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add a comment |
$begingroup$
If $a_n$ is a sequence such that $$a_1 leq a_2 leq a_3 leq ...$$
and has the property that $space$$a_{n+1}-a_n longrightarrow 0$,
Then can we conclude that $a_n$ is convergent?
$$space$$
I know that without the condition that the sequence is increasing, this is not true, as we could consider the sequence given in this answer to a similar question that does not require the sequence to be increasing: $space$ https://math.stackexchange.com/a/1437395/625467
$0, 1, frac12, 0, frac13, frac23, 1, frac34, frac12, frac14, 0, frac15, frac25, frac35, frac45, 1...$
This oscillates between $0$ and $1$, while the difference of consecutive terms approaches $0$ since the difference is always of the form $pmfrac1m$ and $m$ increases the further we go in this sequence.
So how can we use the condition that $a_n$ is increasing to show that $a_n$ must converge? Or is this still not sufficient?
real-analysis sequences-and-series convergence
edited 17 hours ago
José Carlos Santos
162k22130233
asked yesterday
M DM D
1129
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M D is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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3
$begingroup$
Note that while your sequence goes up and down periodically, you could define another sequence with the same step length for each $n$ but with all steps positive. That would be a counterexample to your question.
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– Adayah
yesterday
5
$begingroup$
Have you tried logarithms?
$endgroup$
– Mehrdad
yesterday
1
$begingroup$
Note that by writing $b_1=a_1, b_2=a_2-a_1, b_3=a_3-a_2,...$ the question becomes equivalent to asking whether a positive series with a summation term tending to zero must converge.
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– Bar Alon
21 hours ago
add a comment |
$begingroup$
If $a_n$ is a sequence such that $$a_1 leq a_2 leq a_3 leq ...$$
and has the property that $space$$a_{n+1}-a_n longrightarrow 0$,
Then can we conclude that $a_n$ is convergent?
$$space$$
I know that without the condition that the sequence is increasing, this is not true, as we could consider the sequence given in this answer to a similar question that does not require the sequence to be increasing: $space$ https://math.stackexchange.com/a/1437395/625467
$0, 1, frac12, 0, frac13, frac23, 1, frac34, frac12, frac14, 0, frac15, frac25, frac35, frac45, 1...$
This oscillates between $0$ and $1$, while the difference of consecutive terms approaches $0$ since the difference is always of the form $pmfrac1m$ and $m$ increases the further we go in this sequence.
So how can we use the condition that $a_n$ is increasing to show that $a_n$ must converge? Or is this still not sufficient?
real-analysis sequences-and-series convergence
edited 17 hours ago
José Carlos Santos
162k22130233
asked yesterday
M DM D
1129
New contributor
M D is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
If $a_n$ is a sequence such that $$a_1 leq a_2 leq a_3 leq ...$$
and has the property that $space$$a_{n+1}-a_n longrightarrow 0$,
Then can we conclude that $a_n$ is convergent?
$$space$$
I know that without the condition that the sequence is increasing, this is not true, as we could consider the sequence given in this answer to a similar question that does not require the sequence to be increasing: $space$ https://math.stackexchange.com/a/1437395/625467
$0, 1, frac12, 0, frac13, frac23, 1, frac34, frac12, frac14, 0, frac15, frac25, frac35, frac45, 1...$
This oscillates between $0$ and $1$, while the difference of consecutive terms approaches $0$ since the difference is always of the form $pmfrac1m$ and $m$ increases the further we go in this sequence.
So how can we use the condition that $a_n$ is increasing to show that $a_n$ must converge? Or is this still not sufficient?
real-analysis sequences-and-series convergence
real-analysis sequences-and-series convergence
edited 17 hours ago
José Carlos Santos
162k22130233
asked yesterday
M DM D
1129
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edited 17 hours ago
José Carlos Santos
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asked yesterday
M DM D
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edited 17 hours ago
José Carlos Santos
162k22130233
edited 17 hours ago
José Carlos Santos
162k22130233
edited 17 hours ago
José Carlos Santos
162k22130233
162k22130233
asked yesterday
M DM D
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asked yesterday
M DM D
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asked yesterday
M DM D
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1129
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3
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Note that while your sequence goes up and down periodically, you could define another sequence with the same step length for each $n$ but with all steps positive. That would be a counterexample to your question.
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– Adayah
yesterday
5
$begingroup$
Have you tried logarithms?
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– Mehrdad
yesterday
1
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Note that by writing $b_1=a_1, b_2=a_2-a_1, b_3=a_3-a_2,...$ the question becomes equivalent to asking whether a positive series with a summation term tending to zero must converge.
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– Bar Alon
21 hours ago
add a comment |
3
$begingroup$
Note that while your sequence goes up and down periodically, you could define another sequence with the same step length for each $n$ but with all steps positive. That would be a counterexample to your question.
$endgroup$
– Adayah
yesterday
5
$begingroup$
Have you tried logarithms?
$endgroup$
– Mehrdad
yesterday
1
$begingroup$
Note that by writing $b_1=a_1, b_2=a_2-a_1, b_3=a_3-a_2,...$ the question becomes equivalent to asking whether a positive series with a summation term tending to zero must converge.
$endgroup$
– Bar Alon
21 hours ago
3
3
$begingroup$
Note that while your sequence goes up and down periodically, you could define another sequence with the same step length for each $n$ but with all steps positive. That would be a counterexample to your question.
$endgroup$
– Adayah
yesterday
$begingroup$
Note that while your sequence goes up and down periodically, you could define another sequence with the same step length for each $n$ but with all steps positive. That would be a counterexample to your question.
$endgroup$
– Adayah
yesterday
5
5
$begingroup$
Have you tried logarithms?
$endgroup$
– Mehrdad
yesterday
$begingroup$
Have you tried logarithms?
$endgroup$
– Mehrdad
yesterday
1
1
$begingroup$
Note that by writing $b_1=a_1, b_2=a_2-a_1, b_3=a_3-a_2,...$ the question becomes equivalent to asking whether a positive series with a summation term tending to zero must converge.
$endgroup$
– Bar Alon
21 hours ago
$begingroup$
Note that by writing $b_1=a_1, b_2=a_2-a_1, b_3=a_3-a_2,...$ the question becomes equivalent to asking whether a positive series with a summation term tending to zero must converge.
