$ \sum_{n=2}^\infty \frac{1}{n^3(n^3+1)}. $ The series is:
$$
\sum_{n=2}^\infty \frac{1}{n^3(n^3+1)}.
$$
I tried splitting the whole thing into simple fractions but I don't seem to get anywhere.
Any ideas?
 A: $$
\begin{align}
\sum_{n=2}^\infty\frac1{n^3(n^3+1)}
&=\sum_{n=2}^\infty\left(\frac1{n^3}-\frac1{n^3+1}\right)\\
&=\sum_{n=2}^\infty\left(\frac1{n^3}-\frac1{n^3(1+1/n^3)}\right)\\
&=\sum_{n=2}^\infty\left(\frac1{n^3}-\frac1{n^3}\left(1-\frac1{n^3}+\frac1{n^6}-\dots\right)\right)\\
&=\sum_{n=2}^\infty\left(\frac1{n^6}-\frac1{n^9}+\frac1{n^{12}}-\dots\right)\\
&=\sum_{n=2}^\infty(-1)^n(\zeta(3n)-1)\\
&\stackrel.=0.01555356082097039944\tag{22 terms}
\end{align}
$$
Added Note: Since $\zeta(3n)-1\sim1/8^n$, each term in the sum gives over $0.9$ digits.
A: Hint: Obsereve that 
$$\frac1{n^3(n^3+1)}=\frac1{n^3}-\frac1{n^3+1}$$
And remember that $\zeta(s)=\sum_{n=1}^\infty\frac1{n^s}$.
A: $$\sum_{n=2}^\infty \frac{1}{n^3(n^3+1)} = \sum_{n=2}^\infty \frac{1}{n^3}+\frac{n-2}{3 (n^2-n+1)}-\frac{1}{3 (n+1)} = \\
\zeta{(3)} -1 + \sum_{n=2}^{\infty} \frac{n-2}{3 (n^2-n+1)}-\frac{1}{3 (n+1)} = \\
\zeta{(3)}-1  -\sum_{n=2}^{\infty} \frac{1}{n^3+1} \approx \\
0.015553560820970399435132949324595238582753593120862693333639...$$
