TITLE: Convergence of a series depending on a parameter.
QUESTION [8 upvotes]: 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.

REPLY [2 votes]: We have
$$\frac{\pi}{2} -\arcsin x = \int_x^1 (1-t^2)^{-1/2}\, dt.$$
Now $1-t^2 = (1+t)(1-t),$ so for $x$ close to $1,$ the integrand above looks a lot like $(2(1-t))^{-1/2}.$ And indeed, you can check by L'Hopital and the FTC that
$$\lim_{x\to 1^-}\frac{\int_x^1 (1-t^2)^{-1/2}\, dt}{\int_x^1 (2(1-t))^{-1/2}\, dt} = 1.$$
The nice thing is that the integral in the denominator can be done easily. It equals $\sqrt 2 (1-x)^{1/2}.$ This leads quickly to the answer you obtained.