Minimum integer solution of $\ \ \ \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{55^2} < \frac{N}{500}$ If $\ \ \ \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{55^2} < \frac{N}{500}$
find minimum integer N such that true for this inquality.
or maybe in the other cases for general term of $\ \ \ \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{k^2} < \frac{N}{500}$
P.S. I try to bring them all to    $RMS \geqslant AM \geqslant  GM \geqslant HM $ ,but cannot find the way to solve.
P.S.2  the answer for  $\ \ \ \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{55^2} < \frac{N}{500}$ is$\ \  N=314$
thank you so much for every advices and comments
 A: The RMS-AM-GM-HM inequality is very unlikely to help with this problem as the terms are not close to each other. Here is an alternative approach: 
The solution to the Basel problem tells us that $\displaystyle\sum_{n = 1}^{\infty}\dfrac{1}{n^2} = \dfrac{\pi^2}{6}$.
You can bound $\dfrac{1}{n(n+1)} \le \dfrac{1}{n^2} \le \dfrac{1}{(n-1)n}$ for all positive integers $n$. 
Therefore, $\displaystyle\sum_{n = 56}^{\infty}\dfrac{1}{n^2} \le \sum_{n = 56}^{\infty}\dfrac{1}{(n-1)n} = \sum_{n = 56}^{\infty}\left(\dfrac{1}{n-1}-\dfrac{1}{n}\right) = \dfrac{1}{55}$. 
Similarly, $\displaystyle\sum_{n = 56}^{\infty}\dfrac{1}{n^2} \ge \sum_{n = 56}^{\infty}\dfrac{1}{n(n+1)} = \sum_{n = 56}^{\infty}\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right) = \dfrac{1}{56}$. 
Since $\displaystyle\sum_{n = 1}^{\infty}\dfrac{1}{n^2} = \dfrac{\pi^2}{6}$ and $\dfrac{1}{56} \le \displaystyle\sum_{n = 56}^{\infty}\dfrac{1}{n^2} \le \dfrac{1}{55}$, we have $\dfrac{\pi^2}{6}-\dfrac{1}{55} \le \displaystyle\sum_{n = 1}^{55}\dfrac{1}{n^2} \le \dfrac{\pi^2}{6}-\dfrac{1}{56}$. 
Subtract off the $n = 1$ term to get $\dfrac{\pi^2}{6}-1-\dfrac{1}{55} \le \displaystyle\sum_{n = 2}^{55}\dfrac{1}{n^2} \le \dfrac{\pi^2}{6}-1-\dfrac{1}{56}$.
Can you finish the problem from here?
