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Can $a(n) = frac{n}{n+1}$ be written recursively?

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1

$begingroup$

Take the sequence $$frac{1}{2}, frac{2}{3}, frac{3}{4}, frac{4}{5}, frac{5}{6}, frac{6}{7}, dots$$

Algebraically it can be written as $$a(n) = frac{n}{n + 1}$$

Can you write this as a recursive function as well?

A pattern I have noticed:

• Take $A_{n-1}$ and then inverse it. All you have to do is add two to the denominator. However, it is the denominator increase that causes a problem here.

I am currently in Algebra II Honors and learning sequences

sequences-and-series number-theory recursion

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edited 5 mins ago

user1952500

829712

asked 1 hour ago

Levi KLevi K

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1

$begingroup$

Take the sequence $$frac{1}{2}, frac{2}{3}, frac{3}{4}, frac{4}{5}, frac{5}{6}, frac{6}{7}, dots$$

Algebraically it can be written as $$a(n) = frac{n}{n + 1}$$

Can you write this as a recursive function as well?

A pattern I have noticed:

• Take $A_{n-1}$ and then inverse it. All you have to do is add two to the denominator. However, it is the denominator increase that causes a problem here.

I am currently in Algebra II Honors and learning sequences

sequences-and-series number-theory recursion

share|cite|improve this question

edited 5 mins ago

user1952500

829712

asked 1 hour ago

Levi KLevi K

262

New contributor

Levi K 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 | 

$begingroup$

Take the sequence $$frac{1}{2}, frac{2}{3}, frac{3}{4}, frac{4}{5}, frac{5}{6}, frac{6}{7}, dots$$

Algebraically it can be written as $$a(n) = frac{n}{n + 1}$$

Can you write this as a recursive function as well?

A pattern I have noticed:

• Take $A_{n-1}$ and then inverse it. All you have to do is add two to the denominator. However, it is the denominator increase that causes a problem here.

I am currently in Algebra II Honors and learning sequences

sequences-and-series number-theory recursion

share|cite|improve this question

edited 5 mins ago

user1952500

829712

asked 1 hour ago

Levi KLevi K

262

New contributor

Levi K is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited 5 mins ago

user1952500

829712

asked 1 hour ago

Levi KLevi K

262

New contributor

Levi K is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited 5 mins ago

user1952500

829712

asked 1 hour ago

Levi KLevi K

262

New contributor

Levi K is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 5 mins ago

user1952500

829712

edited 5 mins ago

user1952500

829712

asked 1 hour ago

Levi KLevi K

262

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asked 1 hour ago

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3 Answers
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2

$begingroup$

After some further solving, I was able to come up with an answer

It can be written $${A_{n + 1}} = ({2 - A_{n}})^{-1}$$ where $${A_1 = 1/2}$$

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answered 58 mins ago

Levi KLevi K

262

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2

$begingroup$

begin{align*}
a_{n+1} &= frac{n+1}{n+2} \
&= frac{n+2-1}{n+2} \
&= 1 - frac{1}{n+2} text{, so } \
1 - a_{n+1} &= frac{1}{n+2} text{, } \
frac{1}{1 - a_{n+1}} &= n+2 &[text{and so } frac{1}{1 - a_n} = n+1]\
&= n+1+1 \
&= frac{1}{1- a_n} +1 \
&= frac{1}{1- a_n} + frac{1-a_n}{1-a_n} \
&= frac{2-a_n}{1- a_n} text{, then } \
1 - a_{n+1} &= frac{1-a_n}{2- a_n} text{, and finally } \
a_{n+1} &= 1 - frac{1-a_n}{2- a_n} \
&= frac{2-a_n}{2- a_n} - frac{1-a_n}{2- a_n} \
&= frac{1}{2- a_n} text{.}
end{align*}

share|cite|improve this answer

answered 40 mins ago

Eric TowersEric Towers

33.5k22370

$endgroup$

add a comment | 

0

$begingroup$

Just by playing around with some numbers, I determined a recursive relation to be

$$a_n = frac{na_{n-1} + 1}{n+1}$$

with $a_1 = 1/2$. I mostly derived this by noticing developing this would be easier if I negated the denominator of the previous term (thus multiplying by $n$), adding $1$ to the result (giving the increase in $1$ in the denominator), and then dividing again by the desired denominator ($n+1$).

