# Prove that $1\cdot \frac{1}{2^2}\cdot ...\cdot \frac{1}{n^n}< \left(\frac{2}{n+1}\right)^\frac{n(n+1)}{2}$

Prove that $$1\cdot \frac{1}{2^2}\cdot ...\cdot \frac{1}{n^n}< \left(\frac{2}{n+1}\right)^\frac{n(n+1)}{2}$$ where $$n$$ is a positive integer.
My book suggests using AM-GM, but I couldn't do it. I just applied AM-GM to the numbers in the LHS, but it looks like I need some more upper bounds.

• More precisely, we should have $\geq$ rather than $>$, since we clearly have equality of both LHS and RHS when $n=1$. Commented Apr 22, 2020 at 16:00
• @ ChemistryGeek does my solution make sense to you? Commented Apr 22, 2020 at 16:15
• @ONGSEEHAI yes, it's fine, thanks Commented Apr 22, 2020 at 21:51

Apply AM/GM to the sequence $$a_1,\ldots,a_{n(n+1)/2}$$ which looks lie $$1,1/2,1/2,1/3,1/3,1/3,\ldots,1/n$$ ($$1/k$$ occurs $$k$$ times for $$1\le k\le n$$). The $$AM$$ is $$2/(n+1)$$ etc.

By AM-GM, we have:

$$\frac{1+\frac{1}{2}+\frac{1}{2}+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}+...+(\frac{1}{n}+...+\frac{1}{n})}{\frac{n(n+1)}{2}}$$

$$=\frac{1+1+1+...+1}{\frac{n(n+1)}{2}}$$, where there are $$n$$ copies of $$1$$.

$$=\frac{n}{\frac{n(n+1)}{2}}=\frac{2}{n+1}$$

$$\geq \sqrt[\leftroot{-2}\uproot{2}\frac{n(n+1)}{2}]{1 \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot ... \cdot (\frac{1}{n}...\frac{1}{n})}$$

$$=\sqrt[\leftroot{-2}\uproot{2}\frac{n(n+1)}{2}]{1 \cdot \frac{1}{2^2} \cdot ... \cdot \frac{1}{n^n}}$$

But $$\frac{2}{n+1} \geq \sqrt[\leftroot{-2}\uproot{2}\frac{n(n+1)}{2}]{1 \cdot \frac{1}{2^2} \cdot ... \cdot \frac{1}{n^n}} \Rightarrow 1 \cdot \frac{1}{2^2} \cdot ... \cdot \frac{1}{n^n} \leq (\frac{2}{n+1})^{\frac{n(n+1)}{2}}$$, thus concluding our proof.

• Note that the main trick behind applying AM-GM is to decompose and view each term of the form $\frac{1}{k^k}$ as $k$ equivalent copies of $\frac{1}{k}$. Commented Apr 22, 2020 at 16:04
• Note that there are $\frac{n(n+1)}{2}$ terms in our first line. To understand why, observe that there is $1$ copy of $1$, $2$ copies of $2$, ... , $n$ copies of $n$. And we thus have our standard triangular sum of the first $n$ natural numbers. Commented Apr 22, 2020 at 16:08
• Would appreciate if anyone could provide suggestions (if any) on how to improve my presentation. Thanks in advance! Commented Apr 22, 2020 at 16:10

Also induction works well here.

$$1\cdot \dfrac{1}{2^2}\cdots \dfrac{1}{n^n}\cdot\dfrac{1}{(n+1)^{n+1}} \le \left(\dfrac{2}{n+1}\right)^\frac{n(n+1)}{2}\cdot \dfrac{1}{(n+1)^{n+1}} \\= \Big[ \Big(\dfrac{2}{n+1}\Big)^{\frac{n}{2}}\cdot \dfrac{1}{n+1} \Big]^{n+1} \le(?)\ \left(\dfrac{2}{n+2}\right)^\frac{(n+1)(n+2)}{2}=\Big[ \Big(\dfrac{2}{n+2}\Big)^{\frac{n+2}{2}} \Big]^{n+1}.$$

To conclude you have to show that

$$\Big(\dfrac{2}{n+1}\Big)^{\frac{n}{2}} \cdot \dfrac{1}{n+1} \le \Big(\dfrac{2}{n+2}\Big)^{\frac{n+2}{2}}=\Big(\dfrac{2}{n+2}\Big)^{\frac{n}{2}}\cdot \dfrac{2}{n+2}.$$