Question on calculus I want to compute
$\displaystyle{\sum_{n=4}^{\infty}{1 \over n^{3} + n^{2}\cos\left(n\right)}.\quad}$
Can anyone help me or give a hint?

This is not an answer but it is too long for an answer.
I do not think that a closed form of the summation could be found.
Using the same argument as user125261 in his/her comment, you can write 
$$\sum_{n=4}^{\infty}\frac{1}{n^3+n^2}\lt\sum_{n=4}^{\infty}\frac{1}{n^3+n^2\cos(n)}\lt \sum_{n=4}^{\infty}\frac{1}{n^3-n^2}$$ that is to say $$\frac{3 \pi ^2-29}{18}\lt\sum_{n=4}^{\infty}\frac{1}{n^3+n^2\cos(n)}\lt \frac{61-6 \pi ^2}{36}\approx 0.0495104$$ Now, computing partial sums, one could get the following results for $$S_p=\sum_{n=4}^{p}\frac{1}{n^3+n^2\cos(n)}$$
$$\left(
\begin{array}{cc}
p & S_p \\
 100 & 0.0419069 \\
 200 & 0.0419439 \\
 300 & 0.0419508 \\
 400 & 0.0419533 \\
 500 & 0.0419544 \\
 600 & 0.041955 \\
 700 & 0.0419553 \\
 800 & 0.0419556 \\
 900 & 0.0419558 \\
 1000 & 0.0419559
\end{array}
\right)$$ which do not show a very fast convergence (this was done using Excel).
