How do you determine if the following series converges?Convergence and exponentialsDetermine if the series...
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How do you determine if the following series converges?
Convergence and exponentialsDetermine if the series converges/divergesHow to prove this series converges $sum_{n=2}^infty frac{ln(n)}{n^{3/2}}$?How to determine if this series converges?Divergence test for the series $sum_{n=1}^{infty} (sqrt[n]{2n^2}-1)^n$Determine whether the series converges or diverges?My approach to determine if the following series is convergentHow to prove that $sum_{n=1}^{infty} frac{(log (n))^2}{n^2}$ converges?Determine whether the series converges or diverges.Does $displaystylesum_{n=1}^{infty}sinleft(displaystylefrac{1}{sqrt{n}}right)$ converge?Find all values of $k$ such that the series with terms $k^n / n^k$ converges.
$begingroup$
$$sum_{k=1}^infty mathrm{(1-frac{1}{k})}^{mathrm{k}^{2}}$$
I tried using the limit comparison test with $$sum_{k=1}^infty mathrm{(1-frac{1}{k})}^{mathrm{k}^{}}$$ but this leads to a limit of 0, which doesn't help. I think this may involve some use of
$mathrm{e}^x$, but I don't know where else to start. Any suggestions?
convergence
asked 48 mins ago
JayJay
334
$endgroup$
add a comment |
$begingroup$
$$sum_{k=1}^infty mathrm{(1-frac{1}{k})}^{mathrm{k}^{2}}$$
I tried using the limit comparison test with $$sum_{k=1}^infty mathrm{(1-frac{1}{k})}^{mathrm{k}^{}}$$ but this leads to a limit of 0, which doesn't help. I think this may involve some use of
$mathrm{e}^x$, but I don't know where else to start. Any suggestions?
convergence
asked 48 mins ago
JayJay
334
$endgroup$
$begingroup$
This is susceptible to the same approach as my answer to a different question.
$endgroup$
– T. Bongers
38 mins ago
$begingroup$
Do you know what $left(1-frac{1}{k}right)^{k}$ converges to?
$endgroup$
– JavaMan
32 mins ago
$begingroup$
I cannot believe how horrible my intuition is with this stuff especially given how old I am.
$endgroup$
– Randall
2 mins ago
add a comment |
$begingroup$
$$sum_{k=1}^infty mathrm{(1-frac{1}{k})}^{mathrm{k}^{2}}$$
I tried using the limit comparison test with $$sum_{k=1}^infty mathrm{(1-frac{1}{k})}^{mathrm{k}^{}}$$ but this leads to a limit of 0, which doesn't help. I think this may involve some use of
$mathrm{e}^x$, but I don't know where else to start. Any suggestions?
convergence
asked 48 mins ago
JayJay
334
$endgroup$
$$sum_{k=1}^infty mathrm{(1-frac{1}{k})}^{mathrm{k}^{2}}$$
I tried using the limit comparison test with $$sum_{k=1}^infty mathrm{(1-frac{1}{k})}^{mathrm{k}^{}}$$ but this leads to a limit of 0, which doesn't help. I think this may involve some use of
$mathrm{e}^x$, but I don't know where else to start. Any suggestions?
convergence
convergence
asked 48 mins ago
JayJay
334
asked 48 mins ago
JayJay
334
asked 48 mins ago
JayJay
334
asked 48 mins ago
JayJay
334
asked 48 mins ago
JayJay
334
334
$begingroup$
This is susceptible to the same approach as my answer to a different question.
$endgroup$
– T. Bongers
38 mins ago
$begingroup$
Do you know what $left(1-frac{1}{k}right)^{k}$ converges to?
$endgroup$
– JavaMan
32 mins ago
$begingroup$
I cannot believe how horrible my intuition is with this stuff especially given how old I am.
$endgroup$
– Randall
2 mins ago
add a comment |
$begingroup$
This is susceptible to the same approach as my answer to a different question.
$endgroup$
– T. Bongers
38 mins ago
$begingroup$
Do you know what $left(1-frac{1}{k}right)^{k}$ converges to?
$endgroup$
– JavaMan
32 mins ago
$begingroup$
I cannot believe how horrible my intuition is with this stuff especially given how old I am.
$endgroup$
– Randall
2 mins ago
$begingroup$
This is susceptible to the same approach as my answer to a different question.
$endgroup$
– T. Bongers
38 mins ago
$begingroup$
This is susceptible to the same approach as my answer to a different question.