$endgroup$
– Bar Alon
21 hours ago
add a comment |
6 Answers
6
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votes
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No. Just consider the case in which $a_n=1+frac12+frac13+cdots+frac1n$. Note that then we would have$$lim_{ntoinfty}a_{n+1}-a_n=lim_{ntoinfty}frac1{n+1}=0.$$
answered yesterday
José Carlos SantosJosé Carlos Santos
162k22130233
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Rhys: His sequence is $1, frac32, frac{11}6, frac{25}{12},...$. Essentially the sequence of partial sums associated with the harmonic series. This is still a sequence, not a series, since it is a finite sum.
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– M D
yesterday
5
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@RhysHughes $a_n=1+frac12+cdots+frac1n$ IS increasing and $a_{n+1}-a_n=frac{1}{n+1}to 0$.
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– Robert Z
yesterday
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I see now. Thanks, it just confused me.
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– Rhys Hughes
yesterday
5
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I think adding an explanation from comments into the answer is worth considering and would benefit to the quality of an answer.
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– Ister
yesterday
2
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@wizzwizz4 No, because $a_n neq frac{1}{n}$. Instead, $a_n = 1 + frac{1}{2} + ldots + frac{1}{n}$. So, $$a_{n+1} - a_n = left(1 + frac{1}{2} + ldots + frac{1}{n} + frac{1}{n+1}right) - left(1 + frac{1}{2} + ldots + frac{1}{n}right) = frac{1}{n+1}.$$
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– Theo Bendit
12 hours ago
|
show 3 more comments
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An easy way to visualize why this can't be true is to try putting some points on a number line.
Start with 1 point in [0, 1):
2 points in [1, 2):
And so on:
Now you have a sequence that grows to infinity but keeps getting closer together.
answered yesterday
OwenOwen
744410
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15
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+1 I'm definitely going to steal that. That's a lovely example, easily understandable even by people who don't know the harmonic series diverges.
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– Theo Bendit
yesterday
5
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This should be the accepted answer as it's counterexample's divergence is obvious whereas the harmonic series divergence (though famous) is not
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– gota
22 hours ago
4
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Note that this is (approximately) the same as the sequence $a_n=sqrt{n}$.
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– tomasz
22 hours ago
2
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The one small point to note, though admittedly pretty obvious, is that after inserting all those infinitely many points, each point has only a finite number of points to its left, and therefore a finite index (position) in the overall sequence.
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– Marc van Leeuwen
20 hours ago
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@MarcvanLeeuwen That's a great point. If you think of this as building a set, then you do need to show that each point is preceded by finitely many for it to be a sequence. But if you see it as a recursive definition of a sequence (imagine how you'd write this in code), it follows automatically that each point has an index.
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– Owen
11 hours ago
add a comment |
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Any increasing sequence ${a_n}_{ngeq 1}$ has limit in $mathbb{R}cup{+infty}$. It is $sup_{ngeq 1} a_n$. Such $sup$ or supremum can be a finite number or $+infty$ (even if we know that $a_{n+1}-a_nto 0$).
An example with a finite limit is $a_n=1-1/nto 1$ and $a_{n+1}-a_n=frac{1}{n(n+1)}to 0$.
On the other hand $a_n=sqrt{n}to +infty$ and $a_{n+1}-a_n=sqrt{n+1}-sqrt{n}=frac{1}{sqrt{n+1}+sqrt{n}}to 0$.
So, the answer is NO, the condition $a_{n+1}-a_nto 0$ is not sufficient for an increasing sequence ${a_n}_{ngeq 1}$ to have a FINITE limit.
answered yesterday
Robert ZRobert Z
98.3k1067139
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Robert.A fine answer!+
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– Peter Szilas
yesterday
add a comment |
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Another counterexample is $a_n=ln n$, for $ngeq1$. The difference of successive terms is $ln(n+1)-ln n = ln (1+1/n) rightarrow ln 1 = 0$, as $n rightarrow infty$, yet $ln n$ itself tends to infinity, as $n$ tends to infinity.
answered yesterday
SimonSimon
693512
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1
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Which is, in a way, the same counterexample, because $sum_{k=1}^nfrac1k = ln n + gamma + mathcal Oleft(frac1nright)$.
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– Roman Odaisky
13 hours ago
add a comment |
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The condition $a_{n+1}-a_n to 0$ is not sufficient, as José Carlos Santos pointed out. But, a necessary and sufficient condition, that doesn't require the series to be increasing, is that $limlimits_{ntoinfty}(a_{n+m(n)}-a_n)=0$ for all $m(n)in mathbb{N}$, where $m$ is a function of $n$. Sequences which satisfy this property are called Cauchy sequences.
Also, if you show that a sequence is monotonically increasing and bounded from above, then it converges. The same applies for monotonically decreasing sequences which are bounded from below.
answered yesterday
Haris GusicHaris Gusic
465112
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Your stated condition $lim_{n to infty} (a_{n+m}-a_n) = 0$ for each $m$ is not equivalent to the Cauchy property, and it does not imply that the sequence $a_n$ converges. Consider $a_n = log n$. I think what you would want is that $lim_{n to infty} (a_{n+m}-a_n) = 0$ uniformly in $m$.
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– Nate Eldredge
yesterday
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@NateEldredge But if we take $m=n$, then the condition is not satisfied, is it?
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– Haris Gusic
yesterday
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Okay, the revised condition (where $m$ is a function of $n$) is correct, though it seems awkward to work with in practice.
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– Nate Eldredge
yesterday
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@NateEldredge I formulated it that way because I am more familiar with it. Sorry about any confusion I might have caused.
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– Haris Gusic
yesterday
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I think another way of looking at things would be to say that for any specified positive epsilon, there will be some value of n such that for all i > n, |a[i]-a[n]| will be less than epsilon. Would that be correct?
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– supercat
15 hours ago
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show 2 more comments
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No. Consider the sequence ${a_n}_{n=1}^infty$ given by
$a_n = sumlimits_{k=1}^{n} frac{1}{k}$.
