<s>[INST] 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 $\dfrac{5}{4n}$. 
 [/INST] Since
$$
\frac{1}
{{k^2 }} < \int\limits_{k - 1}^k {\frac{1}
{{x^2 }}} dx\,\,\,\,\,\,\forall k \geqslant 2
$$
we have that
$$
\sum\limits_{k = n}^{3n - 1} {\frac{1}
{{k^2 }} < } \int\limits_{n - 1}^{3n - 1} {\frac{1}
{{x^2 }}} dx = \left[ { - \frac{1}
{x}} \right]_{n - 1}^{3n - 1}  = \frac{1}
{{n - 1}} - \frac{1}
{{3n - 1}}
$$
We have that
$$
\frac{1}
{{n - 1}} - \frac{1}
{{3n - 1}} < \frac{5}
{{4n}}
$$
is true if and only if 
$$
\frac{{2n}}
{{\left( {n - 1} \right)3n - 1}} < \frac{5}
{{4n}}
$$
which is true if and only if 
$$7n^2-20n+5>0$$. This is true for every $n \geq 3$. On the other side, a direct computation shows that the inequality is true for $n=2$.
</s>