Answer
Converges
Work Step by Step
Let $u_n=\Sigma_{n=1}^\infty \dfrac{1}{1^2+2^2+....n^2}$
Re-write $u_n$ as: $\Sigma_{n=1}^\infty \dfrac{1}{1^2+2^2+....n^2}=\Sigma_{n=1}^\infty \dfrac{6}{n(n+1)(2n+1)}$ ; $v_n=\dfrac{1}{ n^3}$
Now, $\lim\limits_{n \to \infty}\dfrac{u_n}{v_n} =\lim\limits_{n \to \infty}\dfrac{\dfrac{6}{n(n+1)(2n+1)}}{1/n^3}$
This implies that $\lim\limits_{n \to \infty}\dfrac{6 n^2}{(n+1)(2n+1)}=\lim\limits_{n \to \infty} \dfrac{6}{(1+\dfrac{1}{n})(2+\dfrac{1}{n})}$
so, $ \dfrac{6}{2}=3$
Here, $\Sigma_{n=1}^\infty \dfrac{1}{n^{(3)}}$ shows a convergent $p$-series test with $p \gt 1$.
Hence, the series converges by the limit comparison test.