A double sum $\sum \limits_{n=1}^{n=\infty}\left(\sum \limits_{k=n}^{k=n^2}\frac{1}{k^2}\right)$

The sum diverges. To see this, lower bound the inner summation by a telescoping sum by writing $\frac{1}{k^2} > \frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}$. Now use the fact that the harmonic series diverges.

$$\begin{align*} \sum_{n=1}^{n=\infty}\left(\sum_{k=n}^{k=n^2}\frac{1}{k^2}\right) &= \sum_{k=1}^\infty\;\sum_{n=\lceil\sqrt{k}\rceil}^k\frac1{k^2}\\ &=\sum_{k=1}^\infty\frac{k-\lceil\sqrt{k}\rceil+1}{k^2}\\ &\ge \sum_{k=1}^\infty\frac{k-\sqrt{k}}{k^2}\\ &=\sum_{k=1}^\infty\left(\frac1k-\frac1{k^{3/2}}\right), \end{align*}$$

which clearly diverges.

By counting how many times a particular $k$ appears, we get $$ \sum_{n=1}^{n=\infty}\left(\sum_{k=n}^{k=n^2}\frac{1}{k^2}\right)=\sum_{k=1}^\infty\frac{k-\left\lceil\sqrt{k}\;\right\rceil+1}{k^2}\ge\sum_{k=1}^\infty\frac{1}{2k} $$ which diverges since the harmonic series diverges and $\left\lceil\sqrt{k}\;\right\rceil-1\le k/2$.