Does it converge? $\sum_{n=1}^{\infty} \sin(\frac{\pi}{2^n})$ Does it converge? $$\sum_{n=1}^{\infty}  \sin(\frac{\pi}{2^n})$$
Well, I've used: $ \sin(\frac{\pi}{2^n}) \le \frac{\pi}{2^n} \le \pi $ And thus it converges. But I'm not sure I can just use that and say it converges.
Anyways, I've used wolfram alpha and it said the limit is $\frac{1}{2}$, can anyone explain how?
The answer I'm looking for is: How can it be solved with $lim |{\frac{a_{n+1}}{a_n}}|$? (the equation test)
 A: The series is clearly convergent since $\sin(x)\sim x$ in a neighbourhood of the origin, and $\sum_{n\geq 1}\frac{\pi}{2^n}=\pi$. By exploiting the Taylor series of the sine function (that is an entire function) we get that
$$ \sum_{n\geq 1}\sin\frac{\pi}{2^n}=\sum_{k\geq 0}\frac{(-1)^k \pi^{2k+1}}{(2k+1)!}\sum_{n\geq 1}\frac{1}{2^{n(2k+1)}}=\sum_{k\geq 0}\frac{(-1)^k \pi^{2k+1}}{(2k+1)!(2\cdot 4^k-1)} $$ 
converting the original series in a series that converges much faster. Numerically,
$$ \sum_{n\geq 1}\sin\frac{\pi}{2^n}\approx 2.48105.$$
A: The limit, as $n$ goes to infinity of
$$
\frac{\sin\left(\frac{\pi}{2^n}\right)}{\sin\left(\frac{\pi}{2^{n+1}}\right)}
$$
is of the form $0/0$; thus in order to compute it, you can consider the function $x\mapsto\sin\left(\frac{\pi}{2^x}\right)$ of continuos variable and use De L'Hopital theorem:
$$
\lim_{x\to+\infty}\frac{\sin\left(\frac{\pi}{2^x}\right)}{\sin\left(\frac{\pi}{2^{x+1}}\right)}
=\lim_{x\to+\infty}\frac{\cos\left(\frac{\pi}{2^x}\right)\frac{\pi}{2^x}(-\log2)}{\cos\left(\frac{\pi}{2^{x+1}}\right)\frac{\pi}{2^{x+1}}(-\log2)}=\frac12<1
$$
thus series converges absolutely.
