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 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 $$1^2+2^2+\ldots +n^2\ge {1\over n}{n^2(n+1)^2\over 4}\ge {n^3\over 4}$$
