For every $n\in N$, we have:$$a_{n}=\frac{1}{1^{2}}+\frac{1}{2^{2}}+\frac{1}{3^{2}}+...+\frac{1}{n^{2}}$$
Then let $\sum$ be: $$\sum =\frac{1}{a_{1}^{2}}+\frac{1}{4a_{2}^{2}}+\frac{1}{9a_{3}^{2}}+...+\frac{1}{n^{2}a_{n}^{2}}$$
Prove that $\sum \leq \frac{3n-2}{2n-1}$, $\vee n\in N,n\neq 0$
My attempts:
First I tried using $\frac{1}{n^{2}}<\frac{1}{n-1}-\frac{1}{n}$ and then I found $a_{n}< 2-\frac{1}{n},\vee n>1$, with $a_{1}=1$
$a_{2}< 2-\frac{1}{2}$ etc.
But then I arrived at $\sum \geq 1^{2}+3^{2}+...+(2n-1)^{2}$, but I need $\sum$ lower than something, not grater.
Then I found that $\frac{1}{n^{2}}>\frac{1}{n}-\frac{1}{n+1}$ , and I arrived at $a_{n}>\frac{n}{n+1}$, but in this case 4,9,..,n^2 don't simplify. Also I found that $a_{n}<\frac{4n}{2n+1},\vee n\geq 1$(with induction) but $a_{n}$ has to be greater than something, so that $\frac{1}{a_{n}}$ and $\sum$ to be lower.
Any help would be appreciated.