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Can $a(n) = fracnn+1$ be written recursively?

Recursive Sequence from Finite SequencesFind another recursive algorithm that is equal to a seriesWhat would describe the following basic sequence?Find the limit of recursive sequence, if it exists: $a_n+1=frac7+3a_n3+a_n$Proving that a recursive sequence convergesCan the Fibonacci sequence be written as an explicit rule?How can I find the Limit of this sequence?Turning a recursively defined sequence into an explicit formulaCan the Nilakantha Series be represented in sigma notation?Prove explicit form of a recursive sequence

1

$begingroup$

Take the sequence $$frac12, frac23, frac34, frac45, frac56, frac67, dots$$

Algebraically it can be written as $$a(n) = fracnn + 1$$

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 recursion

share|cite|improve this question

edited 1 hour ago

Jyrki Lahtonen

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asked 3 hours ago

Levi KLevi K

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Levi K is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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1

$begingroup$

Take the sequence $$frac12, frac23, frac34, frac45, frac56, frac67, dots$$

Algebraically it can be written as $$a(n) = fracnn + 1$$

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 recursion

share|cite|improve this question

edited 1 hour ago

Jyrki Lahtonen

110k13172390

asked 3 hours ago

Levi KLevi K

363

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 $$frac12, frac23, frac34, frac45, frac56, frac67, dots$$

Algebraically it can be written as $$a(n) = fracnn + 1$$

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 recursion

share|cite|improve this question

edited 1 hour ago

Jyrki Lahtonen

110k13172390

asked 3 hours ago

Levi KLevi K

363

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 1 hour ago

Jyrki Lahtonen

110k13172390

asked 3 hours ago

Levi KLevi K

363

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 1 hour ago

Jyrki Lahtonen

110k13172390

asked 3 hours ago

Levi KLevi K

363

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Levi K is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 1 hour ago

Jyrki Lahtonen

110k13172390

edited 1 hour ago

Jyrki Lahtonen

110k13172390

asked 3 hours ago

Levi KLevi K

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Levi KLevi K

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3 Answers
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3

$begingroup$

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

It can be written $$A_n + 1 = frac12 - A_n$$ where $$A_1 = frac12$$

share|cite|improve this answer

edited 1 hour ago

user1952500

833712

answered 3 hours ago

Levi KLevi K

363

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Levi K is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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2

$begingroup$

beginalign*
a_n+1 &= fracn+1n+2 \
&= fracn+2-1n+2 \
&= 1 - frac1n+2 text, so \
1 - a_n+1 &= frac1n+2 text, \
frac11 - a_n+1 &= n+2 &[textand so frac11 - a_n = n+1]\
&= n+1+1 \
&= frac11- a_n +1 \
&= frac11- a_n + frac1-a_n1-a_n \
&= frac2-a_n1- a_n text, then \
1 - a_n+1 &= frac1-a_n2- a_n text, and finally \
a_n+1 &= 1 - frac1-a_n2- a_n \
&= frac2-a_n2- a_n - frac1-a_n2- a_n \
&= frac12- a_n text.
endalign*

share|cite|improve this answer

answered 2 hours 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

$$a_n = fracna_n-1 + 1n+1$$

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 3 hours ago

Eevee TrainerEevee Trainer

9,93831740

$endgroup$

add a comment | 

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3

$begingroup$

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

It can be written $$A_n + 1 = frac12 - A_n$$ where $$A_1 = frac12$$

share|cite|improve this answer

edited 1 hour ago

user1952500

833712

answered 3 hours ago

Levi KLevi K

363

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 | 

3

$begingroup$

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

It can be written $$A_n + 1 = frac12 - A_n$$ where $$A_1 = frac12$$

share|cite|improve this answer

edited 1 hour ago

user1952500

833712

answered 3 hours ago

Levi KLevi K

363

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 $$A_n + 1 = frac12 - A_n$$ where $$A_1 = frac12$$

share|cite|improve this answer

edited 1 hour ago

user1952500

833712

answered 3 hours ago

Levi KLevi K

363

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 answer

edited 1 hour ago

user1952500

833712

answered 3 hours ago

Levi KLevi K

363

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.

