Indirect method (associated with a certain problem of electrostatics) indicates that $$\sum\limits_{j=1}^\infty \frac{(2j-3)!!\,(2j-1)!!}{(2j-2)!!\,(2j+2)!!}=\frac{2}{3\pi}.$$ Is this result known?

## 1 Answer

Using the standard power series for the complete elliptic integral of the second kind $$E(k) = \frac{\pi}{2} \sum_{j=0}^\infty \left(\frac{(2j)!}{2^{2j}(j!)^2}\right)^2 \frac{k^{2j}}{1-2j},$$ we find \begin{align*} \sum\limits_{j=1}^\infty \frac{(2j-3)!!\,(2j-1)!!}{(2j-2)!!\,(2j+2)!!} k^{2j}&=\sum_{j=1}^\infty\frac{-j}{j+1} \left(\frac{(2j)!}{2^{2j}(j!)^2}\right)^2 \frac{k^{2j}}{1-2j} \\ &= -\frac{1}{k^2} \int_0^k\mathrm{d}k\,k^2 \frac{\mathrm{d}}{\mathrm{d}k}\left(\frac{2}{\pi}E(k)\right)\\ &= \frac{2}{3}\frac{k^2-1}{k^2}\frac{2}{\pi}K(k) - \frac{k^2-2}{3k^2}\frac{2}{\pi}E(k). \end{align*} In the limit $k\to 1$ only the second term survives with $E(1)=1$ and therefore \begin{align*} \sum\limits_{j=1}^\infty \frac{(2j-3)!!\,(2j-1)!!}{(2j-2)!!\,(2j+2)!!} &=\frac{2}{3\pi}. \end{align*}