$endgroup$
– T. Bongers
38 mins ago
$begingroup$
Do you know what $left(1-frac{1}{k}right)^{k}$ converges to?
$endgroup$
– JavaMan
32 mins ago
$begingroup$
Do you know what $left(1-frac{1}{k}right)^{k}$ converges to?
$endgroup$
– JavaMan
32 mins ago
$begingroup$
I cannot believe how horrible my intuition is with this stuff especially given how old I am.
$endgroup$
– Randall
2 mins ago
$begingroup$
I cannot believe how horrible my intuition is with this stuff especially given how old I am.
$endgroup$
– Randall
2 mins ago
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
HINT:
Note that $$left( 1-frac1k right)^kle e^{-1}$$
Can you finish?
answered 34 mins ago
Mark ViolaMark Viola
132k1277174
$endgroup$
add a comment |
$begingroup$
The root test works. Consider
$$lim sup sqrt[k]{left(1 - frac{1}{k}right)^{k^2}} = lim sup left(1 - frac{1}{k}right)^k = e^{-1} < 1,$$
hence the series converges.
answered 37 mins ago
Theo BenditTheo Bendit
18.7k12253
$endgroup$
add a comment |
$begingroup$
The ratio test is also interesting
$$a_k=left(1-frac{1}{k}right)^{k^2}implies log(a_k)=k ^2 logleft(1-frac{1}{k}right)$$
$$log(a_{k+1})-log(a_k)=(k+1) ^2 logleft(1-frac{1}{k+1}right)-k ^2 logleft(1-frac{1}{k}right)$$
Develop as a Taylor series for large values of $k$ to get
$$log(a_{k+1})-log(a_k)=-1+frac{1}{3 k^2}+O(left(frac{1}{k^3}right)$$
Continue with Taylor
$$frac{a_{k+1}}{a_k}=e^{log(a_{k+1})-log(a_k)}=frac 1 eleft(1+frac{1}{3 k^2}+Oleft(frac{1}{k^3}right)right)to frac 1 e$$
answered 24 mins ago
Claude LeiboviciClaude Leibovici
122k1157134
$endgroup$
add a comment |
$begingroup$
$begin{array}\
(1-frac1{k})^{k^2}
&=(frac{k-1}{k})^{k^2}\
&=dfrac1{(frac{k}{k-1})^{k^2}}\
&=dfrac1{(1+frac{1}{k-1})^{k^2}}\
&=dfrac1{((1+frac{1}{k-1})^{k})^k}\
&<dfrac1{(1+frac{k}{k-1})^k}
qquadtext{by Bernoulli}\
&=dfrac1{(frac{2k-1}{k-1})^k}\
&<dfrac1{(frac{2k-2}{k-1})^k}\
&=dfrac1{2^k}\
end{array}
$
and the sum of this converges.
answered 13 mins ago
marty cohenmarty cohen
73.8k549128
$endgroup$
add a comment |
Your Answer
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
HINT:
Note that $$left( 1-frac1k right)^kle e^{-1}$$
Can you finish?
answered 34 mins ago
Mark ViolaMark Viola
132k1277174
$endgroup$
add a comment |
$begingroup$
HINT:
Note that $$left( 1-frac1k right)^kle e^{-1}$$
Can you finish?
answered 34 mins ago
Mark ViolaMark Viola
132k1277174
$endgroup$
add a comment |
$begingroup$
HINT:
Note that $$left( 1-frac1k right)^kle e^{-1}$$
Can you finish?
answered 34 mins ago
Mark ViolaMark Viola
132k1277174
$endgroup$
HINT:
Note that $$left( 1-frac1k right)^kle e^{-1}$$
Can you finish?
answered 34 mins ago
Mark ViolaMark Viola
132k1277174
answered 34 mins ago
Mark ViolaMark Viola
132k1277174
answered 34 mins ago
Mark ViolaMark Viola
132k1277174
answered 34 mins ago
Mark ViolaMark Viola
132k1277174
132k1277174
add a comment |
add a comment |
$begingroup$
The root test works. Consider
$$lim sup sqrt[k]{left(1 - frac{1}{k}right)^{k^2}} = lim sup left(1 - frac{1}{k}right)^k = e^{-1} < 1,$$
hence the series converges.
answered 37 mins ago
Theo BenditTheo Bendit
18.7k12253
$endgroup$
add a comment |
$begingroup$
The root test works. Consider
$$lim sup sqrt[k]{left(1 - frac{1}{k}right)^{k^2}} = lim sup left(1 - frac{1}{k}right)^k = e^{-1} < 1,$$
hence the series converges.
