<s>[INST] Convergence of a series depending on a parameter. I was studying the convergence of a series with a parameter and I want to ask you if my conclusion is correct and if there is a better method to do it.
$$\sum_{n=1}^{\infty}\left(\frac{\pi}{2}-\arcsin\frac{n}{n+4} \right)^{\alpha}$$
I'm asked to say for which $\alpha$ the series is convergent. I tried to do this:
$$\sum_{n=1}^{\infty}\left(\frac{\pi}{2}-\arcsin\frac{n}{n+4} \right)^{\alpha}=
\sum_{n=1}^{\infty}\left(\arccos\frac{n}{n+4} \right)^{\alpha}=\sum_{n=1}^{\infty}\left(\arccos\left(1-\frac{4}{n+4}\right)\right)^{\alpha}$$
Now I tried to see the Taylor Series of $\arccos(1-y)$ as $y\rightarrow 0$, $y=\frac{4}{n+4}$.
By deriving $g(y)=\arccos(1-y)$ I obtain:
$$g'(y)=\frac{1}{\sqrt{1-(1-y)^2}}=(-x^2+2x)^{\frac{1}{2}}=(2x)^{\frac{1}{2}}(1-\frac{x}{2})^{\frac{1}{2}}=\sqrt{2y}+o(\sqrt{2y})$$
So I see that:
$$g(y)=\int\frac{1}{\sqrt{2y}}+o\left(\frac{1}{\sqrt{2x}}\right)dx=\sqrt{2x}+o(\sqrt{2x})$$
From this, I can conclude that my series is:
$$\sum_{n=1}^{\infty}\left(\arccos\left(1-\frac{4}{n+4}\right)\right)^{\alpha} \sim \left(\sqrt{\frac{8}{n+4}}\right)^{\alpha}$$
So, for asymptotic comparison to $\left(\frac{1}{n}\right)^{\frac{\alpha}{2}}$ the series converges if and only if $\alpha > 2$.
Am I right? If yes, may I ask you if there is a better method to do this computation? And, expecially, if there is a better method (if possible without involving integrals) to compute the Taylor series of inverse trigonometric functions!
Thanks in advance.
 [/INST] $\sin(x) = \sin(\frac{\pi}{2}) + \cos\left(\frac{\pi}{2}\right)\left(x - \frac{\pi}{2}\right) - \frac{1}{2}\sin\left(\frac{\pi}{2}\right)\left(\frac{\pi}{2} - x\right)^2 + o_{x \rightarrow \frac{\pi}{2}}\left(\left(\frac{\pi}{2} - x\right)^2\right) \\ = 1 - \frac{1}{2}\left(\frac{\pi}{2} - x\right)^2 + o_{x \rightarrow \frac{\pi}{2}}\left(\left(\frac{\pi}{2} - x\right)^2\right)$
Then
\begin{align}
\lim_{x\rightarrow \frac{\pi}{2}^-} \frac{1 - \sin(x)}{\left(\frac{\pi}{2} - x\right)^2} = \frac{1}{2} 
\end{align}
Using  $u = \sin(x)$ we have
\begin{align}
\lim_{u \rightarrow 1^-} \frac{1 - u}{\left(\frac{\pi}{2} - \arcsin(u)\right)^2} = \frac{1}{2}
\end{align}
Taking the root we have $\frac{\pi}{2} - \arcsin(u) \sim_{u\rightarrow 1} \sqrt{2(1-u)}$.
</s>