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Determine for what values of $x$ the series converges.

Here is the series:

$$\sum_{n = 1}^{\infty} \frac{x^n}{\sin^n n}.$$

My trial: Considering $a_{n} = \frac{1}{\sin^n n} $

I calculated $|\frac{a_{n+1} x^{n+1}}{a_n x^n}|$ which is equal to $|x||\frac{\sin^n n}{\sin^{n+1}(n+1)}|$ then I want to take the limit as $n \rightarrow \infty$ but then what shall I do? I do not know how to solve this limit

Intuition
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1 Answers1

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(Big) hint: use the root test, along with the fact that $(\sin n)_n$ is dense in $[-1,1]$.

In more detail (put your mouse on the area to reveal it):

Since $(\sin n)_n$ is dense in $[-1,1]$, there exists a sequence of integers $(n_k)_k$ such that $\sin n_k \xrightarrow[k\to\infty]{} 0^+$, from which $\sqrt[n_k]{a_{n_k}} = \frac{1}{\sin n_k} \xrightarrow[k\to\infty]{} +\infty$. This means that $\limsup_n \sqrt[n]{a_n} = +\infty$: applying the root test (with $C\stackrel{\rm def}{=}\infty$), you therefore get that the radius of convergence is $\frac{1}{\limsup_n a_n} = 0$.

Clement C.
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  • could you write the detailed steps for me as I am confused about |\sin n| how I will write its limit? – Intuition Jan 29 '19 at 07:26
  • @hopefully Done, see my edit. Once again, the root test asks you to consider the $\limsup$ of $\sqrt[n]{a_n}$ (which exists), not the limit (which may not exist, and here indeed does not exist). – Clement C. Jan 29 '19 at 07:33
  • I need a proof that the limtsup of sin n is +infinity – Intuition Jan 29 '19 at 07:35
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    It's in the answer (in the "spoiler"). It's not of $\sin n$, it's of $\sqrt[n]{a_n}$. – Clement C. Jan 29 '19 at 07:37
  • but the problem that we have absolute value of sin n not only sin n ..... so the limit sup is 1 if we take the sequence $(4m +1)\pi/2$ – Intuition Jan 29 '19 at 07:39
  • This doesn't change anything. We are using the existence of a subsequence such that $\sin n_k \to 0^+$; the same subsequence has $|\sin n_k |\to 0^+$. – Clement C. Jan 29 '19 at 07:41
  • why did you choose the sequence that makes the sin converges to zero ..... there are other sequences as I mentioned above that make the sign did not converge to zero – Intuition Jan 29 '19 at 07:41
  • These other sequences would not help me prove the limsup is infinity... You want to show that to conclude. – Clement C. Jan 29 '19 at 07:43
  • what do you mean by sin being dense in this interval ? what is the importance of the existence of that subsequence if there are many others that doen not converge to zero? – Intuition Jan 29 '19 at 07:44
  • Those other sequences would only show the limsup is "at least some value". – Clement C. Jan 29 '19 at 07:44
  • does not the limit sup defined as the largest accumulation point? – Intuition Jan 29 '19 at 07:45
  • If you want to show there exist blue fishes, you find a blue fish and show it to your friends. You don't care about the red, yellow, or green fish. They do exist, but they're not useful to show the existence of blue fishes. – Clement C. Jan 29 '19 at 07:45
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    Yes. The largest. But we want the limsup of 1/sin n, NOT of sin n. – Clement C. Jan 29 '19 at 07:46
  • could you give me an explicit form of that sequence please? – Intuition Jan 29 '19 at 07:46
  • what do you mean by sin being dense in this interval? – Intuition Jan 29 '19 at 07:49
  • That every point of [-1,1] is an accumulation point of the sequence (sin n). We only use it for the point 0, though. – Clement C. Jan 29 '19 at 07:51
  • could you give me an explicit form of that sequence please? – Intuition Jan 29 '19 at 07:53
  • what is C? is it the limit of the root test? – Intuition Jan 29 '19 at 08:02
  • Yes. I defined C that way for you to read and easily apply the theorem from the Wikipedia link, where it's the notation they use. – Clement C. Jan 29 '19 at 13:31