[Math] Proof that a recursively defined sequence is monotonically decreasing.

induction

I am wanting to prove that the following recursive sequence is monotonic decreasing via proof by induction.

Let $ S_1 = 1, ~ S_{n+1} = \frac{n}{n+1} (S_n)^2;~ n \geq 1. $

Here is what I have so far but I feel the proof fails at my last statement and I am unsure how to correct it.

$$ \text{Base:} ~ S_1 = 1 > \frac{1}{2} = S_2 \\ $$ $$ \text{Assumption:} ~ S_{k+1} > S_{k+2} \\ $$ $$ \text{Proof for:} ~ k+2: $$ $$ \text{ie:} ~ S_{k+1} > S_{k+2} \\ \Rightarrow S_{k+2} = \frac{k+1}{k+2} (S_{k+1})^2 \\ \Rightarrow S_{k+2} =\frac{k+1}{k+2}(\frac{k}{k+1})^2S_{k}^4 \\ \Rightarrow S_{k+2} = \frac{k^2}{(k+1)(k+2)}S_{k}^4 < \frac{k^2}{(k+1)(k+1)}S_{k}^4 = [(\frac{k}{k+1})S_k^2]^2 = S_{k+1}^2 \\ \text{Since}~ S_{k+1}^2 > S_{k+2} \Rightarrow S_{k+1} > S_{k+2} $$

Is this fine or have I messed up badly?

Any help/hints is/are appreciated.

Best Answer

HINT: Use the fact that $x^2<x$ whenever $0<x<1$. I’ve left the details spoiler-protected.

Note that $0<S_2<1$. Suppose that $0<S_n<1$, where $n\ge 2$. Then $0<S_n^2<S_n<1$, so $0<S_{n+1}=\frac{n}{n+1}S_n^2<S_n^2<S_n<1$; this not only tells you that $S_{n+1}<S_n$, it also establishes that $S_{n+1}$ satisfies the condition $0<S_{n+1}<1$, allowing the induction to continue.

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