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I need to find the radius of convergence and the behavior at the endpoints of $$\sum_{n=0}^{\infty}\frac{x^n}{(\cos^{-1}(\sqrt{n+1}-\sqrt{n}))^n}. $$ I would like that you check if my results are correct, and if not, where I made a mistake. Thank you.

Using the root test, I get $$\lim_{n \to \infty}\left|\frac{x^n}{(\cos^{-1}(\sqrt{n+1}-\sqrt{n}))^n}\right|^{\frac{1}{n}}\;\text{ if and only if }\; |x|< \frac{\pi}{2}, $$ by using the fact that $\lim_{n\to \infty} \sqrt{n+1}-\sqrt{n}=0$ and $|\cos^{-1}(0)|=\frac{\pi}{2}$

Now, on the endpoints, we have $x=\frac{\pi}{2}$ and $x=\frac{-\pi}{2}$ which, using the root test again, seems to result in $1$ and $-1$. So we sum up infinitely many $1$ resp. $-1$, so it diverges on both endpoints.


Are my results correct? Thanks for your help!

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Edit: in order to answer as comprehensively as possible to the question, following the OP's comments, I have reworked and extended my former answer.

The first one of your conclusions is right, while you are misinterpreting the second one: when $x=\pm\frac{\pi}{2}$ you are on the boundary of the circle of convergence, therefore the power series may or may not converge at that points: the root test does to work when $$ \lim_{n \to \infty} \sqrt[n]{a_nx^n}=\lim_{n \to \infty}\left|\frac{1}{2^n}\frac{\pi^n}{(\cos^{-1}(\sqrt{n+1}-\sqrt{n}))^n}\right|^{\frac{1}{n}}=1 $$
The Cauchy-Hadamard theorem states simply than the root test applied to the coefficients $\langle a_n \rangle_{n \in \mathbb{N}}$ (in general $a_n \in \mathbb{C}$ for all $n \in \mathbb{N}$) of a given power series $\sum_{n=0}^{\infty} a_n x^n$, determines the values $R$ of the radius of convergence $$ R=\frac{1}{\lim_{n \to \infty} \sqrt[n]{a_n}} $$ for which $|x|<R$ implies the convergence of your power series, but cannot say anything when $|x|=R$: thus you have correctly determined the radius of convergence $R=\frac{\pi}{2}$ of your power series, but this does not give any knowledge on its behavior path the extremes of its interval (disk) of convergence.

How to find the behavior of $\sum_{n=0}^{\infty} a_n x^n$ for $x=R$?

Assuming without loss of generality $R=1$ (you can always use the transformation $x\mapsto Rx$ and consider a new power series $\sum_{n=0}^{\infty} \hat{a}_n x^n$ with $\hat{a}_n=a_nR^n$ for all $n\in\mathbb{N}$), apart from the standard Cauchy convergence criterion which Dèö cites in the comments to his answer, the only necessary and sufficient condition for the convergence of power series on the boundary points $|x|=1$ I am aware of is Tauber's second theorem. In the current situation, it states that the power series is summable and the value of its sum is $s_\pm$ ($s_+$ for $x=1$ and $s_-$ for $x=-1$; note that the original statement of Tauber considers each boundary point singularly) if and only if

  1. $\lim_{z\to \pm 1^-}\sum_{n=0}^\infty a_nx^n=s_\pm$ and
  2. $a_1+2a_2+\cdots+na_n=o(n)\quad \forall n\in\mathbb{N}_+$.

However, like Cauchy criterion, this is not easily applicable to real problems: the fulfillment of condition 1 means requiring the series $\sum_{n=0}^\infty a_n$ to be Abel summable, while the fulfillment of condition 2 means that the Cesaro mean of its partial sums vanishes as $n\to \infty$. Definitely not the easiest properties to check, even if in some case they can be quite effective.

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  • $\begingroup$ But then how should I proceed to find the behavior at the endpoints ? I mean, if I plug in $\frac{\pm \pi}{2}$, what should I do next ? Given that it's raised to the nth power, how can I define the convergence there in another way ? $\endgroup$
    – Ryukyu
    Dec 29 '18 at 17:02
  • $\begingroup$ In another way, how should I proceed in order to evaluate the convergence/ divergence at the boundary if I can't use the root test ? $\endgroup$
    – Ryukyu
    Dec 29 '18 at 17:13
  • $\begingroup$ This is a difficult matter: I'll add something to my answer later. $\endgroup$ Dec 29 '18 at 17:50
  • $\begingroup$ @Poujh: thanks for accepting my answer. I corrected it a bit and added a few considerations. $\endgroup$ Dec 29 '18 at 22:53
  • $\begingroup$ Okay, thank you ! $\endgroup$
    – Ryukyu
    Dec 30 '18 at 11:30
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You can directly find the radius of convergence for the power series $\sum_{n=0}^{\infty}a_nx^n$ by $\textit{Hadamard's formula}$, given by $$\frac{1}{R}=\limsup_{n\rightarrow \infty } \sqrt[n]{a_n}.$$ In the given series, $a_n=\frac{1}{(\cos^{-1}(\sqrt{n+1}-\sqrt{n}))^n} \implies \frac{1}{R}=\limsup_{n\rightarrow \infty } \frac{1}{(\cos^{-1}(\sqrt{n+1}-\sqrt{n}))}$, since in the interval $(-\frac{\pi}{2},\frac{\pi}{2})$ containing $0$ the function $\cos^{-1}$ is continuous and $\lim_{n\rightarrow \infty}(\sqrt{n+1}-\sqrt{n})=0$, hence $R=\frac{\pi}{2}$. At the end point say $x=\frac{\pi}{2}$, the series $\sum_{n=0}^{\infty}\frac{(\frac{\pi}{2})^n}{(\cos^{-1}(\sqrt{n+1}-\sqrt{n}))^n} $ diverges as the $n^{th}$ terms goes to $1$ as $n \rightarrow \infty$, similarly for $x=-\frac{\pi}{2}$.

The above calculation is valid if the convergence is asked to be in $\mathbb{R}$. If the convergence is asked in $\mathbb{C}$, the disc of convergence would be $|z|<\frac{\pi}{2}$ and we have to look for all the points in $|z|=R$ , if it converges or not.

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  • $\begingroup$ Just a question : what method did you exactly use to prove the divergence at the endpoints ? $\endgroup$
    – Ryukyu
    Dec 29 '18 at 17:24
  • $\begingroup$ If the series $\sum a_n$ converges then the $\lim_{n\rightarrow \infty} a_n=0$, which trivially follows from $S_n-S_{n-1}=a_n$, where $S_n$ is the partial sum of first $n$ terms. $\endgroup$
    – Dèö
    Dec 29 '18 at 17:29
  • $\begingroup$ Yeah, but so you used the fact that we get pi over two over pi over two as n tends toward infinity which yields 1 (resp. -1). You didn't use the root test or ratio test because we can't use them at the borders. Basically, we just ask us at the endpoints if we get 0 as n approaches infinity by plugging in the endpoint values, we don't use another "test" as before, is that correct ? $\endgroup$
    – Ryukyu
    Dec 29 '18 at 17:33
  • $\begingroup$ But $\lim_{n\rightarrow \infty}a_n =0$ doesn't guarantee the convergence of the series, but if it is not 0 we can certainly say that the series does not converge. Judging if the series converges or not needs a few more arguments which are illustrated by @ Daniele Tampieri. $\endgroup$
    – Dèö
    Dec 30 '18 at 16:46

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