Q:

Convergence of a series involving trigonometric ratios

How do I go about investigating the series
$$
\sum _{n=1}^{\infty }\left(1+\frac{2n^{2}+n}{3n^3-n\sin (n)}\right)^{\tfrac{n^2+n\cos(n)}{n+\sin(n)}}
$$
for convergence or divergence? 
What I don't understand is how to evaluate when faced with trigonometric ratios.  This is a new concept for me and I'd appreciate any hints. 

A:

$\frac {n^{2}+n\cos \, n} {n+\sin \, n} \log (1+\frac {2n^{2}+n} {3n^{3}-n\sin \, n}) \to 1$ as $n \to \infty$ so the n-th term of the series tends to $e^{1}=e$. Hence the series is divergent. [ I have used the fact that $\frac {log (1+x)} x \to 1$ as $ x\to 0$. (Take $x=\frac  {2n^{2}+n} {3n^{3}-n\sin \, n})$; note that if you divide numerator and denominator by $n^{2}$ you get  $x=\frac {2+n^{-1}} {3n-n^{-2}\sin \, n})$ and this quantity converges to $0$ Thus, $x \to 0$ as $n \to \infty$]. 