answered 37 mins ago
Theo BenditTheo Bendit
18.7k12253
$endgroup$
add a comment |
$begingroup$
The root test works. Consider
$$lim sup sqrt[k]{left(1 - frac{1}{k}right)^{k^2}} = lim sup left(1 - frac{1}{k}right)^k = e^{-1} < 1,$$
hence the series converges.
answered 37 mins ago
Theo BenditTheo Bendit
18.7k12253
$endgroup$
The root test works. Consider
$$lim sup sqrt[k]{left(1 - frac{1}{k}right)^{k^2}} = lim sup left(1 - frac{1}{k}right)^k = e^{-1} < 1,$$
hence the series converges.
answered 37 mins ago
Theo BenditTheo Bendit
18.7k12253
answered 37 mins ago
Theo BenditTheo Bendit
18.7k12253
answered 37 mins ago
Theo BenditTheo Bendit
18.7k12253
answered 37 mins ago
Theo BenditTheo Bendit
18.7k12253
18.7k12253
add a comment |
add a comment |
$begingroup$
The ratio test is also interesting
$$a_k=left(1-frac{1}{k}right)^{k^2}implies log(a_k)=k ^2 logleft(1-frac{1}{k}right)$$
$$log(a_{k+1})-log(a_k)=(k+1) ^2 logleft(1-frac{1}{k+1}right)-k ^2 logleft(1-frac{1}{k}right)$$
Develop as a Taylor series for large values of $k$ to get
$$log(a_{k+1})-log(a_k)=-1+frac{1}{3 k^2}+O(left(frac{1}{k^3}right)$$
Continue with Taylor
$$frac{a_{k+1}}{a_k}=e^{log(a_{k+1})-log(a_k)}=frac 1 eleft(1+frac{1}{3 k^2}+Oleft(frac{1}{k^3}right)right)to frac 1 e$$
answered 24 mins ago
Claude LeiboviciClaude Leibovici
122k1157134
$endgroup$
add a comment |
$begingroup$
The ratio test is also interesting
$$a_k=left(1-frac{1}{k}right)^{k^2}implies log(a_k)=k ^2 logleft(1-frac{1}{k}right)$$
$$log(a_{k+1})-log(a_k)=(k+1) ^2 logleft(1-frac{1}{k+1}right)-k ^2 logleft(1-frac{1}{k}right)$$
Develop as a Taylor series for large values of $k$ to get
$$log(a_{k+1})-log(a_k)=-1+frac{1}{3 k^2}+O(left(frac{1}{k^3}right)$$
Continue with Taylor
$$frac{a_{k+1}}{a_k}=e^{log(a_{k+1})-log(a_k)}=frac 1 eleft(1+frac{1}{3 k^2}+Oleft(frac{1}{k^3}right)right)to frac 1 e$$
answered 24 mins ago
Claude LeiboviciClaude Leibovici
122k1157134
$endgroup$
add a comment |
$begingroup$
The ratio test is also interesting
$$a_k=left(1-frac{1}{k}right)^{k^2}implies log(a_k)=k ^2 logleft(1-frac{1}{k}right)$$
$$log(a_{k+1})-log(a_k)=(k+1) ^2 logleft(1-frac{1}{k+1}right)-k ^2 logleft(1-frac{1}{k}right)$$
Develop as a Taylor series for large values of $k$ to get
$$log(a_{k+1})-log(a_k)=-1+frac{1}{3 k^2}+O(left(frac{1}{k^3}right)$$
Continue with Taylor
$$frac{a_{k+1}}{a_k}=e^{log(a_{k+1})-log(a_k)}=frac 1 eleft(1+frac{1}{3 k^2}+Oleft(frac{1}{k^3}right)right)to frac 1 e$$
answered 24 mins ago
Claude LeiboviciClaude Leibovici
122k1157134
$endgroup$
The ratio test is also interesting
$$a_k=left(1-frac{1}{k}right)^{k^2}implies log(a_k)=k ^2 logleft(1-frac{1}{k}right)$$
$$log(a_{k+1})-log(a_k)=(k+1) ^2 logleft(1-frac{1}{k+1}right)-k ^2 logleft(1-frac{1}{k}right)$$
Develop as a Taylor series for large values of $k$ to get
