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

Formula for a sequenceUse of Recursively Defined FunctionsRecursive Sequence from Finite SequencesFinding the lowest common value in repeating sequencesUnexplanied pattern from increasing rational sequencesWhat would describe the following basic sequence?Understanding sub-sequencesCan the Fibonacci sequence be written as an explicit rule?Turning a recursively defined sequence into an explicit formulaWhat's the formula for producing these series?

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 number-theory recursion

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edited 6 mins ago

user1952500

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asked 1 hour ago

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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 number-theory recursion

share|cite|improve this question

edited 6 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 $$frac12, frac23, frac34, frac45, frac56, frac67, dots$$

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 number-theory recursion

share|cite|improve this question

edited 6 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 6 mins ago

user1952500

829712

asked 1 hour ago

Levi KLevi K

262

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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 6 mins ago

user1952500

829712

asked 1 hour ago

Levi KLevi K

262

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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 6 mins ago

user1952500

829712

edited 6 mins ago

user1952500

829712

asked 1 hour ago

Levi KLevi K

262

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asked 1 hour ago

Levi KLevi K

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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 $$A_1 = 1/2$$

share|cite

answered 59 mins ago

Levi KLevi K

262

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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 41 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

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

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 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 59 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 59 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 59 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 59 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 59 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 59 mins ago

Levi KLevi K

262

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 41 mins 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 41 mins 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 41 mins ago

Eric TowersEric Towers

33.5k22370

$endgroup$

share|cite|improve this answer

answered 41 mins ago

Eric TowersEric Towers

33.5k22370

answered 41 mins ago

Eric TowersEric Towers

33.5k22370

answered 41 mins ago

Eric TowersEric Towers

33.5k22370

0

$begingroup$

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

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

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 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

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

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 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

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

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 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