Convergence of a trig series Prove the convergence of the series:$$\sum\limits_{n = 1}^\infty  {\left(n\sin \left(\frac{1}{n}\right)\right)^{n^3 } }.$$
I managed to rewrite it using $
\sin (\frac{1}{n}) \approx \frac{1}{n} - \frac{1}{{6n^3 }}$ but to no avail .
Any ideas ? I've been working on it for a while now, but I can't see the solution.
 A: We can use the root test.
Define $a_n:=\left(n\sin(1/n)\right)^{n^3}$. Then 
$$|a_n|^{\frac 1n}=\left(n\sin\left(\frac 1n\right)\right)^{n^2}=\left(1-\frac {1+\varepsilon_n}{6n^2}\right)^{n^2},$$
where $\varepsilon_n\to 0$. For $n$ large enough, we have $|\varepsilon_n|\lt 1/2$, hence $$0\leqslant 1-\frac {1+\varepsilon_n}{6n^2}\leqslant 1-\frac 1{12n^2},$$
and we obtain 
$$\limsup_{n\to \infty}|a_n|^{\frac 1n}\leqslant e^{-1/12}\lt 1.$$
A: We have
$$n\sin\frac1n=n\left(\frac1n-\frac1{6n^3}+O\left(\frac1{n^5}\right)\right)=1-\frac1{6n^2}+O\left(\frac1{n^4}\right)$$
so
$$\left(n\sin\frac1n\right)^{n^3}=\exp\left(n^3\ln\left(1-\frac1{6n^2}+O\left(\frac1{n^4}\right)\right)\right)\\=\exp\left(-\frac n{6}+O\left(\frac1{n}\right)\right)\sim_\infty e^{-n/6}$$
and the geometric series $\sum e^{-n/6}$ is convergent. Conclude.
A: Let $u_n = \left( n\sin \frac 1n
\right)^{n^2}$.
$$
\log u_n = n^2 \log \left(n\sin \frac 1n\right)
\sim n^2  \left(n\sin \frac 1n - 1\right) =
n^3  \left(\sin \frac 1n - \frac1n\right) \sim -n^3 \frac 1{3n^3}
 = -\frac 13$$
So, what is the conclusion?