$$log(a_{k+1})-log(a_k)=-1+frac{1}{3 k^2}+O(left(frac{1}{k^3}right)$$
Continue with Taylor
$$frac{a_{k+1}}{a_k}=e^{log(a_{k+1})-log(a_k)}=frac 1 eleft(1+frac{1}{3 k^2}+Oleft(frac{1}{k^3}right)right)to frac 1 e$$
answered 24 mins ago
Claude LeiboviciClaude Leibovici
122k1157134
answered 24 mins ago
Claude LeiboviciClaude Leibovici
122k1157134
answered 24 mins ago
Claude LeiboviciClaude Leibovici
122k1157134
answered 24 mins ago
Claude LeiboviciClaude Leibovici
122k1157134
122k1157134
add a comment |
add a comment |
$begingroup$
$begin{array}\
(1-frac1{k})^{k^2}
&=(frac{k-1}{k})^{k^2}\
&=dfrac1{(frac{k}{k-1})^{k^2}}\
&=dfrac1{(1+frac{1}{k-1})^{k^2}}\
&=dfrac1{((1+frac{1}{k-1})^{k})^k}\
&<dfrac1{(1+frac{k}{k-1})^k}
qquadtext{by Bernoulli}\
&=dfrac1{(frac{2k-1}{k-1})^k}\
&<dfrac1{(frac{2k-2}{k-1})^k}\
&=dfrac1{2^k}\
end{array}
$
and the sum of this converges.
answered 13 mins ago
marty cohenmarty cohen
73.8k549128
$endgroup$
add a comment |
$begingroup$
$begin{array}\
(1-frac1{k})^{k^2}
&=(frac{k-1}{k})^{k^2}\
&=dfrac1{(frac{k}{k-1})^{k^2}}\
&=dfrac1{(1+frac{1}{k-1})^{k^2}}\
&=dfrac1{((1+frac{1}{k-1})^{k})^k}\
&<dfrac1{(1+frac{k}{k-1})^k}
qquadtext{by Bernoulli}\
&=dfrac1{(frac{2k-1}{k-1})^k}\
&<dfrac1{(frac{2k-2}{k-1})^k}\
&=dfrac1{2^k}\
end{array}
$
and the sum of this converges.
answered 13 mins ago
marty cohenmarty cohen
73.8k549128
$endgroup$
add a comment |
$begingroup$
$begin{array}\
(1-frac1{k})^{k^2}
&=(frac{k-1}{k})^{k^2}\
&=dfrac1{(frac{k}{k-1})^{k^2}}\
&=dfrac1{(1+frac{1}{k-1})^{k^2}}\
&=dfrac1{((1+frac{1}{k-1})^{k})^k}\
&<dfrac1{(1+frac{k}{k-1})^k}
qquadtext{by Bernoulli}\
&=dfrac1{(frac{2k-1}{k-1})^k}\
&<dfrac1{(frac{2k-2}{k-1})^k}\
&=dfrac1{2^k}\
end{array}
$
and the sum of this converges.
answered 13 mins ago
marty cohenmarty cohen
73.8k549128
$endgroup$
$begin{array}\
(1-frac1{k})^{k^2}
&=(frac{k-1}{k})^{k^2}\
&=dfrac1{(frac{k}{k-1})^{k^2}}\
&=dfrac1{(1+frac{1}{k-1})^{k^2}}\
&=dfrac1{((1+frac{1}{k-1})^{k})^k}\
&<dfrac1{(1+frac{k}{k-1})^k}
qquadtext{by Bernoulli}\
&=dfrac1{(frac{2k-1}{k-1})^k}\
&<dfrac1{(frac{2k-2}{k-1})^k}\
&=dfrac1{2^k}\
end{array}
$
and the sum of this converges.
answered 13 mins ago
marty cohenmarty cohen
73.8k549128
answered 13 mins ago
marty cohenmarty cohen
73.8k549128
answered 13 mins ago
marty cohenmarty cohen
73.8k549128
answered 13 mins ago
marty cohenmarty cohen
73.8k549128
73.8k549128
add a comment |
add a comment |
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$begingroup$
This is susceptible to the same approach as my answer to a different question.
$endgroup$
– T. Bongers
38 mins ago
$begingroup$
Do you know what $left(1-frac{1}{k}right)^{k}$ converges to?
$endgroup$
– JavaMan
32 mins ago
$begingroup$
I cannot believe how horrible my intuition is with this stuff especially given how old I am.
$endgroup$
– Randall
2 mins ago