Answer
$\Sigma^{\infty}_{n=1}\frac{3n^{2}+3n+1}{(n^{2}+n)^{3}} = 1$
Work Step by Step
$\frac{3n^{2}+3n+1}{(n^{2}+n)^{3}}=\frac{1}{n^{3}}-\frac{1}{(n+1)^{3}}$
$\Sigma^{\infty}_{n=1} \frac{1}{n^{3}}- \frac{1}{(n+1)^{3}} = \lim\limits_{N \to \infty}\Sigma^{N}_{n=1} \frac{1}{n^{3}}- \frac{1}{(n+1)^{3}}$
$= \lim\limits_{N \to \infty} \frac{1}{1^{3}}-\frac{1}{2^{3}}+\frac{1}{2^{3}}-\frac{1}{3^{3}}+\frac{1}{3^{3}}-\frac{1}{4^{3}}+...+\frac{1}{(N-1)^{3}}-\frac{1}{N^{3}}+\frac{1}{N^{3}}-\frac{1}{(N+1)^{3}}$
$\lim\limits_{N \to \infty}\frac{1}{1^{3}} - \frac{1}{(N+1)^{3}}=1-0=1$
$\Sigma^{\infty}_{n=1}\frac{3n^{2}+3n+1}{(n^{2}+n)^{3}} = 1$
