Answer
$\Sigma^{\infty}_{n=3} \frac{1}{n^{5}-5n^{3}+4n}= \frac{1}{96}$
Work Step by Step
$\Sigma^{\infty}_{n=3} \frac{1}{n^{5}-5n^{3}+4n} = \Sigma^{\infty}_{n=3} \frac{1}{(n-2)(n-1)n(n+1)(n+2)}$
$\frac{1}{4}\Sigma^{\infty}_{n=3} \frac{(n+2)-(n-2)}{(n-2)(n-1)n(n+1)(n+2)}$
$\frac{1}{4}\Sigma^{\infty}_{n=3} [\frac{(n+2)}{(n-2)(n-1)n(n+1)(n+2)}-\frac{(n-2)}{(n-2)(n-1)n(n+1)(n+2)}]$
$\frac{1}{4}\Sigma^{\infty}_{n=3} [\frac{1}{(n-2)(n-1)n(n+1)}-\frac{1}{(n-1)n(n+1)(n+2)}]$
$\lim\limits_{N \to \infty}\frac{1}{4}\Sigma^{\infty}_{n=3} [\frac{(n+2)}{(n-2)(n-1)n(n+1)(n+2)}-\frac{(n-2)}{(n-2)(n-1)n(n+1)(n+2)}]$
$=\lim\limits_{N \to \infty} \frac{1}{4} [\frac{1}{1 \times 2 \times 3\times 4}-\frac{1}{2 \times 3 \times 4 \times 5}] +\frac{1}{4} [\frac{1}{2 \times 3 \times 4\times 5}-\frac{1}{3 \times 4 \times 5\times 6}] + \frac{1}{4}[\frac{1}{3 \times 4 \times 5\times 6} -\frac{1}{4 \times 5 \times 6 \times 7}] +...+ \frac{1}{4}[\frac{1}{(N-3) \times (N-2) \times (N-1)(N)} - \frac{1}{(N-2) \times (N-1) \times (N) \times (N+1)}] + \frac{1}{4} [\frac{1}{(N-2) \times (N-1) \times (N)(N+1)} - \frac{1}{(N-1) \times (N) \times (N+1)\times (N+2)}]$
$=\lim\limits_{N \to \infty} \frac{1}{4} [\frac{1}{1 \times 2 \times 3 \times 4}-\frac{1}{(N-1) \times (N) \times (N+1) \times (N+2)}]$
$=\frac{1}{4} [\frac{1}{1 \times 2 \times 3 \times 4} -0] = \frac{1}{96}$
