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

Question

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?

Answer 1

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 $0$.

Answer 2

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.