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Check if $$\sum_{n = 1}^{\infty}2^n \sin\left(\frac{\pi}{3^n}\right)$$ converges.

I tried to solve this by using the ratio test - I have ended up with the following limit to evaluate: $$\lim_{n \to \infty} \left(\frac{2\sin\left(\frac{\pi}{3 \cdot 3^n} \right)}{\sin \left(\frac{\pi}{3^n} \right)} \right)$$ And now - I am stuck and don't know how to proceed with this limit. Any hints?

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  • $\begingroup$ Use the fact that $\sin(x) \leq x$ $\endgroup$ – TheOscillator Jan 21 '18 at 22:00
  • $\begingroup$ L'Hospital's rule will help you with the limit you pose, but you can show convergence directly as Olivier outlines. $\endgroup$ – Malcolm Jan 21 '18 at 22:01
  • $\begingroup$ @Malcolm This is a sequence, not a function. A sequence is not differentiable. $\endgroup$ – Aemilius Jan 21 '18 at 22:01
  • $\begingroup$ If $\lim_{x \to \infty} \left(\frac{2\sin\left(\frac{\pi}{3 \cdot 3^x} \right)}{\sin \left(\frac{\pi}{3^x} \right)} \right)$ exists (which it does) then $\lim_{n \to \infty} \left(\frac{2\sin\left(\frac{\pi}{3 \cdot 3^n} \right)}{\sin \left(\frac{\pi}{3^n} \right)} \right)$ also exists and equals the same value. $\endgroup$ – Malcolm Jan 21 '18 at 22:05
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Hint. One has $$ \left|\sin\left(\frac{\pi}{3^n}\right)\right|\le\frac{\pi}{3^n},\quad n=1,2,\cdots. $$ Can you take it from here?

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  • $\begingroup$ Your observation implies that this limit in question - say, L: $$0 \le L \le 2/3$$ and so this series does converge? $\endgroup$ – Aemilius Jan 21 '18 at 22:04
  • $\begingroup$ You then have $$\left|\sum_{n = 1}^{\infty}2^n \sin \frac{\pi}{3^n} \right| \le \sum_{n = 1}^{\infty}\left|2^n \sin \frac{\pi}{3^n} \right| \le \pi \sum_{n = 1}^{\infty}\frac{2^n}{3^n}<\infty$$ and the given series is absolutely convergent then it is convergent. $\endgroup$ – Olivier Oloa Jan 21 '18 at 22:07
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Note that

$$2^n \sin\left(\frac{\pi}{3^n}\right) \sim \pi\frac{2^n}{3^n}$$

then use comparison test with $$\sum \frac{2^n}{3^n}$$

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If you want to stick with your approach using the ratio test, you can (although Olivier's answer is more direct). Caveat: I'll detail every step of the derivation, which is not actually necessary for a proof.

We will only rely on elementary arguments, specifically the fact that $\sin'(0)=\cos 0 = 1$ — which is equivalent to $$\lim_{x\to 0} \frac{\sin x}{x} = 1 \,.\tag{1} $$ From there, you can write $$ \lim_{n \to \infty} \frac{2\sin\left(\frac{\pi}{3 \cdot 3^n} \right)}{\sin \left(\frac{\pi}{3^n} \right)} = \lim_{n \to \infty} 2\cdot \frac{1}{3}\cdot \frac{\sin\left(\frac{\pi}{3 \cdot 3^n} \right)}{\frac{\pi}{3 \cdot 3^n}}\cdot\frac{\frac{\pi}{3^n}}{\sin \left(\frac{\pi}{3^n} \right)} = \frac{2}{3}\lim_{n \to \infty} \frac{\sin\left(\frac{\pi}{3 \cdot 3^n} \right)}{\frac{\pi}{3 \cdot 3^n}}\cdot\left(\frac{\sin \left(\frac{\pi}{3^n} \right)}{\frac{\pi}{3^n}}\right)^{-1} \tag{2} $$ and, by (1), we get $$\begin{align*} \lim_{n \to \infty} \frac{2\sin\left(\frac{\pi}{3 \cdot 3^n} \right)}{\sin \left(\frac{\pi}{3^n} \right)} &= \frac{2}{3}\lim_{n \to \infty} \frac{\sin\left(\frac{\pi}{3 \cdot 3^n} \right)}{\frac{\pi}{3 \cdot 3^n}}\cdot \lim_{n \to \infty} \left(\frac{\sin \left(\frac{\pi}{3^n} \right)}{\frac{\pi}{3^n}}\right)^{-1} =\frac{2}{3}\lim_{n \to \infty} \frac{\sin\left(\frac{\pi}{3 \cdot 3^n} \right)}{\frac{\pi}{3 \cdot 3^n}}\cdot \left(\lim_{n \to \infty} \frac{\sin \left(\frac{\pi}{3^n} \right)}{\frac{\pi}{3^n}}\right)^{-1} \\ &= \frac{2}{3}\cdot 1\cdot 1^{-1} = \boxed{\frac{2}{3}} \end{align*}$$ and you can conclude with the ratio test.

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Same idea as Olivier express in a different way, you know that $$ \sin\left(x\right)\underset{(0)}{=}x+o\left(x\right) $$ Hence

$$ 2^n\sin\left(\frac{\pi}{3^n}\right) \underset{(+\infty)}{\sim}\pi \left(\frac{2}{3}\right)^n $$

What can you say about $\displaystyle \sum_{n \geq 0}\left(\frac{2}{3}\right)^n$ ?

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