Show that $\lim_{n\to\infty}S_n$ exists 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 $n$
 A: 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 $n \geq 2$,
$$
S_n \leq \frac{1}{n} \quad \Longrightarrow \quad S_{n+1} = \frac{n}{n+1} S_n^2 \leq \frac{n}{n+1} \bigg(\frac{1}{n}\bigg)^2 = \frac{1}{n(n+1)} \leq \frac{1}{n+1}
$$
A: 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.
A: 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 $S_n\to 0$.
A: 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.
