Can someone please help me prove that this series is convergent?
The problem is I don't know what to do with sin.
$$\sum_{n=1}^{\infty} 2^n \sin{\frac{\pi}{3^n}} $$
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UfmdFkiF
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As is, this question does not appear to be bad and does not seem as though it should have downvotes – Simply Beautiful Art Nov 29 '16 at 20:55
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As $\sin x<x$ for $x>0$ we can estimate $$ \sum_{n=1}^\infty2^n\sin(\pi/3^n)\leq\sum_{n=1}^\infty2^n\frac{\pi}{3^n}. $$ The RHS is finite by the geometric series so our (positive) series is convergent.
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1@SimpleArt Thank you! ^^ It's from Steins;Gate. I won't be able to answer with my profile pic next time someone ask for a nice proof of the sum of the first $n$ natural numbers, but I love that anime! ;) – Nov 29 '16 at 21:14
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1XD If you are interested, I have a post concerning the sum of the first $n^p$ numbers, if you think you'll be interested. – Simply Beautiful Art Nov 29 '16 at 21:17
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@SimpleArt Indeed I am interested!! 0.0 I happen to be these days studying summation methods, and I didn't know about Faulhaber's formula beyond the natural-exponent case. I've got hooked on the subject (apparently) the last 4 hours, so thank you for inviting me into it. =) I'll keep track of your question! – Nov 30 '16 at 01:49
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lol, you're welcomed any time, and I do hope you find some amazing answers. (and I don't think we do Faulhaber's formula beyond the natural-exponent case...? O_o) – Simply Beautiful Art Nov 30 '16 at 01:52
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@SimpleArt Check this paper. It is an asymptotic result but still seemed like a good starting point (and after having a view at the proof of the main theorem and notice that it used some of the theory I am studying these days I couldn't help to work it out). – Nov 30 '16 at 02:02
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You can make any number of asymptotic formulas for such. If I may point out, a combination of different zeta functions should do the job. Mainly the Riemann zeta and Hurwitz zeta. – Simply Beautiful Art Nov 30 '16 at 02:14
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@SimpleArt I am aware that asymptotic estimations often don't give much information to come up with a useful formula for the finite case. In fact before starting reading that article I skipped some sources that exactly --as you say-- used zeta functions to approach the subject, but the one I read seemed to me like a good way to get-to-know the topic. If it turns out to be a good starting point to think about, then perfect, but anyway It was fun to work out! =) – Nov 30 '16 at 02:35
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Consider the root test.
Let $L = \lim_{n \to \infty} |2^n\sin(\frac{\pi}{3^n})|^{\frac{1}{n}}.$ Then $L = \lim_{n \to \infty} (2^n\sin(\frac{\pi}{3^n}))^{\frac{1}{n}} = \lim_{n \to \infty} 2\sin(\frac{\pi}{3^n})^{\frac{1}{n}} = 0$, so the series converges by root test.
user389056
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We have $$\sin(X)\sim X \;(X\to 0)$$
$$\implies 2^n\sin(\frac{\pi}{3^n})\sim (\frac{2}{3})^n\;(n\to+\infty)$$
$\implies \sum 2^n\sin(\frac{\pi}{3^n})$ is convergent since the geometric series $\sum (\frac{2}{3})^n$ is positive and convergent.
hamam_Abdallah
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