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Find the radius of convergence of the series $$ \sum_{k=1}^{\infty}\left(\frac{x}{\sin k}\right)^k $$

I tried to use the following formula: $r^{-1}=\lim_{k\rightarrow\infty}\sup|c_k|^{1/n}$, where $c_k=\left(\frac{1}{\sin k}\right)^k$.

So, I got something like this: $$ r^{-1}=\lim_{k\rightarrow\infty}\sup\left|\frac{1}{\sin k}\right|= \lim_{k_1\rightarrow\infty}\left|\frac{1}{\sin(\pi k_1)}\right|=\infty\Rightarrow\\ \Rightarrow r=0 $$ Is my solution correct?

Bonrey
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2 Answers2

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No, it is not correct, since you did not justify the strange equality $\limsup_{k\to\infty}\left\lvert\dfrac1{\sin(k)}\right\rvert=\lim_{k_1\to\infty}\left\lvert\dfrac1{\sin(\pi k_1)}\right\rvert=0$. On the other hand, it is true (although not obvious) that $\limsup_{k\to\infty}\left\lvert\dfrac1{\sin(k)}\right\rvert=\infty$ (which is equivalent to the assertion that there is a sequence $(k_n)_{n\in\mathbb N}$ of natural numbers such that $\lim_{n\to\infty}\sin(k_n)=0$). Therefore, the radius of convergence of your power series is $0$.

José Carlos Santos
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  • But why is this equality strange? I just meant that $k=\pi k_1$ because that way $\sin(\pi k_1)=0$ and $\frac{1}{\sin(\pi k_1)}=\infty$ – Bonrey Nov 09 '19 at 15:26
  • How did a sequence if sines of natural numbers become a sequence of sines of multiples of $\pi$? – José Carlos Santos Nov 09 '19 at 15:28
  • I meant that there is such a sequence $k_n=\pi n$ that $\lim_{n\rightarrow\infty}\sin k_n=0$ – Bonrey Nov 09 '19 at 21:20
  • Sure, there is such a sequence. What has that to do with the problem of studying the sequence $(\sin k)_{k\in\mathbb N}$? – José Carlos Santos Nov 09 '19 at 21:24
  • Yes, what I wrote doesn't make much sense. I got a bit confused. Could you, please, explain what I should have done here? And I will accept your answer. – Bonrey Nov 10 '19 at 08:52
  • I think that my answer contains what you should have done here. With one exception: I did not prove that $\limsup_{k\to\infty}\left\lvert\frac1{\sin k}\right\rvert=0$. But\begin{align}\limsup_{k\to\infty}\left\lvert\frac1{\sin k}\right\rvert=0&\iff\liminf_{k\to\infty}\lvert\sin k\vert=0\&\iff\lim_{n\to\infty}\sin k_n=0\text{ for some sequence of natural numbers.}\end{align}And this last statement is proved here. – José Carlos Santos Nov 10 '19 at 09:12
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Completing Jose's answer: Claim: there exists a subsequence of natural numbers such that $\lim \sin(n_k) = 0$.

This will force $\limsup$ of the root to go to infinity, and hence ROC is zero.

Proof of claim: integer radians along circle are dense in it.

Behnam Esmayli
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