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 $\dfrac{5}{4n}$. 

The telescoping sum trick, with ${1\over k^2-k}={1\over k-1}-{1\over k}$, works, but you need to split off the first term:
$$\sum_{k=n}^{3n-1}{1\over k^2}={1\over n^2}+\sum_{k=n+1}^{3n-1}{1\over k^2}\lt{1\over n^2}+\sum_{k=n+1}^{3n-1}{1\over k^2-k}={1\over n^2}+\left({1\over n}-{1\over3n-1}\right)={5\over4n}-\left({1\over4n}+{1\over3n-1}-{1\over n^2} \right)$$
and
$${1\over4n}+{1\over3n-1}-{1\over n^2}={(3n-1)n+4n^2-4(3n-1)\over4(3n-1)n^2}={7n^2-13n+4\over4(3n-1)n^2}\gt{7n(n-2)+4\over4(3n-1)n^2}\gt0$$
for $n\ge2$.