share|cite|improve this answer

answered 1 hour ago

Eevee TrainerEevee Trainer

9,91631740

$endgroup$

add a comment | 

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2

$begingroup$

After some further solving, I was able to come up with an answer

It can be written $${A_{n + 1}} = ({2 - A_{n}})^{-1}$$ where $${A_1 = 1/2}$$

share|cite

answered 58 mins ago

Levi KLevi K

262

New contributor

Levi K 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 | 

2

$begingroup$

After some further solving, I was able to come up with an answer

It can be written $${A_{n + 1}} = ({2 - A_{n}})^{-1}$$ where $${A_1 = 1/2}$$

share|cite

answered 58 mins ago

Levi KLevi K

262

New contributor

Levi K 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 | 

$begingroup$

After some further solving, I was able to come up with an answer

It can be written $${A_{n + 1}} = ({2 - A_{n}})^{-1}$$ where $${A_1 = 1/2}$$

share|cite

answered 58 mins ago

Levi KLevi K

262

New contributor

Levi K is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite

answered 58 mins ago

Levi KLevi K

262

New contributor

Levi K is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered 58 mins ago

Levi KLevi K

262

New contributor

Levi K is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered 58 mins ago

Levi KLevi K

262

2

$begingroup$

begin{align*}
a_{n+1} &= frac{n+1}{n+2} \
&= frac{n+2-1}{n+2} \
&= 1 - frac{1}{n+2} text{, so } \
1 - a_{n+1} &= frac{1}{n+2} text{, } \
frac{1}{1 - a_{n+1}} &= n+2 &[text{and so } frac{1}{1 - a_n} = n+1]\
&= n+1+1 \
&= frac{1}{1- a_n} +1 \
&= frac{1}{1- a_n} + frac{1-a_n}{1-a_n} \
&= frac{2-a_n}{1- a_n} text{, then } \
1 - a_{n+1} &= frac{1-a_n}{2- a_n} text{, and finally } \
a_{n+1} &= 1 - frac{1-a_n}{2- a_n} \
&= frac{2-a_n}{2- a_n} - frac{1-a_n}{2- a_n} \
&= frac{1}{2- a_n} text{.}
end{align*}

share|cite|improve this answer

answered 40 mins ago

Eric TowersEric Towers

33.5k22370

$endgroup$

add a comment | 

2

$begingroup$

begin{align*}
a_{n+1} &= frac{n+1}{n+2} \
&= frac{n+2-1}{n+2} \
&= 1 - frac{1}{n+2} text{, so } \
1 - a_{n+1} &= frac{1}{n+2} text{, } \
frac{1}{1 - a_{n+1}} &= n+2 &[text{and so } frac{1}{1 - a_n} = n+1]\
&= n+1+1 \
&= frac{1}{1- a_n} +1 \
&= frac{1}{1- a_n} + frac{1-a_n}{1-a_n} \
&= frac{2-a_n}{1- a_n} text{, then } \
1 - a_{n+1} &= frac{1-a_n}{2- a_n} text{, and finally } \
a_{n+1} &= 1 - frac{1-a_n}{2- a_n} \
&= frac{2-a_n}{2- a_n} - frac{1-a_n}{2- a_n} \
&= frac{1}{2- a_n} text{.}
end{align*}

share|cite|improve this answer

answered 40 mins ago

Eric TowersEric Towers

33.5k22370

$endgroup$

add a comment | 

$begingroup$

begin{align*}
a_{n+1} &= frac{n+1}{n+2} \
&= frac{n+2-1}{n+2} \
&= 1 - frac{1}{n+2} text{, so } \
1 - a_{n+1} &= frac{1}{n+2} text{, } \
frac{1}{1 - a_{n+1}} &= n+2 &[text{and so } frac{1}{1 - a_n} = n+1]\
&= n+1+1 \
&= frac{1}{1- a_n} +1 \
&= frac{1}{1- a_n} + frac{1-a_n}{1-a_n} \
&= frac{2-a_n}{1- a_n} text{, then } \
1 - a_{n+1} &= frac{1-a_n}{2- a_n} text{, and finally } \
a_{n+1} &= 1 - frac{1-a_n}{2- a_n} \
&= frac{2-a_n}{2- a_n} - frac{1-a_n}{2- a_n} \
&= frac{1}{2- a_n} text{.}
end{align*}

share|cite|improve this answer

answered 40 mins ago

Eric TowersEric Towers

33.5k22370

$endgroup$

share|cite|improve this answer

answered 40 mins ago

Eric TowersEric Towers

33.5k22370

answered 40 mins ago

Eric TowersEric Towers

33.5k22370

answered 40 mins ago

Eric TowersEric Towers

33.5k22370

0

$begingroup$

Just by playing around with some numbers, I determined a recursive relation to be

$$a_n = frac{na_{n-1} + 1}{n+1}$$

with $a_1 = 1/2$. I mostly derived this by noticing developing this would be easier if I negated the denominator of the previous term (thus multiplying by $n$), adding $1$ to the result (giving the increase in $1$ in the denominator), and then dividing again by the desired denominator ($n+1$).

share|cite|improve this answer

answered 1 hour ago

Eevee TrainerEevee Trainer

9,91631740

$endgroup$

add a comment | 

0

$begingroup$

Just by playing around with some numbers, I determined a recursive relation to be

$$a_n = frac{na_{n-1} + 1}{n+1}$$

with $a_1 = 1/2$. I mostly derived this by noticing developing this would be easier if I negated the denominator of the previous term (thus multiplying by $n$), adding $1$ to the result (giving the increase in $1$ in the denominator), and then dividing again by the desired denominator ($n+1$).

share|cite|improve this answer

answered 1 hour ago

Eevee TrainerEevee Trainer

9,91631740

$endgroup$

add a comment | 

$begingroup$

Just by playing around with some numbers, I determined a recursive relation to be

$$a_n = frac{na_{n-1} + 1}{n+1}$$

with $a_1 = 1/2$. I mostly derived this by noticing developing this would be easier if I negated the denominator of the previous term (thus multiplying by $n$), adding $1$ to the result (giving the increase in $1$ in the denominator), and then dividing again by the desired denominator ($n+1$).

share|cite|improve this answer

answered 1 hour ago

Eevee TrainerEevee Trainer

9,91631740

$endgroup$

share|cite|improve this answer

answered 1 hour ago

Eevee TrainerEevee Trainer

9,91631740

answered 1 hour ago

Eevee TrainerEevee Trainer

9,91631740

answered 1 hour ago

Eevee TrainerEevee Trainer

9,91631740