Summing the series $(-1)^k \frac{(2k)!!}{(2k+1)!!} a^{2k+1}$ How does one sum the series $$ S = a -\frac{2}{3}a^{3} + \frac{2 \cdot 4}{3 \cdot 5} a^{5} - \frac{ 2 \cdot 4 \cdot 6}{ 3 \cdot 5 \cdot 7}a^{7} + \cdots $$
This was asked to me by a high school student, and I am embarrassed that I couldn't solve it. Can anyone give me a hint?!

You can use the formula
$\displaystyle \int_{0}^{\frac{\pi}{2}}{\sin^{2k+1}(x) dx} = \frac{2 \cdot 4 \cdot 6 \cdots 2k}{3\cdot 5 \cdots (2k+1)}$
This is called Wallis's product.
So we have $\displaystyle S(a) = \sum_{k=0}^{\infty} (-1)^k a^{2k+1} \int_{0}^{\frac{\pi}{2}}{\sin^{2k+1}(x) dx}$ 
Interchanging the sum and the integral
$\displaystyle S(a) =   \int_{0}^{\frac{\pi}{2}}{\sum_{k=0}^{\infty}{(-1)^{k}(a\sin x)^{2k+1}} dx}$
The sum inside the integral is a geometric series of the form
$\displaystyle x - x^3 + x^5 - \cdots = x(1 - x^2 + x^4 - \cdots)  = \frac{x}{1+x^2}$
Hence,
$\displaystyle S(a) =   \int_{0}^{\frac{\pi}{2}}{\frac{a\sin x}{1 + (a\sin x)^2}}dx$
Now substitute $\displaystyle t = a \cos x$
The integral becomes
$\displaystyle \int_{0}^{a}{\frac{1}{1+a^2 - t^2}}dt = \frac{1}{2\sqrt{a^2+1}}\ln \left(\frac{\sqrt{a^2+1}+a}{\sqrt{a^2+1}-a} \right)$
Now $\displaystyle \ln \left(\frac{\sqrt{a^2+1}+a}{\sqrt{a^2+1}-a}\right) = \ln \left(\frac{\left(\sqrt{a^2+1}+a \right)^2}{\left(\sqrt{a^2+1}-a \right)\left(\sqrt{a^2+1}+a \right)}\right) = 2\ln \left(\sqrt{a^2+1}+a \right)$
So
$\displaystyle S(a) = \frac{1}{\sqrt{a^2+1}}\ln \left(\sqrt{a^2+1}+a \right) = \frac{\sinh^{-1}(a)}{\sqrt{a^2+1}}$
