Limit of this sequence? $a_1 = 1,a_{n+1} = \left(1 - 1/(n+1)^2\right)a_n$

Question

$a_1 = 1$

$a_{n+1} = \left[1 - \dfrac{1}{(n+1)^2}\right]a_n$ for $n \in \mathbb{N}$.

I've proved it converges by bounding it below (by 0) and showing it is monotonically decreasing.

How can I find its limit?

I've tried $\displaystyle\lim_{n \to \infty} a_n = \lim_{n \to \infty} a_{n+1}$ but ran into problems.

Maybe show by induction that $\displaystyle a_n = \frac{n+1}{2n}$. — Alma Arjuna, Oct 30 '23 at 01:51

copper.hat · Answer 1 · 2023-10-30T02:17:44.007

3

$a_{n+1} = {n^2 +2n \over (n+1)^2} a_n = {n+2 \over n+1} { n \over n+1} a_n$ and so ${n+1 \over n+2} a_{n+1} = { n \over n+1} a_n$. In particular, ${ n \over n+1} a_n =C$ for some constant $C$.

edited Oct 30 '23 at 02:17

answered Oct 30 '23 at 02:10

copper.hat

172,524

Patterns like this appear in convergence & speed of convergence proofs. More usually as inequalities, but the underlying idea is to look for some 'constant'. – copper.hat Oct 30 '23 at 02:13
I don't follow your work. Can you start with the definition of $a_{n + 1}$ I gave? – user129393192 Oct 30 '23 at 02:14
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I did, write $1 -{1 \over (1+n^2)} = { (n+1)^2-1 \over (n+1)^2} = {n^2+2n \over (n+1)^2}$. (I also had a typo.) – copper.hat Oct 30 '23 at 02:16
How do you have the last bit? The $= C$ part? – user129393192 Oct 30 '23 at 02:18
You need to do some work. ${ n \over n+1} a_n = { n-1 \over n+1-1} a_{n-1} = ... = {1 \over 2} a_1$. – copper.hat Oct 30 '23 at 02:22
Alas, this was a multiduplicate. So, better not edit a new answer (at least not here). – Anne Bauval Oct 30 '23 at 03:04
@copper.hat I get what you've said now, but then how does that get us the limit? – user129393192 Oct 30 '23 at 03:28
It gives you a formula for $a_n$ and the limit is easy to compute. – copper.hat Oct 30 '23 at 03:56

mof · Answer 2 · 2023-10-30T02:14:00.703

1

Something you could try is rewriting it as the following $$ a_{n} = (1 - \frac{1}{n^2})a_{n-1} = (\frac{(n)^2 -1}{n^2})a_{n-1}= (\frac{(n+1)(n-1)}{n^2})a_{n-1} $$ and then the next steps become a little more obvious, notably using the recurrence relationship $$ a_n = \frac{(n+1)(n-1)}{n^2} \cdot \frac{(n)(n-2)}{(n-1)^2}\cdot \frac{(n-1)(n-3)}{(n-2)^2} ... \\ $$ from here you can notice that all the terms will cancel expect the first term and the last term which will leave you with: $$ a_n = \frac{n+1}{2n} $$ from there the you can apply the limit as you would to any other series to show that it converges to $\frac{1}{2}$

edited Oct 30 '23 at 02:14

answered Oct 30 '23 at 02:08

mof

46
4

At the bottom, we have $n + 1$, not $n$. – user129393192 Oct 30 '23 at 02:16
the definitions are equivalent im looking at $a_n$ instead of $a_{n+1}$, its easier for me to see in that way but if you would prefer you could just add one everywhere you see n – mof Oct 30 '23 at 02:18
Alas, this was a multiduplicate. So, better not edit a new answer (at least not here). – Anne Bauval Oct 30 '23 at 03:05

Limit of this sequence? $a_1 = 1,a_{n+1} = \left(1 - 1/(n+1)^2\right)a_n$

2 Answers2