$\sum\limits_{n=1}^{\infty}\sin ( \frac{n}{2^n})$ converges? [closed]

I was trying to determine weather or not $\sum\limits_{n=1}^{\infty}\sin ( \frac{n}{2^n})$ converges using perhaps the D'Alembert test, but given the sine I cant really see it happening..are there other ways?

closed as off-topic by Did, Adam Hughes, user91500, kjetil b halvorsen, Tom-TomJun 15 '15 at 9:38

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• If you want to use the D'Alembert test. You should first note that $\sin(\frac{n}{2^n})\approx\frac{n}{2^n}$, since the term inside the $\sin$ tends to $0$. Calculating the limit of the ratio should be simple if you do this. – Kitegi Jun 13 '15 at 17:00

$$\left|\sin\left(\frac{n}{2^n}\right)\right|\leq \frac{n}{2^n}.$$
• @Bak1139: $\sin x\leq x$ for any $x\geq 0$ is really trivial. – Jack D'Aurizio Jun 13 '15 at 16:31
• Le $f(x)=\sin x-x$, you have that $f'(x)=\cos x-1\leq 0$ for all $x$, and thus it's a decreasing function. Moreover, $f(0)=0$, therefore $f(x)\leq 0$ for all $x\geq 0$ what prove that $\sin x\leq x$ if $x\geq 0$. I let you prove that $|\sin x|\leq |x|$ for all $x$. – idm Jun 13 '15 at 16:41