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 $p=4$.
 A: For $p=4$ each term is $\frac {n(n+1)(2n+1)}{6n^4} \gt \frac 1{3n}$  The sum of these is bounded below by $\frac 13$ of the harmonic series, so the sum diverges.
A: We do not need to apply the explicit formula for $1^2+2^2+\ldots +n^2.$ Instead observe that
$$n^3-(n-1)^3=3n^2-3n+1\ge 3n^2$$
Hence $$1^2+2^2+\ldots +n^2\ge {1\over 3}\left ([1^3-0^3]+[2^3-1^3]+\ldots +[n^3-(n-1)^3]\right ]={1\over 3}n^3$$
Hence the $n$th term of the series for $p=4$ is greater or equal $\displaystyle{1\over 3n}.$
Another approach is to apply the Cauchy-Schwarz inequality $$(1+2+\ldots +n)^2\le (1^2+2^2+\ldots +n^2)n$$ hence $$1^2+2^2+\ldots +n^2\ge {1\over n}{n^2(n+1)^2\over 4}\ge {n^3\over 4}$$
A: 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 $\frac 1{n^2}$.
