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

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2

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 recursion

share|cite|improve this question

edited 2 days ago

Jyrki Lahtonen

110k13172390

asked 2 days ago

Levi KLevi K

435

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

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 recursion

share|cite|improve this question

edited 2 days ago

Jyrki Lahtonen

110k13172390

asked 2 days ago

Levi KLevi K

435

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 recursion

share|cite|improve this question

edited 2 days ago

Jyrki Lahtonen

110k13172390

asked 2 days ago

Levi KLevi K

435

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 2 days ago

Jyrki Lahtonen

110k13172390

asked 2 days ago

Levi KLevi K

435

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 2 days ago

Jyrki Lahtonen

110k13172390

asked 2 days ago

Levi KLevi K

435

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 2 days ago

Jyrki Lahtonen

110k13172390

edited 2 days ago

Jyrki Lahtonen

110k13172390

asked 2 days ago

Levi KLevi K

435

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asked 2 days ago

Levi KLevi K

435

5 Answers
5

active

oldest

votes

6

$begingroup$

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 2 days ago

Eric TowersEric Towers

33.6k22370

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Brilliant! I got the same answer, but I appreciate all the work you put in showing how to get there. I just found it by noticing a pattern with 2's, but your solution is a great mathematical approach. Thank you!
  $endgroup$
  – Levi K
  2 days ago
  

  
  
  
  

add a comment | 

3

$begingroup$

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

It can be written $${A_{n + 1}} = frac{1}{2 - A_{n}}$$ where $$A_1 = frac{1}{2}$$

share|cite|improve this answer

edited 2 days ago

user1952500

915712

answered 2 days ago

Levi KLevi K

435

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 | 

1

$begingroup$

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 2 days ago

Eevee TrainerEevee Trainer

10.2k31742

$endgroup$

add a comment | 

1

$begingroup$

Perhaps a bit simpler is to note that
$$
overbrace{ frac1{1-a_n} }^{n+1}+1=overbrace{frac1{1-a_{n+1}}}^{n+2}\
$$
solving for $a_{n+1}$ yields
$$
frac1{2-a_n}=a_{n+1}\
$$

share|cite|improve this answer

answered yesterday

robjohn♦robjohn

270k27313642

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I like the way you thought about it!
  $endgroup$
  – Levi K
  yesterday
  

  
  
  
  

add a comment | 

0

$begingroup$

You can, and in multiple ways. Such as:$$a_n=frac{1}{a_{n-1}+frac{2}{n}}tag{1}$$

or

$$a_n=frac{na_{n-1}+1}{n+1}tag{2}$$

etc.

equation 1, is simply noting: $$frac{n-1}{n}+frac{2}{n}=frac{n+1}{n}=frac{1}{a_n}$$ Where the first fraction in the sum is $a_{n-1}$

and equation 2, simply notes:

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

etc.

share|cite|improve this answer

answered 2 days ago

Roddy MacPheeRoddy MacPhee

742118

$endgroup$

add a comment | 

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6

$begingroup$

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 2 days ago

Eric TowersEric Towers

33.6k22370

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Brilliant! I got the same answer, but I appreciate all the work you put in showing how to get there. I just found it by noticing a pattern with 2's, but your solution is a great mathematical approach. Thank you!
  $endgroup$
  – Levi K
  2 days ago
  

  
  
  
  

add a comment | 

6

$begingroup$

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 2 days ago

Eric TowersEric Towers

33.6k22370

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Brilliant! I got the same answer, but I appreciate all the work you put in showing how to get there. I just found it by noticing a pattern with 2's, but your solution is a great mathematical approach. Thank you!
  $endgroup$
  – Levi K
  2 days ago
  

  
  
  
  

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 2 days ago

Eric TowersEric Towers

33.6k22370

$endgroup$

share|cite|improve this answer

answered 2 days ago

Eric TowersEric Towers

33.6k22370

answered 2 days ago

Eric TowersEric Towers

33.6k22370

answered 2 days ago

Eric TowersEric Towers

33.6k22370

3

$begingroup$

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

It can be written $${A_{n + 1}} = frac{1}{2 - A_{n}}$$ where $$A_1 = frac{1}{2}$$

share|cite|improve this answer

edited 2 days ago

user1952500

915712

answered 2 days ago

Levi KLevi K

435

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 $${A_{n + 1}} = frac{1}{2 - A_{n}}$$ where $$A_1 = frac{1}{2}$$

share|cite|improve this answer

edited 2 days ago

user1952500

915712

answered 2 days ago

Levi KLevi K

435

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 $${A_{n + 1}} = frac{1}{2 - A_{n}}$$ where $$A_1 = frac{1}{2}$$

