Convergence of the series $\sum\limits_{n=1}^\infty n\sin(2^{-n})$ 
I'm studying the series
  $$\sum_{n=1}^\infty n\sin(2^{-n})$$
  and I have to find this limit
  $$\lim_{n \to +\infty}n\sin(2^{-n})$$

I can say that:
$$n\sin(2^{-n})= \frac {sen( \frac {1}{2^n})}{\frac {1}{n}}<\frac {sen( \frac {1}{n})}{\frac {1}{n}} \rightarrow 1 $$ for $n \rightarrow \infty$ and then $0<\lim_{n \to +\infty}n\sin(2^{-n})<1$ but nothing more and the limit should be $0$ if it converges
 A: As $x\rightarrow 0$, $\sin x \approx x$, thus $n \sin 2^{-n}\approx \frac{n}{2^n}\rightarrow 0$.
A bit more rigorously, using the Taylor's theorem with Peano form of the remainder:
$$\sin x = x + o(x)$$
$$n \sin 2^{-n} = n (2^{-n} + o(2^{-n})) \rightarrow 0 $$
A: Try to use the inequality $|sin(x)| \leq |x|$
A: $$L = \lim_{n \to \infty} n \sin \left( 2^{-n}\right) = \frac{n \cdot 2^{-n} \cdot\sin \left( 2^{-n}\right)}{2^{-n}}$$
$$n=1/u, n \to \infty, u \to 0^+$$
$$L = \lim_{u \to 0^+}\frac{2^{-1/u} \cdot\sin \left( 2^{-1/u}\right)}{u \cdot 2^{-1/u}} = \lim_{u \to 0^+}\frac{1}{u\cdot 2^{1/u}}$$
Reversing our substitution
$$L = \lim_{n \to \infty} \frac{n}{2^n} = 0$$
A: Note that
$$n\sin(2^{-n})=\frac{n}{2^n}\frac{\sin\left(\frac{1}{2^n}\right)}{\frac{1}{2^n}}\to 0\cdot 1=0$$
A: Since $|\sin a =\int_0^a\cos tdt|\le |a|$ we have 
$$|n\sin (2^{-n})|\le \frac{n}{2^n}\to 0$$
And hence also we have  $$\bigg|\sum_{n=1}^\infty n\sin(2^{-n})\bigg|\le \sum_{n=1}^\infty \frac{n}{2^n} <\infty$$
since, 
$$\sum_{n=1}^\infty\frac{n}{2^n} =\frac12 \left(\sum_{n=1}^\infty x^{n}\right)'\bigg|_{x=1/2} = \frac12 \left(\frac{x}{1-x}\right)'\bigg|_{x=1/2} =2$$