It follows that
- $a_n > a_{n-1}$
$a_n - a_{n-1} = frac{1}{n} rightarrow 0$ as $n rightarrow infty$, but
$a_n = sumlimits_{k=1}^{n} frac{1}{k} rightarrow infty$ as $n rightarrow infty$ (by, e.g., integral test).
answered yesterday
24thAlchemist24thAlchemist
212
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
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votes
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$begingroup$
No. Just consider the case in which $a_n=1+frac12+frac13+cdots+frac1n$. Note that then we would have$$lim_{ntoinfty}a_{n+1}-a_n=lim_{ntoinfty}frac1{n+1}=0.$$
answered yesterday
José Carlos SantosJosé Carlos Santos
162k22130233
$endgroup$
5
$begingroup$
Rhys: His sequence is $1, frac32, frac{11}6, frac{25}{12},...$. Essentially the sequence of partial sums associated with the harmonic series. This is still a sequence, not a series, since it is a finite sum.
$endgroup$
– M D
yesterday
5
$begingroup$
@RhysHughes $a_n=1+frac12+cdots+frac1n$ IS increasing and $a_{n+1}-a_n=frac{1}{n+1}to 0$.
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– Robert Z
yesterday
$begingroup$
I see now. Thanks, it just confused me.
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– Rhys Hughes
yesterday
5
$begingroup$
I think adding an explanation from comments into the answer is worth considering and would benefit to the quality of an answer.
$endgroup$
– Ister
yesterday
2
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@wizzwizz4 No, because $a_n neq frac{1}{n}$. Instead, $a_n = 1 + frac{1}{2} + ldots + frac{1}{n}$. So, $$a_{n+1} - a_n = left(1 + frac{1}{2} + ldots + frac{1}{n} + frac{1}{n+1}right) - left(1 + frac{1}{2} + ldots + frac{1}{n}right) = frac{1}{n+1}.$$
$endgroup$
– Theo Bendit
12 hours ago
|
show 3 more comments
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No. Just consider the case in which $a_n=1+frac12+frac13+cdots+frac1n$. Note that then we would have$$lim_{ntoinfty}a_{n+1}-a_n=lim_{ntoinfty}frac1{n+1}=0.$$
answered yesterday
José Carlos SantosJosé Carlos Santos
162k22130233
$endgroup$
5
$begingroup$
Rhys: His sequence is $1, frac32, frac{11}6, frac{25}{12},...$. Essentially the sequence of partial sums associated with the harmonic series. This is still a sequence, not a series, since it is a finite sum.
$endgroup$
– M D
yesterday
5
$begingroup$
@RhysHughes $a_n=1+frac12+cdots+frac1n$ IS increasing and $a_{n+1}-a_n=frac{1}{n+1}to 0$.
$endgroup$
– Robert Z
yesterday
$begingroup$
I see now. Thanks, it just confused me.
$endgroup$
– Rhys Hughes
yesterday
5
$begingroup$
I think adding an explanation from comments into the answer is worth considering and would benefit to the quality of an answer.
$endgroup$
– Ister
yesterday
2
$begingroup$
@wizzwizz4 No, because $a_n neq frac{1}{n}$. Instead, $a_n = 1 + frac{1}{2} + ldots + frac{1}{n}$. So, $$a_{n+1} - a_n = left(1 + frac{1}{2} + ldots + frac{1}{n} + frac{1}{n+1}right) - left(1 + frac{1}{2} + ldots + frac{1}{n}right) = frac{1}{n+1}.$$
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– Theo Bendit
12 hours ago
|
show 3 more comments
$begingroup$
No. Just consider the case in which $a_n=1+frac12+frac13+cdots+frac1n$. Note that then we would have$$lim_{ntoinfty}a_{n+1}-a_n=lim_{ntoinfty}frac1{n+1}=0.$$
answered yesterday
José Carlos SantosJosé Carlos Santos
162k22130233
$endgroup$
No. Just consider the case in which $a_n=1+frac12+frac13+cdots+frac1n$. Note that then we would have$$lim_{ntoinfty}a_{n+1}-a_n=lim_{ntoinfty}frac1{n+1}=0.$$
answered yesterday
José Carlos SantosJosé Carlos Santos
162k22130233
edited yesterday
answered yesterday
José Carlos SantosJosé Carlos Santos
162k22130233
answered yesterday
José Carlos SantosJosé Carlos Santos
162k22130233
answered yesterday
José Carlos SantosJosé Carlos Santos
162k22130233
162k22130233
5
$begingroup$
Rhys: His sequence is $1, frac32, frac{11}6, frac{25}{12},...$. Essentially the sequence of partial sums associated with the harmonic series. This is still a sequence, not a series, since it is a finite sum.
$endgroup$
– M D
yesterday
5
$begingroup$
@RhysHughes $a_n=1+frac12+cdots+frac1n$ IS increasing and $a_{n+1}-a_n=frac{1}{n+1}to 0$.
$endgroup$
– Robert Z
yesterday
$begingroup$
I see now. Thanks, it just confused me.
$endgroup$
– Rhys Hughes
yesterday
5
$begingroup$
I think adding an explanation from comments into the answer is worth considering and would benefit to the quality of an answer.
$endgroup$
– Ister
yesterday
2
$begingroup$
@wizzwizz4 No, because $a_n neq frac{1}{n}$. Instead, $a_n = 1 + frac{1}{2} + ldots + frac{1}{n}$. So, $$a_{n+1} - a_n = left(1 + frac{1}{2} + ldots + frac{1}{n} + frac{1}{n+1}right) - left(1 + frac{1}{2} + ldots + frac{1}{n}right) = frac{1}{n+1}.$$
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– Theo Bendit
12 hours ago
|
show 3 more comments
5
$begingroup$
Rhys: His sequence is $1, frac32, frac{11}6, frac{25}{12},...$. Essentially the sequence of partial sums associated with the harmonic series. This is still a sequence, not a series, since it is a finite sum.
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– M D
yesterday
5
$begingroup$
@RhysHughes $a_n=1+frac12+cdots+frac1n$ IS increasing and $a_{n+1}-a_n=frac{1}{n+1}to 0$.
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– Robert Z
yesterday
$begingroup$
I see now. Thanks, it just confused me.
