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:


So I see that:


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.

  • $\begingroup$ Hyperbolic functions have nothing to do here. This being said, Taylor's expansion of inverse trigonometric and hyperbolic functions are standard near $0$. $\endgroup$
    – Bernard
    Commented Aug 11, 2017 at 16:42
  • $\begingroup$ Oh, sorry, I was asking for inverse trigonometric functions, not inverse hyperbolic ones. Is there any method to expand those functions without involving integrals? $\endgroup$ Commented Aug 11, 2017 at 16:46

3 Answers 3


$$\begin{eqnarray*}\arcsin(1)-\arcsin\left(\frac{n}{n+4}\right) &=& \int_{\frac{n}{n+4}}^{1}\frac{dz}{\sqrt{1-z^2}}\\&\stackrel{z\mapsto 1-x}{=}&\int_{0}^{\frac{4}{n+4}}\frac{dx}{\sqrt{x(2-x)}}\\&\stackrel{x\to u^2}{=}&2\int_{0}^{\frac{2}{\sqrt{n+4}}}\frac{du}{\sqrt{2-u^2}}&\end{eqnarray*}$$ leads to $\frac{\pi}{2}-\arcsin\left(\frac{n}{n+4}\right)\sim\sqrt{\frac{8}{n+4}}$ in a more efficient way, but your solution is just fine.


$\sin(x) = \sin(\frac{\pi}{2}) + \cos\left(\frac{\pi}{2}\right)\left(x - \frac{\pi}{2}\right) - \frac{1}{2}\sin\left(\frac{\pi}{2}\right)\left(\frac{\pi}{2} - x\right)^2 + o_{x \rightarrow \frac{\pi}{2}}\left(\left(\frac{\pi}{2} - x\right)^2\right) \\ = 1 - \frac{1}{2}\left(\frac{\pi}{2} - x\right)^2 + o_{x \rightarrow \frac{\pi}{2}}\left(\left(\frac{\pi}{2} - x\right)^2\right)$


\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 $\frac{\pi}{2} - \arcsin(u) \sim_{u\rightarrow 1} \sqrt{2(1-u)}$.


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.


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