Convergence of series $\sum_{n \geq 1} {\frac{1^2+2^2+ \cdots + n^2}{n^4}}$ In the study of the following series
$$
\sum_{n \geq 1} {\frac{1^2+2^2+ \cdots + n^2}{n^p}}
$$
it is not hard to prove that it diverges for $p \leq 3$, since the sequence itself does not converge to 0. You can also conclude that the series converges for $p > 4$ by comparison with Riemann series. Raabe's test yields that the series diverges for p between 3 and 4. However it does not give any information for the case $p=4$.

We know that $\displaystyle \sum _{i=1}^{n} i^{2} \ =\ \frac{n( n+1)( 2n+1)}{6}$
So $\displaystyle \frac{\sum _{i=1}^{n} i^{2}}{n^{p}} \ =\ \frac{( n+1)( 2n+1)}{6n^{p-1}}$
Now if we take the limit $\displaystyle n\rightarrow \infty $ and apply L'Hospital's Rule, we can see that
$\displaystyle \frac{4n+3}{6( p-1) n^{p-2}} \ =\ \frac{4}{6( p-1)( p-2) n^{p-3}}$
Edit: As correctly pointed out in the comments, this series diverges for $p = 4$ and later on converges for $p = 5$ (the famous Basel problem).
This is because when $p = 4$, each term effectively becomes of the form $\frac 1n$ and this we know to be diverging. On the other hand for $p = 5$ it becomes $\frac 1{n^2}$.