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– Rhys Hughes
yesterday
5
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I think adding an explanation from comments into the answer is worth considering and would benefit to the quality of an answer.
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– Ister
yesterday
2
$begingroup$
@wizzwizz4 No, because $a_n neq frac{1}{n}$. Instead, $a_n = 1 + frac{1}{2} + ldots + frac{1}{n}$. So, $$a_{n+1} - a_n = left(1 + frac{1}{2} + ldots + frac{1}{n} + frac{1}{n+1}right) - left(1 + frac{1}{2} + ldots + frac{1}{n}right) = frac{1}{n+1}.$$
$endgroup$
– Theo Bendit
12 hours ago
5
5
$begingroup$
Rhys: His sequence is $1, frac32, frac{11}6, frac{25}{12},...$. Essentially the sequence of partial sums associated with the harmonic series. This is still a sequence, not a series, since it is a finite sum.
$endgroup$
– M D
yesterday
$begingroup$
Rhys: His sequence is $1, frac32, frac{11}6, frac{25}{12},...$. Essentially the sequence of partial sums associated with the harmonic series. This is still a sequence, not a series, since it is a finite sum.
$endgroup$
– M D
yesterday
5
5
$begingroup$
@RhysHughes $a_n=1+frac12+cdots+frac1n$ IS increasing and $a_{n+1}-a_n=frac{1}{n+1}to 0$.
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– Robert Z
yesterday
$begingroup$
@RhysHughes $a_n=1+frac12+cdots+frac1n$ IS increasing and $a_{n+1}-a_n=frac{1}{n+1}to 0$.
$endgroup$
– Robert Z
yesterday
$begingroup$
I see now. Thanks, it just confused me.
$endgroup$
– Rhys Hughes
yesterday
$begingroup$
I see now. Thanks, it just confused me.
$endgroup$
– Rhys Hughes
yesterday
5
5
$begingroup$
I think adding an explanation from comments into the answer is worth considering and would benefit to the quality of an answer.
$endgroup$
– Ister
yesterday
$begingroup$
I think adding an explanation from comments into the answer is worth considering and would benefit to the quality of an answer.
$endgroup$
– Ister
yesterday
2
2
$begingroup$
@wizzwizz4 No, because $a_n neq frac{1}{n}$. Instead, $a_n = 1 + frac{1}{2} + ldots + frac{1}{n}$. So, $$a_{n+1} - a_n = left(1 + frac{1}{2} + ldots + frac{1}{n} + frac{1}{n+1}right) - left(1 + frac{1}{2} + ldots + frac{1}{n}right) = frac{1}{n+1}.$$
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– Theo Bendit
12 hours ago
$begingroup$
@wizzwizz4 No, because $a_n neq frac{1}{n}$. Instead, $a_n = 1 + frac{1}{2} + ldots + frac{1}{n}$. So, $$a_{n+1} - a_n = left(1 + frac{1}{2} + ldots + frac{1}{n} + frac{1}{n+1}right) - left(1 + frac{1}{2} + ldots + frac{1}{n}right) = frac{1}{n+1}.$$
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– Theo Bendit
12 hours ago
|
show 3 more comments
$begingroup$
An easy way to visualize why this can't be true is to try putting some points on a number line.
Start with 1 point in [0, 1):
2 points in [1, 2):
And so on:
Now you have a sequence that grows to infinity but keeps getting closer together.
answered yesterday
OwenOwen
744410
$endgroup$
15
$begingroup$
+1 I'm definitely going to steal that. That's a lovely example, easily understandable even by people who don't know the harmonic series diverges.
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– Theo Bendit
yesterday
5
$begingroup$
This should be the accepted answer as it's counterexample's divergence is obvious whereas the harmonic series divergence (though famous) is not
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– gota
22 hours ago
4
$begingroup$
Note that this is (approximately) the same as the sequence $a_n=sqrt{n}$.
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– tomasz
22 hours ago
2
$begingroup$
The one small point to note, though admittedly pretty obvious, is that after inserting all those infinitely many points, each point has only a finite number of points to its left, and therefore a finite index (position) in the overall sequence.
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– Marc van Leeuwen
20 hours ago
$begingroup$
@MarcvanLeeuwen That's a great point. If you think of this as building a set, then you do need to show that each point is preceded by finitely many for it to be a sequence. But if you see it as a recursive definition of a sequence (imagine how you'd write this in code), it follows automatically that each point has an index.
$endgroup$
– Owen
11 hours ago
add a comment |
$begingroup$
An easy way to visualize why this can't be true is to try putting some points on a number line.
Start with 1 point in [0, 1):
2 points in [1, 2):
And so on:
Now you have a sequence that grows to infinity but keeps getting closer together.
answered yesterday
OwenOwen
744410
$endgroup$
15
$begingroup$
+1 I'm definitely going to steal that. That's a lovely example, easily understandable even by people who don't know the harmonic series diverges.
$endgroup$
– Theo Bendit
yesterday
5
$begingroup$
This should be the accepted answer as it's counterexample's divergence is obvious whereas the harmonic series divergence (though famous) is not
$endgroup$
– gota
22 hours ago
4
$begingroup$
Note that this is (approximately) the same as the sequence $a_n=sqrt{n}$.
$endgroup$
– tomasz
22 hours ago
2
$begingroup$
The one small point to note, though admittedly pretty obvious, is that after inserting all those infinitely many points, each point has only a finite number of points to its left, and therefore a finite index (position) in the overall sequence.
$endgroup$
– Marc van Leeuwen
20 hours ago
$begingroup$
@MarcvanLeeuwen That's a great point. If you think of this as building a set, then you do need to show that each point is preceded by finitely many for it to be a sequence. But if you see it as a recursive definition of a sequence (imagine how you'd write this in code), it follows automatically that each point has an index.
$endgroup$
– Owen
11 hours ago
add a comment |
$begingroup$
An easy way to visualize why this can't be true is to try putting some points on a number line.
Start with 1 point in [0, 1):
2 points in [1, 2):
And so on:
Now you have a sequence that grows to infinity but keeps getting closer together.
answered yesterday
OwenOwen
744410
$endgroup$
An easy way to visualize why this can't be true is to try putting some points on a number line.
