$s_1 = 1$ and $s_{n+1}=(\frac{n}{n+1})s_n^2$ monotonically decreases? Hi I came across this question on page 65 of Elementary Analysis by Kenneth A.Ross.
I am given that $s_1 = 1$ and I need to show that $s_{n+1} = (\frac{n}{n+1}) s_n^2$ monotonically decreases.
I'm thinking about using induction but I got stuck here:
Assume that $s_{n+1}<s_{n}$, let's consider $s_{n+2}$:
$s_{n+2} = (\frac{n+1}{n+2})s_{n+1}^2=(\frac{n+1}{n+2}s_{n+1})s_{n+1}$
There is actually hint behind the book, but it says "apparently $\frac{n+1}{n+2}s_{n+1}$ <1", which to be honest, isn't so obvious to me. Any idea?
Also, another question is that, when using the ratio to decide whether $s_{n+1}$ is larger than $s_n$ or not, I feel like we are already assuming that all terms of the sequence must be positive, which seems obvious, but is it necessary to "prove" it first?
 A: Edit: Looking at this now and others, I've decided to change it to showing $s_n \leq 1$ for all $n$.
You don't really need that $s_{n}$ is positive. You only need that it is bounded above by one for the proof.
To show this, use induction. In fact, $s_{n+1} < s_{n}$ will also fall out of this proof. For $n = 1$, $s_1 = 1 \leq 1$, so it holds in this case obviously. Now assume that $s_{n} \leq 1$ for some $n \geq 1$ and we show the $n +1$ case. We have:
$$
s_{n+1} = \dfrac{n}{n+1} s_{n}^2 = \bigg(\dfrac{n}{n+1} \cdot s_{n}\bigg) s_{n} < (1\cdot 1) s_{n} = s_{n} \leq 1.
$$
Note that this shows both that $s_{n+1} < s_{n}$ and $s_{n+1} \leq 1$. Thus by induction, it holds for all $n$, concluding the proof.
A: I think it's more important to see where $\frac{n+1}{n+2}s_{n+1} < 1$ came from.
Note that 
\begin{align}
s_{n+1} - s_n <0 &\iff \frac{s_n}{n+1}(ns_n - n - 1)<0\\
&\iff (ns_n - n - 1) < 0\\
&\iff \frac{n}{n+1}s_n<1
\end{align}
So your job is to show that $\frac{n}{n+1}s_n<1$ is true and you have to perform induction to prove this. Then you can conclude that $\frac{n+1}{n+2}s_{n+1}<1$ must also be true by induction.
Now as for boundness, note that $s_1 = 1$, so we claim $s_n > 0$ and assuming this holds for the case $n$, we note that $s_{n+1} > 0 \iff s_{n+1}=(\frac{n}{n+1})s_n^2 > 0$, which is true by the inductive hypothesis. From there on, I think you are on the right track.
A: First observe that it is easy to prove by induction that $s_{n+1} < 1$ for $n>1$ . An easy way is by contradiction: For $s_k$ to be greater or equal to unity the $s_{k-1}$ should also be $>=1$. But by induction it contradicts the fact that $s_2<1$.
Next from
$$s_{n+2} = (\frac{n+1}{n+2} s_{n+1}) s_{n+1}$$
and noting that $(\frac{n+1}{n+2} s_{n+1})<1 $ we get that $s_{n+2}<s_{n+1}$
A: First, observe that $s_n > 0, \forall n$. Thus $s_{n+1} - s_n = \dfrac{ns_n^2 - (n+1)s_n}{n+1} = s_n\cdot \dfrac{ns_n - n - 1}{n+1}$
Claim: $s_n \le \dfrac{n+1}{n} ,\forall n \geq 1$
Proof: First show that $s_n \leq 1, \forall n \geq 1$ by induction. For $n = 1$, s$_1 = 1 \leq 1$ which is true. So assume $s_n \leq 1$, then $s_{n+1} = \dfrac{n}{n+1}\cdot s_n^2 \leq \dfrac{n}{n+1} \cdot 1 < 1$. So $s_n \leq 1 < \dfrac{n+1}{n}, \forall n \geq 1$. Thus: $s_n \leq \dfrac{n+1}{n}$, and the claim is proved. This implies $s_{n+1} - s_n \leq 0$, and $(s_n)$ is a decreasing sequence.
