Does the series $\sum 2^n \sin(\frac{\pi}{3^n})$ converge?
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?
$\endgroup$ 44 Answers
$\begingroup$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?
$\endgroup$ 2 $\begingroup$Note that
$$2^n \sin\left(\frac{\pi}{3^n}\right) \sim \pi\frac{2^n}{3^n}$$
then use comparison test with $$\sum \frac{2^n}{3^n}$$
$\endgroup$ $\begingroup$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.
$\endgroup$ $\begingroup$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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