Start with 1 point in [0, 1):
2 points in [1, 2):
And so on:
Now you have a sequence that grows to infinity but keeps getting closer together.
answered yesterday
OwenOwen
744410
answered yesterday
OwenOwen
744410
answered yesterday
OwenOwen
744410
answered yesterday
OwenOwen
744410
744410
15
$begingroup$
+1 I'm definitely going to steal that. That's a lovely example, easily understandable even by people who don't know the harmonic series diverges.
$endgroup$
– Theo Bendit
yesterday
5
$begingroup$
This should be the accepted answer as it's counterexample's divergence is obvious whereas the harmonic series divergence (though famous) is not
$endgroup$
– gota
22 hours ago
4
$begingroup$
Note that this is (approximately) the same as the sequence $a_n=sqrt{n}$.
$endgroup$
– tomasz
22 hours ago
2
$begingroup$
The one small point to note, though admittedly pretty obvious, is that after inserting all those infinitely many points, each point has only a finite number of points to its left, and therefore a finite index (position) in the overall sequence.
$endgroup$
– Marc van Leeuwen
20 hours ago
$begingroup$
@MarcvanLeeuwen That's a great point. If you think of this as building a set, then you do need to show that each point is preceded by finitely many for it to be a sequence. But if you see it as a recursive definition of a sequence (imagine how you'd write this in code), it follows automatically that each point has an index.
$endgroup$
– Owen
11 hours ago
add a comment |
15
$begingroup$
+1 I'm definitely going to steal that. That's a lovely example, easily understandable even by people who don't know the harmonic series diverges.
$endgroup$
– Theo Bendit
yesterday
5
$begingroup$
This should be the accepted answer as it's counterexample's divergence is obvious whereas the harmonic series divergence (though famous) is not
$endgroup$
– gota
22 hours ago
4
$begingroup$
Note that this is (approximately) the same as the sequence $a_n=sqrt{n}$.
$endgroup$
– tomasz
22 hours ago
2
$begingroup$
The one small point to note, though admittedly pretty obvious, is that after inserting all those infinitely many points, each point has only a finite number of points to its left, and therefore a finite index (position) in the overall sequence.
$endgroup$
– Marc van Leeuwen
20 hours ago
$begingroup$
@MarcvanLeeuwen That's a great point. If you think of this as building a set, then you do need to show that each point is preceded by finitely many for it to be a sequence. But if you see it as a recursive definition of a sequence (imagine how you'd write this in code), it follows automatically that each point has an index.
$endgroup$
– Owen
11 hours ago
15
15
$begingroup$
+1 I'm definitely going to steal that. That's a lovely example, easily understandable even by people who don't know the harmonic series diverges.
$endgroup$
– Theo Bendit
yesterday
$begingroup$
+1 I'm definitely going to steal that. That's a lovely example, easily understandable even by people who don't know the harmonic series diverges.
$endgroup$
– Theo Bendit
yesterday
5
5
$begingroup$
This should be the accepted answer as it's counterexample's divergence is obvious whereas the harmonic series divergence (though famous) is not
$endgroup$
– gota
22 hours ago
$begingroup$
This should be the accepted answer as it's counterexample's divergence is obvious whereas the harmonic series divergence (though famous) is not
$endgroup$
– gota
22 hours ago
4
4
$begingroup$
Note that this is (approximately) the same as the sequence $a_n=sqrt{n}$.
$endgroup$
– tomasz
22 hours ago
$begingroup$
Note that this is (approximately) the same as the sequence $a_n=sqrt{n}$.
$endgroup$
– tomasz
22 hours ago
2
2
$begingroup$
The one small point to note, though admittedly pretty obvious, is that after inserting all those infinitely many points, each point has only a finite number of points to its left, and therefore a finite index (position) in the overall sequence.
$endgroup$
– Marc van Leeuwen
20 hours ago
$begingroup$
The one small point to note, though admittedly pretty obvious, is that after inserting all those infinitely many points, each point has only a finite number of points to its left, and therefore a finite index (position) in the overall sequence.
$endgroup$
– Marc van Leeuwen
20 hours ago
$begingroup$
@MarcvanLeeuwen That's a great point. If you think of this as building a set, then you do need to show that each point is preceded by finitely many for it to be a sequence. But if you see it as a recursive definition of a sequence (imagine how you'd write this in code), it follows automatically that each point has an index.
$endgroup$
– Owen
11 hours ago
$begingroup$
@MarcvanLeeuwen That's a great point. If you think of this as building a set, then you do need to show that each point is preceded by finitely many for it to be a sequence. But if you see it as a recursive definition of a sequence (imagine how you'd write this in code), it follows automatically that each point has an index.
$endgroup$
– Owen
11 hours ago
add a comment |
$begingroup$
Any increasing sequence ${a_n}_{ngeq 1}$ has limit in $mathbb{R}cup{+infty}$. It is $sup_{ngeq 1} a_n$. Such $sup$ or supremum can be a finite number or $+infty$ (even if we know that $a_{n+1}-a_nto 0$).
An example with a finite limit is $a_n=1-1/nto 1$ and $a_{n+1}-a_n=frac{1}{n(n+1)}to 0$.
On the other hand $a_n=sqrt{n}to +infty$ and $a_{n+1}-a_n=sqrt{n+1}-sqrt{n}=frac{1}{sqrt{n+1}+sqrt{n}}to 0$.
So, the answer is NO, the condition $a_{n+1}-a_nto 0$ is not sufficient for an increasing sequence ${a_n}_{ngeq 1}$ to have a FINITE limit.
answered yesterday
Robert ZRobert Z
98.3k1067139
$endgroup$
$begingroup$
Robert.A fine answer!+
$endgroup$
– Peter Szilas
yesterday
add a comment |
$begingroup$
Any increasing sequence ${a_n}_{ngeq 1}$ has limit in $mathbb{R}cup{+infty}$. It is $sup_{ngeq 1} a_n$. Such $sup$ or supremum can be a finite number or $+infty$ (even if we know that $a_{n+1}-a_nto 0$).
An example with a finite limit is $a_n=1-1/nto 1$ and $a_{n+1}-a_n=frac{1}{n(n+1)}to 0$.
