Show that $\pi-\left(90\sum_{n=1}^mn^{-4}\right)^{\frac14}<\pi-\left(6\sum_{n=1}^mn^{-2}\right)^{\frac12}$

Question

I was bored and I found the inequality $$\pi-\left(90\sum_{n=1}^mn^{-4}\right)^{\frac14}<\pi-\left(6\sum_{n=1}^mn^{-2}\right)^{\frac12},$$ where $m$ is a positive integer. Which is basically derived from $\zeta(4)=\frac{\pi^4}{90}$ and $\zeta(2)=\frac{\pi^2}6$.

This is equivalent to $$5\sum_{n=1}^mn^{-4}>2\left(\sum_{n=1}^mn^{-2}\right)^2.$$ At first I tried a direct proof, and simplified it down to $$3\sum_{n=1}^mn^{-4}>4\sum_{\substack{1\le i,j\le m\\i\neq j}}(ij)^{-2}$$ but there wasn't an obvious way for me to proceed from here. My second attempt was to use induction, and the case $m=1$ was trivial, but the IH was too weak. I used the hypothesis to get $$5(m+1)^{-4}+5\sum_{n=1}^mn^{-4}>5(m+1)^{-4}+2\left(\sum_{n=1}^mn^{-2}\right)^2,$$ but $2\left(\sum_{n=1}^mn^{-2}+(m+1)^{-2}\right)^2\ge RHS$ (by desmos), so that didn't work as well. I don't think Cauchy works either, though not that sure.

If anyone could provide a solution, then I'd greatly appreciate it! I preferably want an elementary solution.

Answer 1

$5\sum_{n=1}^mn^{-4}>2\left(\sum_{n=1}^mn^{-2}\right)^2. $

$5(\frac{\pi^4}{90}-\sum_{n=m+1}^{\infty}n^{-4})>2(\frac{\pi^2}{6}-\left(\sum_{n=m+1}^{\infty}n^{-2})\right)^2). $

$5(\frac{\pi^4}{90}-t_4(m))>2(\frac{\pi^2}{6}-t_2(m))^2. $

$\frac{\pi^4}{18}-5t_4(m) \gt 2(\frac{\pi^4}{36}-2\frac{\pi^2}{6}t_2(m)-t_2^2(m))\\ = \frac{\pi^4}{18}-\frac{2\pi^2}{3}t_2(m)-2t_2^2(m)\\ $

Interesting the $\pi^2/18$ cancels out.

$5t_4(m) \lt \frac{2\pi^2}{3}t_2(m)+2t_2^2(m)\\ $

According to Computing the tail of the zeta function $\sum_{n>x}n^{-s}$,

$t_s(n) =\frac{n^{1-s}}{s-1}\left(1-\frac{s-1}{2n}+O\left(\frac1{n^2}\right)\right). $

Therefore $t_2(n) =\frac{n^{-1}}{1}\left(1-\frac{1}{2n}+O\left(\frac1{n^2}\right)\right) =\frac1{n}\left(1-\frac{1}{2n}+O\left(\frac1{n^2}\right)\right) $ and $t_4(n) =\frac{n^{-3}}{3}\left(1-\frac{3}{2n}+O\left(\frac1{n^2}\right)\right) $.

Using the first term of each, $t_2(m) \approx \dfrac1{m} $ and $t_4(m) \approx \dfrac1{3n^3} $.

Putting these in, ths inequality becomes

$5 \dfrac1{3m^3} \lt \dfrac{2\pi^2}{3}\dfrac1{m}+2\dfrac1{m^2} $ or $\dfrac53 \lt \dfrac{2\pi^2m^2}{3}+2m $ and this is true for all $m$.

I'll leave it at this.

Answer 2

Let $$f(m) = 5\sum_{n=1}^m n^{-4} - 2\left(\sum_{n=1}^m n^{-2}\right)^2.$$ We have $$f(m+1) = 5\sum_{n=1}^m n^{-4} + \frac{5}{(m+1)^4} - 2\left(\sum_{n=1}^m n^{-2} + \frac{1}{(m+1)^2}\right)^2 $$ and \begin{align*} f(m) - f(m+1) &= \frac{4}{(m+1)^2}\sum_{n=1}^m n^{-2} - \frac{3}{(m+1)^4}\\ &\ge \frac{4}{(m+1)^2} m^{-2} - \frac{3}{(m+1)^4}\\ & > 0. \end{align*} Also, $\lim_{m\to \infty} f(m) = 5\zeta(4) - 2[\zeta(2)]^2 = 5\cdot \frac{\pi^2}{90} - 2\cdot (\frac{\pi^2}{6})^2 = 0$.

Thus, $f(m) > 0$ for all $m\ge 1$.

Answer 3

At leats, for large $m$, it is easy to show $$f(m) = 5\sum_{n=1}^m n^{-4} - 2\left(\sum_{n=1}^m n^{-2}\right)^2=5 H_m^{(4)}-2 \left(H_m^{(2)}\right){}^2$$ Using asymptotics $$f(m)=\frac{2 \pi ^2}{3 m}-\frac{6+{\pi ^2}}{3m^2}+\frac{3+\pi ^2}{9 m^3}+O\left(\frac{1}{m^4}\right)$$ which is positive for all $m$.