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)

• You should provide more context. If you've just been through a few calculus classes, not all of these limits' closed forms will make any sense to you, but convergence can still be proven with simple techniques. – Adam Hughes Dec 17 '16 at 18:54
• @AdamHughes I've already finished Calculus 2, but who counts? Edit: I've edited the post. You may answer now :) – Ilan Aizelman WS Dec 17 '16 at 18:55
• $\pi/2^n=0?\,\,$ – zhw. Dec 17 '16 at 19:03
• @zhw. I didn't use the $n \rightarrow \infty$. I thought it needed to be used, but nevermind. – Ilan Aizelman WS Dec 17 '16 at 19:05
• Are you interested in the question of convergence or of the value? – Steven Stadnicki Dec 17 '16 at 19:09

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, $$\sum_{n\geq 1}\sin\frac{\pi}{2^n}\approx 2.48105.$$
• How can it be solved with $lim |{\frac{a_{n+1}}{a_n}}|$ sir? – Ilan Aizelman WS Dec 17 '16 at 19:03
• @IlanAizelmanWS: if we set $a_n=\sin\frac{\pi}{2^n}$, we have that $a_n$ is positive and $\lim_{n\to +\infty}\frac{a_{n+1}}{a_n}=\frac{1}{2}$, hence the series is convergent by the ratio test, if you like. – Jack D'Aurizio Dec 17 '16 at 19:15
• @IlanAizelmanWS: since $\lim_{x\to 0}\frac{\sin x}{x}=1$, we have: $$\lim_{n\to +\infty}\frac{\sin(\pi/2^{n+1})}{\sin(\pi/2^n)}=\lim_{n\to +\infty}\frac{\pi/2^{n+1}}{\pi/2^n}=\frac{1}{2}.$$ – Jack D'Aurizio Dec 17 '16 at 19:19
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.