On the other hand $a_n=sqrt{n}to +infty$ and $a_{n+1}-a_n=sqrt{n+1}-sqrt{n}=frac{1}{sqrt{n+1}+sqrt{n}}to 0$.
So, the answer is NO, the condition $a_{n+1}-a_nto 0$ is not sufficient for an increasing sequence ${a_n}_{ngeq 1}$ to have a FINITE limit.
answered yesterday
Robert ZRobert Z
98.3k1067139
$endgroup$
$begingroup$
Robert.A fine answer!+
$endgroup$
– Peter Szilas
yesterday
add a comment |
$begingroup$
Any increasing sequence ${a_n}_{ngeq 1}$ has limit in $mathbb{R}cup{+infty}$. It is $sup_{ngeq 1} a_n$. Such $sup$ or supremum can be a finite number or $+infty$ (even if we know that $a_{n+1}-a_nto 0$).
An example with a finite limit is $a_n=1-1/nto 1$ and $a_{n+1}-a_n=frac{1}{n(n+1)}to 0$.
On the other hand $a_n=sqrt{n}to +infty$ and $a_{n+1}-a_n=sqrt{n+1}-sqrt{n}=frac{1}{sqrt{n+1}+sqrt{n}}to 0$.
So, the answer is NO, the condition $a_{n+1}-a_nto 0$ is not sufficient for an increasing sequence ${a_n}_{ngeq 1}$ to have a FINITE limit.
answered yesterday
Robert ZRobert Z
98.3k1067139
$endgroup$
Any increasing sequence ${a_n}_{ngeq 1}$ has limit in $mathbb{R}cup{+infty}$. It is $sup_{ngeq 1} a_n$. Such $sup$ or supremum can be a finite number or $+infty$ (even if we know that $a_{n+1}-a_nto 0$).
An example with a finite limit is $a_n=1-1/nto 1$ and $a_{n+1}-a_n=frac{1}{n(n+1)}to 0$.
On the other hand $a_n=sqrt{n}to +infty$ and $a_{n+1}-a_n=sqrt{n+1}-sqrt{n}=frac{1}{sqrt{n+1}+sqrt{n}}to 0$.
So, the answer is NO, the condition $a_{n+1}-a_nto 0$ is not sufficient for an increasing sequence ${a_n}_{ngeq 1}$ to have a FINITE limit.
answered yesterday
Robert ZRobert Z
98.3k1067139
edited yesterday
answered yesterday
Robert ZRobert Z
98.3k1067139
answered yesterday
Robert ZRobert Z
98.3k1067139
answered yesterday
Robert ZRobert Z
98.3k1067139
98.3k1067139
$begingroup$
Robert.A fine answer!+
$endgroup$
– Peter Szilas
yesterday
add a comment |
$begingroup$
Robert.A fine answer!+
$endgroup$
– Peter Szilas
yesterday
$begingroup$
Robert.A fine answer!+
$endgroup$
– Peter Szilas
yesterday
$begingroup$
Robert.A fine answer!+
$endgroup$
– Peter Szilas
yesterday
add a comment |
$begingroup$
Another counterexample is $a_n=ln n$, for $ngeq1$. The difference of successive terms is $ln(n+1)-ln n = ln (1+1/n) rightarrow ln 1 = 0$, as $n rightarrow infty$, yet $ln n$ itself tends to infinity, as $n$ tends to infinity.
answered yesterday
SimonSimon
693512
$endgroup$
1
$begingroup$
Which is, in a way, the same counterexample, because $sum_{k=1}^nfrac1k = ln n + gamma + mathcal Oleft(frac1nright)$.
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– Roman Odaisky
13 hours ago
add a comment |
$begingroup$
Another counterexample is $a_n=ln n$, for $ngeq1$. The difference of successive terms is $ln(n+1)-ln n = ln (1+1/n) rightarrow ln 1 = 0$, as $n rightarrow infty$, yet $ln n$ itself tends to infinity, as $n$ tends to infinity.
answered yesterday
SimonSimon
693512
$endgroup$
1
$begingroup$
Which is, in a way, the same counterexample, because $sum_{k=1}^nfrac1k = ln n + gamma + mathcal Oleft(frac1nright)$.
$endgroup$
– Roman Odaisky
13 hours ago
add a comment |
$begingroup$
Another counterexample is $a_n=ln n$, for $ngeq1$. The difference of successive terms is $ln(n+1)-ln n = ln (1+1/n) rightarrow ln 1 = 0$, as $n rightarrow infty$, yet $ln n$ itself tends to infinity, as $n$ tends to infinity.
answered yesterday
SimonSimon
693512
$endgroup$
Another counterexample is $a_n=ln n$, for $ngeq1$. The difference of successive terms is $ln(n+1)-ln n = ln (1+1/n) rightarrow ln 1 = 0$, as $n rightarrow infty$, yet $ln n$ itself tends to infinity, as $n$ tends to infinity.
answered yesterday
SimonSimon
693512
answered yesterday
SimonSimon
693512
answered yesterday
SimonSimon
693512
answered yesterday
SimonSimon
693512
693512
1
$begingroup$
Which is, in a way, the same counterexample, because $sum_{k=1}^nfrac1k = ln n + gamma + mathcal Oleft(frac1nright)$.
$endgroup$
– Roman Odaisky
13 hours ago
add a comment |
1
$begingroup$
Which is, in a way, the same counterexample, because $sum_{k=1}^nfrac1k = ln n + gamma + mathcal Oleft(frac1nright)$.
$endgroup$
– Roman Odaisky
13 hours ago
1
1
$begingroup$
Which is, in a way, the same counterexample, because $sum_{k=1}^nfrac1k = ln n + gamma + mathcal Oleft(frac1nright)$.
$endgroup$
– Roman Odaisky
13 hours ago
$begingroup$
Which is, in a way, the same counterexample, because $sum_{k=1}^nfrac1k = ln n + gamma + mathcal Oleft(frac1nright)$.
$endgroup$
– Roman Odaisky
13 hours ago
add a comment |
$begingroup$
The condition $a_{n+1}-a_n to 0$ is not sufficient, as José Carlos Santos pointed out. But, a necessary and sufficient condition, that doesn't require the series to be increasing, is that $limlimits_{ntoinfty}(a_{n+m(n)}-a_n)=0$ for all $m(n)in mathbb{N}$, where $m$ is a function of $n$. Sequences which satisfy this property are called Cauchy sequences.
