TITLE: An interesting series to test convergence
QUESTION [5 upvotes]: I have another series in mind, today it is 
$$\sum^\infty_{n=10}\sin\left(\frac{1}{n^3}+\frac{\cos(n)}{n^2}\right)$$
I have tried to investigate the argument:
it is basically $\frac{1+n\cos(n)}{n^3}$ or we could say that $1/n^3$ converges, $1/n^2$ also and $\cos(n)$ have bounded partial sums, so the argument should converge. However, we are not talking about a sum of sines.
I tried also dividing the sum into two sums using summation rule for sine:
$\sin(...)=\sin(1/n^3)\cos(\cos(n)/n^2)+\sin(\cos(n)/n^2)\cos(1/n^3)$, however, $\cos(1/n^3)$ does not converge, so it is not bounded and I am afraid that $\sin(\cos(n)/n^2)$ is not bounded either.

REPLY [6 votes]: $|n^{-3}+n^{-2}\cos n|\le 2n^{-2}$, and $|\sin x|\le |x|$; hence the terms of the series are bounded in absolute value by the respective terms of the convergent series $\sum 2n^{-2}$, and the series converges absolutely.