Convergence of the series $\sum_{n = 1}^{+\infty}{\left(n\sin{\frac{1}{n}}\right)^n}$ I have to study the convergence of the series 
$$
\sum_{n = 1}^{+\infty}{\left(n\sin{\frac{1}{n}}\right)^n}
$$
and 
$$
\sum_{n = 1}^{+\infty}{\left(\left(n\sin{\frac{1}{n}}\right)^n - 1\right)}.
$$
I know I should study the limit
$$
\lim_{n\to +\infty}{\left(n\sin{\frac{1}{n}}\right)^n}
$$
and that 
$$
\lim_{n\to +\infty}{n\sin{\frac{1}{n}}} = 1
$$
but I don't see how it helps. Any ideas ?
Thank you in advance !
 A: On the interval $(0,1)$ we have
$$ 1-\frac{x^2}{6} \leq \frac{\sin x}{x}\leq e^{-x^2/6} $$
hence $\left(n\sin\frac{1}{n}\right)^n$ behaves like $e^{-\frac{1}{6n}}=1-\frac{1}{6n}+O\left(\frac{1}{n^2}\right)$ for large values of $n$ and the given series are divergent.
A: It is easy to show from the mean-value theorem that $x-\frac16 x^3\le \sin(x)\le x$ for $0\le x\le 1$.  Therefore,
$$n\sin(1/n)\ge 1 -\frac{1}{6n^2}$$
Using Bernoulli's Inequality, we find that
$$(n\sin(1/n))^n\ge \left(1 -\frac{1}{6n^2}\right)^n\ge 1-\frac{1}{6n}$$
Inasmuch as the general terms of the series do not approach $0$ as $n\to \infty$, the series diverges.
A: Since $\sin\frac1n=\frac1n-\frac1{6n^3}+O(1/n^5)$, it follows that $$\lim \left(n\sin\frac1n\right)^n=\left(1-\frac{1}{6n^2}\right)^n=1,$$whence the divergence of the first series. We now also know that the second series has indeed a tending-to-zero general term, but $$\left(n\sin\frac1n\right)^n-1<\left(1-\frac{1}{7n^2}\right)^n -1<-\frac{1}{8n}$$ and thus we have divergence to $-\infty$.