Also, if you show that a sequence is monotonically increasing and bounded from above, then it converges. The same applies for monotonically decreasing sequences which are bounded from below.
answered yesterday
Haris GusicHaris Gusic
465112
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2
$begingroup$
Your stated condition $lim_{n to infty} (a_{n+m}-a_n) = 0$ for each $m$ is not equivalent to the Cauchy property, and it does not imply that the sequence $a_n$ converges. Consider $a_n = log n$. I think what you would want is that $lim_{n to infty} (a_{n+m}-a_n) = 0$ uniformly in $m$.
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– Nate Eldredge
yesterday
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@NateEldredge But if we take $m=n$, then the condition is not satisfied, is it?
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– Haris Gusic
yesterday
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Okay, the revised condition (where $m$ is a function of $n$) is correct, though it seems awkward to work with in practice.
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– Nate Eldredge
yesterday
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@NateEldredge I formulated it that way because I am more familiar with it. Sorry about any confusion I might have caused.
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– Haris Gusic
yesterday
$begingroup$
I think another way of looking at things would be to say that for any specified positive epsilon, there will be some value of n such that for all i > n, |a[i]-a[n]| will be less than epsilon. Would that be correct?
$endgroup$
– supercat
15 hours ago
|
show 2 more comments
$begingroup$
The condition $a_{n+1}-a_n to 0$ is not sufficient, as José Carlos Santos pointed out. But, a necessary and sufficient condition, that doesn't require the series to be increasing, is that $limlimits_{ntoinfty}(a_{n+m(n)}-a_n)=0$ for all $m(n)in mathbb{N}$, where $m$ is a function of $n$. Sequences which satisfy this property are called Cauchy sequences.
Also, if you show that a sequence is monotonically increasing and bounded from above, then it converges. The same applies for monotonically decreasing sequences which are bounded from below.
answered yesterday
Haris GusicHaris Gusic
465112
$endgroup$
2
$begingroup$
Your stated condition $lim_{n to infty} (a_{n+m}-a_n) = 0$ for each $m$ is not equivalent to the Cauchy property, and it does not imply that the sequence $a_n$ converges. Consider $a_n = log n$. I think what you would want is that $lim_{n to infty} (a_{n+m}-a_n) = 0$ uniformly in $m$.
$endgroup$
– Nate Eldredge
yesterday
$begingroup$
@NateEldredge But if we take $m=n$, then the condition is not satisfied, is it?
$endgroup$
– Haris Gusic
yesterday
$begingroup$
Okay, the revised condition (where $m$ is a function of $n$) is correct, though it seems awkward to work with in practice.
$endgroup$
– Nate Eldredge
yesterday
$begingroup$
@NateEldredge I formulated it that way because I am more familiar with it. Sorry about any confusion I might have caused.
$endgroup$
– Haris Gusic
yesterday
$begingroup$
I think another way of looking at things would be to say that for any specified positive epsilon, there will be some value of n such that for all i > n, |a[i]-a[n]| will be less than epsilon. Would that be correct?
$endgroup$
– supercat
15 hours ago
|
show 2 more comments
$begingroup$
The condition $a_{n+1}-a_n to 0$ is not sufficient, as José Carlos Santos pointed out. But, a necessary and sufficient condition, that doesn't require the series to be increasing, is that $limlimits_{ntoinfty}(a_{n+m(n)}-a_n)=0$ for all $m(n)in mathbb{N}$, where $m$ is a function of $n$. Sequences which satisfy this property are called Cauchy sequences.
Also, if you show that a sequence is monotonically increasing and bounded from above, then it converges. The same applies for monotonically decreasing sequences which are bounded from below.
answered yesterday
Haris GusicHaris Gusic
465112
$endgroup$
The condition $a_{n+1}-a_n to 0$ is not sufficient, as José Carlos Santos pointed out. But, a necessary and sufficient condition, that doesn't require the series to be increasing, is that $limlimits_{ntoinfty}(a_{n+m(n)}-a_n)=0$ for all $m(n)in mathbb{N}$, where $m$ is a function of $n$. Sequences which satisfy this property are called Cauchy sequences.
Also, if you show that a sequence is monotonically increasing and bounded from above, then it converges. The same applies for monotonically decreasing sequences which are bounded from below.
answered yesterday
Haris GusicHaris Gusic
465112
edited yesterday
answered yesterday
Haris GusicHaris Gusic
465112
answered yesterday
Haris GusicHaris Gusic
465112
answered yesterday
Haris GusicHaris Gusic
465112
465112
2
$begingroup$
Your stated condition $lim_{n to infty} (a_{n+m}-a_n) = 0$ for each $m$ is not equivalent to the Cauchy property, and it does not imply that the sequence $a_n$ converges. Consider $a_n = log n$. I think what you would want is that $lim_{n to infty} (a_{n+m}-a_n) = 0$ uniformly in $m$.
$endgroup$
– Nate Eldredge
yesterday
$begingroup$
@NateEldredge But if we take $m=n$, then the condition is not satisfied, is it?
$endgroup$
– Haris Gusic
yesterday
$begingroup$
Okay, the revised condition (where $m$ is a function of $n$) is correct, though it seems awkward to work with in practice.
$endgroup$
– Nate Eldredge
yesterday
$begingroup$
@NateEldredge I formulated it that way because I am more familiar with it. Sorry about any confusion I might have caused.
$endgroup$
– Haris Gusic
yesterday
$begingroup$
I think another way of looking at things would be to say that for any specified positive epsilon, there will be some value of n such that for all i > n, |a[i]-a[n]| will be less than epsilon. Would that be correct?
$endgroup$
– supercat
15 hours ago
|
show 2 more comments
2
$begingroup$
Your stated condition $lim_{n to infty} (a_{n+m}-a_n) = 0$ for each $m$ is not equivalent to the Cauchy property, and it does not imply that the sequence $a_n$ converges. Consider $a_n = log n$. I think what you would want is that $lim_{n to infty} (a_{n+m}-a_n) = 0$ uniformly in $m$.
