Evaluate the sum


Totally no idea. I think this series may related to the $\sin x$ series because of those missing even powers. Another way of writing this series:


  • $\begingroup$ Those missing even powers simply mean that your function is an odd function; that is, $f(-x)=-f(x)$. $\endgroup$ – robjohn Oct 21 '13 at 21:34

In this answer, I mention this identity, which can be proven by repeated integration by parts: $$ \int_0^{\pi/2}\sin^{2k+1}(x)\;\mathrm{d}x=\frac{2k}{2k+1}\frac{2k-2}{2k-1}\cdots\frac{2}{3}=\frac{1}{2k+1}\frac{4^k}{\binom{2k}{k}}\tag{1} $$ Your sum can be rewritten as $$ f(x)=\sum_{k=0}^\infty\frac1{(2k+1)}\frac{4^k}{\binom{2k}{k}}x^{2k+1}\tag{2} $$ Combining $(1)$ and $(2)$, we get $$ \begin{align} f(x) &=\int_0^{\pi/2}\sum_{k=0}^\infty\sin^{2k+1}(t)x^{2k+1}\,\mathrm{d}t\\ &=\int_0^{\pi/2}\frac{x\sin(t)\,\mathrm{d}t}{1-x^2\sin^2(t)}\\ &=\int_0^{\pi/2}\frac{-\,\mathrm{d}x\cos(t)}{1-x^2+x^2\cos^2(t)}\\ &=-\frac1{\sqrt{1-x^2}}\left.\tan^{-1}\left(\frac{x\cos(t)}{\sqrt{1-x^2}}\right)\right]_0^{\pi/2}\\ &=\frac1{\sqrt{1-x^2}}\tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)\\ &=\frac{\sin^{-1}(x)}{\sqrt{1-x^2}}\tag{3} \end{align} $$

Radius of Convergence

This doesn't appear to be part of the question, but since some other answers have touched on it, I might as well add something regarding it.

A corollary of Cauchy's Integral Formula is that the radius of convergence of a complex analytic function is the distance from the center of the power series expansion to the nearest singularity. The nearest singularity of $f(z)$ to $z=0$ is $z=1$. Thus, the radius of convergence is $1$.

  • $\begingroup$ This is awesome! Thank you! $\endgroup$ – Frank Lu Oct 21 '13 at 22:22

I refer you to this solution. The function $f(x)$ represented by the sum in question is

$$f(x) = \sum_{n=0}^{\infty} \frac{2^{2 n}}{\displaystyle (2 n+1) \binom{2 n}{n}} x^{2 n+1} = \frac{\arcsin{x}}{\sqrt{1-x^2}}$$

You may deduce that the radius of convergence is $1$ from the relation

$$\frac{1}{2^{2 n}} \binom{2 n}{n} \sim \frac{1}{\sqrt{\pi n}} \quad (n \to \infty)$$

which may be shown as a result of, e.g., Stirling's approximation.


Try to find out $\lim {a_{n+1}\over a_n}$, $a_n={1.2.\dots 2n \over 1.3.5\dots 2n+1}x^{2n+1}$

  • $\begingroup$ That would help you find the radius of convergence. But the sum? $\endgroup$ – Daniel Fischer Oct 21 '13 at 17:46
  • $\begingroup$ I think it means to find some $f(x)$ which coincides with this power series within the radius of convergence. $\endgroup$ – Frank Lu Oct 21 '13 at 17:47

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