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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 $lim_{nto infty} sqrt{n+1}-sqrt{n}=0$ 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.

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 $x=pmfrac{pi}{2}$ 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 $xmapsto Rx$ and consider a new power series
$sum_{n=0}^{infty} hat{a}_n x^n$ with $hat{a}_n=a_nR^n$ for all $ninmathbb{N}$), 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. 
   $lim_{zto pm 1^-}sum_{n=0}^infty a_nx^n=s_pm$ and


2. 
   $a_1+2a_2+cdots+na_n=o(n)quad forall ninmathbb{N}_+$.

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 $nto infty$. Definitely not the easiest properties to check, even if in some case they can be quite effective.

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 $$frac{1}{R}=limsup_{nrightarrow infty } sqrt[n]{a_n}.$$
In the given series, $a_n=frac{1}{(cos^{-1}(sqrt{n+1}-sqrt{n}))^n} implies frac{1}{R}=limsup_{nrightarrow infty } frac{1}{(cos^{-1}(sqrt{n+1}-sqrt{n}))}$, since in the interval $(-frac{pi}{2},frac{pi}{2})$ containing $0$ the function $cos^{-1}$ is continuous and $lim_{nrightarrow infty}(sqrt{n+1}-sqrt{n})=0$, 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.