edited 1 hour ago

user1952500

833712

edited 1 hour ago

user1952500

833712

answered 3 hours ago

Levi KLevi K

363

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 3 hours ago

Levi KLevi K

363

2

$begingroup$

beginalign*
a_n+1 &= fracn+1n+2 \
&= fracn+2-1n+2 \
&= 1 - frac1n+2 text, so \
1 - a_n+1 &= frac1n+2 text, \
frac11 - a_n+1 &= n+2 &[textand so frac11 - a_n = n+1]\
&= n+1+1 \
&= frac11- a_n +1 \
&= frac11- a_n + frac1-a_n1-a_n \
&= frac2-a_n1- a_n text, then \
1 - a_n+1 &= frac1-a_n2- a_n text, and finally \
a_n+1 &= 1 - frac1-a_n2- a_n \
&= frac2-a_n2- a_n - frac1-a_n2- a_n \
&= frac12- a_n text.
endalign*

share|cite|improve this answer

answered 2 hours ago

Eric TowersEric Towers

33.5k22370

$endgroup$

add a comment | 

2

$begingroup$

beginalign*
a_n+1 &= fracn+1n+2 \
&= fracn+2-1n+2 \
&= 1 - frac1n+2 text, so \
1 - a_n+1 &= frac1n+2 text, \
frac11 - a_n+1 &= n+2 &[textand so frac11 - a_n = n+1]\
&= n+1+1 \
&= frac11- a_n +1 \
&= frac11- a_n + frac1-a_n1-a_n \
&= frac2-a_n1- a_n text, then \
1 - a_n+1 &= frac1-a_n2- a_n text, and finally \
a_n+1 &= 1 - frac1-a_n2- a_n \
&= frac2-a_n2- a_n - frac1-a_n2- a_n \
&= frac12- a_n text.
endalign*

share|cite|improve this answer

answered 2 hours ago

Eric TowersEric Towers

33.5k22370

$endgroup$

add a comment | 

$begingroup$

beginalign*
a_n+1 &= fracn+1n+2 \
&= fracn+2-1n+2 \
&= 1 - frac1n+2 text, so \
1 - a_n+1 &= frac1n+2 text, \
frac11 - a_n+1 &= n+2 &[textand so frac11 - a_n = n+1]\
&= n+1+1 \
&= frac11- a_n +1 \
&= frac11- a_n + frac1-a_n1-a_n \
&= frac2-a_n1- a_n text, then \
1 - a_n+1 &= frac1-a_n2- a_n text, and finally \
a_n+1 &= 1 - frac1-a_n2- a_n \
&= frac2-a_n2- a_n - frac1-a_n2- a_n \
&= frac12- a_n text.
endalign*

share|cite|improve this answer

answered 2 hours ago

Eric TowersEric Towers

33.5k22370

$endgroup$

share|cite|improve this answer

answered 2 hours ago

Eric TowersEric Towers

33.5k22370

answered 2 hours ago

Eric TowersEric Towers

33.5k22370

answered 2 hours ago

Eric TowersEric Towers

33.5k22370

0

$begingroup$

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

$$a_n = fracna_n-1 + 1n+1$$

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 3 hours ago

Eevee TrainerEevee Trainer

9,93831740

$endgroup$

add a comment | 

0

$begingroup$

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

$$a_n = fracna_n-1 + 1n+1$$

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 3 hours ago

Eevee TrainerEevee Trainer

9,93831740

$endgroup$

add a comment | 

$begingroup$

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

$$a_n = fracna_n-1 + 1n+1$$

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 3 hours ago

Eevee TrainerEevee Trainer

9,93831740

$endgroup$

share|cite|improve this answer

answered 3 hours ago

Eevee TrainerEevee Trainer

9,93831740

answered 3 hours ago

Eevee TrainerEevee Trainer

9,93831740

answered 3 hours ago

Eevee TrainerEevee Trainer

9,93831740