$endgroup$
– Nate Eldredge
yesterday
$begingroup$
@NateEldredge But if we take $m=n$, then the condition is not satisfied, is it?
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– Haris Gusic
yesterday
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Okay, the revised condition (where $m$ is a function of $n$) is correct, though it seems awkward to work with in practice.
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– Nate Eldredge
yesterday
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@NateEldredge I formulated it that way because I am more familiar with it. Sorry about any confusion I might have caused.
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– Haris Gusic
yesterday
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I think another way of looking at things would be to say that for any specified positive epsilon, there will be some value of n such that for all i > n, |a[i]-a[n]| will be less than epsilon. Would that be correct?
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– supercat
15 hours ago
2
2
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Your stated condition $lim_{n to infty} (a_{n+m}-a_n) = 0$ for each $m$ is not equivalent to the Cauchy property, and it does not imply that the sequence $a_n$ converges. Consider $a_n = log n$. I think what you would want is that $lim_{n to infty} (a_{n+m}-a_n) = 0$ uniformly in $m$.
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– Nate Eldredge
yesterday
$begingroup$
Your stated condition $lim_{n to infty} (a_{n+m}-a_n) = 0$ for each $m$ is not equivalent to the Cauchy property, and it does not imply that the sequence $a_n$ converges. Consider $a_n = log n$. I think what you would want is that $lim_{n to infty} (a_{n+m}-a_n) = 0$ uniformly in $m$.
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– Nate Eldredge
yesterday
$begingroup$
@NateEldredge But if we take $m=n$, then the condition is not satisfied, is it?
$endgroup$
– Haris Gusic
yesterday
$begingroup$
@NateEldredge But if we take $m=n$, then the condition is not satisfied, is it?
$endgroup$
– Haris Gusic
yesterday
$begingroup$
Okay, the revised condition (where $m$ is a function of $n$) is correct, though it seems awkward to work with in practice.
$endgroup$
– Nate Eldredge
yesterday
$begingroup$
Okay, the revised condition (where $m$ is a function of $n$) is correct, though it seems awkward to work with in practice.
$endgroup$
– Nate Eldredge
yesterday
$begingroup$
@NateEldredge I formulated it that way because I am more familiar with it. Sorry about any confusion I might have caused.
$endgroup$
– Haris Gusic
yesterday
$begingroup$
@NateEldredge I formulated it that way because I am more familiar with it. Sorry about any confusion I might have caused.
$endgroup$
– Haris Gusic
yesterday
$begingroup$
I think another way of looking at things would be to say that for any specified positive epsilon, there will be some value of n such that for all i > n, |a[i]-a[n]| will be less than epsilon. Would that be correct?
$endgroup$
– supercat
15 hours ago
$begingroup$
I think another way of looking at things would be to say that for any specified positive epsilon, there will be some value of n such that for all i > n, |a[i]-a[n]| will be less than epsilon. Would that be correct?
$endgroup$
– supercat
15 hours ago
|
show 2 more comments
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No. Consider the sequence ${a_n}_{n=1}^infty$ given by
$a_n = sumlimits_{k=1}^{n} frac{1}{k}$.
It follows that
- $a_n > a_{n-1}$
$a_n - a_{n-1} = frac{1}{n} rightarrow 0$ as $n rightarrow infty$, but
$a_n = sumlimits_{k=1}^{n} frac{1}{k} rightarrow infty$ as $n rightarrow infty$ (by, e.g., integral test).
answered yesterday
24thAlchemist24thAlchemist
212
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add a comment |
$begingroup$
No. Consider the sequence ${a_n}_{n=1}^infty$ given by
$a_n = sumlimits_{k=1}^{n} frac{1}{k}$.
It follows that
- $a_n > a_{n-1}$
$a_n - a_{n-1} = frac{1}{n} rightarrow 0$ as $n rightarrow infty$, but
$a_n = sumlimits_{k=1}^{n} frac{1}{k} rightarrow infty$ as $n rightarrow infty$ (by, e.g., integral test).
answered yesterday
24thAlchemist24thAlchemist
212
New contributor
24thAlchemist is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
No. Consider the sequence ${a_n}_{n=1}^infty$ given by
$a_n = sumlimits_{k=1}^{n} frac{1}{k}$.
It follows that
- $a_n > a_{n-1}$
$a_n - a_{n-1} = frac{1}{n} rightarrow 0$ as $n rightarrow infty$, but
$a_n = sumlimits_{k=1}^{n} frac{1}{k} rightarrow infty$ as $n rightarrow infty$ (by, e.g., integral test).
answered yesterday
24thAlchemist24thAlchemist
212
New contributor
24thAlchemist is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
No. Consider the sequence ${a_n}_{n=1}^infty$ given by
$a_n = sumlimits_{k=1}^{n} frac{1}{k}$.
It follows that
- $a_n > a_{n-1}$
$a_n - a_{n-1} = frac{1}{n} rightarrow 0$ as $n rightarrow infty$, but
$a_n = sumlimits_{k=1}^{n} frac{1}{k} rightarrow infty$ as $n rightarrow infty$ (by, e.g., integral test).
answered yesterday
24thAlchemist24thAlchemist
212
New contributor
24thAlchemist is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday
24thAlchemist24thAlchemist
212
New contributor
24thAlchemist is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday
24thAlchemist24thAlchemist
212
answered yesterday
24thAlchemist24thAlchemist
212
212
New contributor
24thAlchemist is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
24thAlchemist is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
24thAlchemist is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
M D is a new contributor. Be nice, and check out our Code of Conduct.
M D is a new contributor. Be nice, and check out our Code of Conduct.
M D is a new contributor. Be nice, and check out our Code of Conduct.
M D is a new contributor. Be nice, and check out our Code of Conduct.
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3
$begingroup$
Note that while your sequence goes up and down periodically, you could define another sequence with the same step length for each $n$ but with all steps positive. That would be a counterexample to your question.
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– Adayah
yesterday
5
$begingroup$
Have you tried logarithms?
$endgroup$
– Mehrdad
yesterday
1
$begingroup$
Note that by writing $b_1=a_1, b_2=a_2-a_1, b_3=a_3-a_2,...$ the question becomes equivalent to asking whether a positive series with a summation term tending to zero must converge.
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– Bar Alon
21 hours ago