Show that $\lim_{n\to\infty}S_n$ exists

Question

Let $S_1 = 1$ and $S_{n+1} = (\frac{n}{n+1})S_n^2$ for $n \geq 1$

I am asked to show that $\lim_{n\to\infty}S_n$ exists

I am wondering whether or not I solve for $S_n$ and then let $S_{n+2} \leq S_{n+1}$ by plugging in $n+2$ for $n$

E W H Lee · Answer 1 · 2014-09-20T23:15:54.473

2

It is quite easy to see that $S_n > 0$ for all $n$. Prove, by induction on $n \geq 2$, that $\displaystyle S_n \leq \frac{1}{n}$. It follows that $\displaystyle \lim_{n \rightarrow \infty} S_n = 0$.

ADDED LATER: For $n \geq 2$,

$$ S_n \leq \frac{1}{n} \quad \Longrightarrow \quad S_{n+1} = \frac{n}{n+1} S_n^2 \leq \frac{n}{n+1} \bigg(\frac{1}{n}\bigg)^2 = \frac{1}{n(n+1)} \leq \frac{1}{n+1} $$

edited Sep 20 '14 at 23:15

answered Sep 20 '14 at 21:19

E W H Lee

2,336

Could you please show how to do the induction step to show that $S_n\le \frac{1}{n}$ - thanks. – user84413 Sep 20 '14 at 23:07
@user84413: Answer updated. – E W H Lee Sep 20 '14 at 23:17
Thank you - I was getting confused when I tried this, and now I can see it's pretty straightforward. – user84413 Sep 20 '14 at 23:19

user84413 · Answer 2 · 2014-09-20T23:20:55.300

1

Hint:

a) First show by induction that $0<S_n\le1$ for all $n$.

b) Then show by induction that $S_{n+1}<S_n$ for all $n$.

Now you can conclude that you have a decreasing sequence that is bounded below.

edited Sep 20 '14 at 23:20

answered Sep 20 '14 at 22:46

user84413

27,211

can you explain more? thank you – Pasie15 Sep 20 '14 at 23:01
The inequality $0<S_n\le1$ is clear when $n=1$, and if it's true for an integer $n\ge1$, then it is true for $n+1$ since it is not hard to show that $S_{n+1}=\frac{n}{n+1}S_{n}^{2}$ satisfies $0<S_{n+1}\le1$ if $0<S_n\le 1$. Then you can use the inequality from part a) to show by induction that $S_{n+1}<S_{n}$ for all n by induction. Once you have shown you have a bounded monotonic sequence, you can conclude that it converges. – user84413 Sep 20 '14 at 23:13

score 1 · Answer 3 · answered Sep 20 '14 at 23:10

We have that $$ S_1=1,\,\,S_2=\frac{1}{2}, $$ and $S_{n+1}<S_n^2$, for all $n$. Hence $$ 0<S_n<S_{n-1}^2<S_{n-2}^{2^2}<S_{n-3}^{2^3}<S_2^{2^{n-2}}=2^{-2^{n-2}}. $$ Now, $2^{n-2}>n-2$, for all $n$, and hence $$ 0<S_n<\frac{1}{2^{n-2}}, $$ which implies that $S_n\to 0$.

score 1 · Answer 4 · answered Sep 20 '14 at 23:25

Here is a particularly fast way to do it: $0\leq S_n\leq 1$ for all $n$, so $S_{n+1}\leq \frac{n}{n+1} S_n \leq \frac{n}{n+1} \cdot \frac{n-1}{n} S_{n-1} \leq \cdots \leq \frac{n}{n+1} \cdot \frac{n-1}{n} \cdots \frac{1}{2} \cdot 1 = \frac{1}{n+1}\to 0$

Also, since you were asked to show that the limit exists, rather than to calculate it, you can use the approach you suggested. We have $S_{n+1} \leq \frac{n}{n+1} S_n \leq S_n$, so the sequence is nonincreasing, and bounded from below by $0$. This implies that it has a limit.

Show that $\lim_{n\to\infty}S_n$ exists

4 Answers4