For $n>1,$ prove that, for $\sum_{n}^{3n-1}\frac{1}{x^2} < \frac{5}{4n}$ For $n>1$ prove that for $$\sum_{n}^{3n-1}\frac{1}{x^2} < \frac{5}{4n}$$ 
I know that $\dfrac{1}{x}>\dfrac{1}{(x+1)}$ and I'm trying to break the summation into smaller sums to work with, but I'm just not making that final bridge to the $\dfrac{5}{4n}$. 

For $n=2$, we have:
$$\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2} < \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{3^2}+\frac{1}{3^2}=\frac{1}{4}+\frac{1}{3}=\frac{7}{12}<\frac{5}{8}$$
Now assume $n\geq 3$. Since:
$$\frac{1}{x^2} < \frac{1}{x(x-1)}=\frac{1}{x-1}-\frac{1}{x}$$
we have
$$\sum_{n}^{3n-1}\frac{1}{x^2} < \sum_{n}^{3n-1}\left(\frac{1}{x-1}-\frac{1}{x}\right)=\frac{1}{n-1}-\frac{1}{3n-1}=\frac{2n}{(n-1)(3n-1)}$$
It remains to prove:
$$\frac{2n}{(n-1)(3n-1)}< \frac{5}{4n}\Leftrightarrow 7n^2-20n+5 > 0$$
which holds when $n \geq 3$.