share|cite|improve this answer

edited 2 days ago

user1952500

915712

answered 2 days ago

Levi KLevi K

435

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 2 days ago

user1952500

915712

answered 2 days ago

Levi KLevi K

435

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 2 days ago

user1952500

915712

edited 2 days ago

user1952500

915712

answered 2 days ago

Levi KLevi K

435

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 2 days ago

Levi KLevi K

435

1

$begingroup$

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 2 days ago

Eevee TrainerEevee Trainer

10.2k31742

$endgroup$

add a comment | 

1

$begingroup$

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 2 days ago

Eevee TrainerEevee Trainer

10.2k31742

$endgroup$

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 2 days ago

Eevee TrainerEevee Trainer

10.2k31742

$endgroup$

share|cite|improve this answer

answered 2 days ago

Eevee TrainerEevee Trainer

10.2k31742

answered 2 days ago

Eevee TrainerEevee Trainer

10.2k31742

answered 2 days ago

Eevee TrainerEevee Trainer

10.2k31742

1

$begingroup$

Perhaps a bit simpler is to note that
$$
overbrace{ frac1{1-a_n} }^{n+1}+1=overbrace{frac1{1-a_{n+1}}}^{n+2}\
$$
solving for $a_{n+1}$ yields
$$
frac1{2-a_n}=a_{n+1}\
$$

share|cite|improve this answer

answered yesterday

robjohn♦robjohn

270k27313642

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I like the way you thought about it!
  $endgroup$
  – Levi K
  yesterday
  

  
  
  
  

add a comment | 

1

$begingroup$

Perhaps a bit simpler is to note that
$$
overbrace{ frac1{1-a_n} }^{n+1}+1=overbrace{frac1{1-a_{n+1}}}^{n+2}\
$$
solving for $a_{n+1}$ yields
$$
frac1{2-a_n}=a_{n+1}\
$$

share|cite|improve this answer

answered yesterday

robjohn♦robjohn

270k27313642

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I like the way you thought about it!
  $endgroup$
  – Levi K
  yesterday
  

  
  
  
  

add a comment | 

$begingroup$

Perhaps a bit simpler is to note that
$$
overbrace{ frac1{1-a_n} }^{n+1}+1=overbrace{frac1{1-a_{n+1}}}^{n+2}\
$$
solving for $a_{n+1}$ yields
$$
frac1{2-a_n}=a_{n+1}\
$$

share|cite|improve this answer

answered yesterday

robjohn♦robjohn

270k27313642

$endgroup$

share|cite|improve this answer

answered yesterday

robjohn♦robjohn

270k27313642

answered yesterday

robjohn♦robjohn

270k27313642

answered yesterday

robjohn♦robjohn

270k27313642

0

$begingroup$

You can, and in multiple ways. Such as:$$a_n=frac{1}{a_{n-1}+frac{2}{n}}tag{1}$$

or

$$a_n=frac{na_{n-1}+1}{n+1}tag{2}$$

etc.

equation 1, is simply noting: $$frac{n-1}{n}+frac{2}{n}=frac{n+1}{n}=frac{1}{a_n}$$ Where the first fraction in the sum is $a_{n-1}$

and equation 2, simply notes:

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

etc.

share|cite|improve this answer

answered 2 days ago

Roddy MacPheeRoddy MacPhee

742118

$endgroup$

add a comment | 

0

$begingroup$

You can, and in multiple ways. Such as:$$a_n=frac{1}{a_{n-1}+frac{2}{n}}tag{1}$$

or

$$a_n=frac{na_{n-1}+1}{n+1}tag{2}$$

etc.

equation 1, is simply noting: $$frac{n-1}{n}+frac{2}{n}=frac{n+1}{n}=frac{1}{a_n}$$ Where the first fraction in the sum is $a_{n-1}$

and equation 2, simply notes:

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

etc.

share|cite|improve this answer

answered 2 days ago

Roddy MacPheeRoddy MacPhee

742118

$endgroup$

add a comment | 

$begingroup$

You can, and in multiple ways. Such as:$$a_n=frac{1}{a_{n-1}+frac{2}{n}}tag{1}$$

or

$$a_n=frac{na_{n-1}+1}{n+1}tag{2}$$

etc.

equation 1, is simply noting: $$frac{n-1}{n}+frac{2}{n}=frac{n+1}{n}=frac{1}{a_n}$$ Where the first fraction in the sum is $a_{n-1}$

and equation 2, simply notes:

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

etc.

share|cite|improve this answer

answered 2 days ago

Roddy MacPheeRoddy MacPhee

742118

$endgroup$

share|cite|improve this answer

answered 2 days ago

Roddy MacPheeRoddy MacPhee

742118

answered 2 days ago

Roddy MacPheeRoddy MacPhee

742118

answered 2 days ago

Roddy MacPheeRoddy MacPhee

742118