$\sum_{k=0} ^\infty (-1)^k \frac{(2k)!!}{(2k+1)!!} a^{2k+1}$ to differential equation

Question

I want to use $S = \sum_{k=0} ^\infty (-1)^k \frac{(2k)!!}{(2k+1)!!} a^{2k+1}$ and get the relation $(a^2+1) S'=1−aS$. So far I am just getting $\frac{dS}{da} = \sum_{k=0} ^\infty (-1)^k \frac{(2k)!!}{(2k+1)!!} a^{2k} (2k+1)$, which I am not seeing how to use with S to get that relation. What should I do to progress and get the desired relation?

Answer 1

Turns out to be quite straightforward once the double factorials are replaced.

$S(a) = \sum_{k=0} ^\infty (-1)^k \dfrac{(2k)!!}{(2k+1)!!} a^{2k+1} $.

$\begin{array}\\ \dfrac{(2k)!!}{(2k+1)!!} &=\dfrac{\prod_{j=0}^{k-1}(2k-2j)}{\prod_{j=0}^{k-1}(2k+1-2j)}\\ &=\dfrac{\prod_{j=1}^{k}(2j)}{\prod_{j=1}^{k}(2j+1)}\\ &=\dfrac{\prod_{j=1}^{k}(2j)^2}{\prod_{j=1}^{k}(2j)(2j+1)}\\ &=\dfrac{4^k(k!)^2}{(2k+1)!}\\ \end{array} $

$\begin{array}\\ S(a) &= \sum_{k=0} ^\infty (-1)^k \dfrac{(2k)!!}{(2k+1)!!} a^{2k+1}\\ &= \sum_{k=0} ^\infty (-1)^k \dfrac{4^k(k!)^2}{(2k+1)!} a^{2k+1}\\ S'(a) &= \sum_{k=0} ^\infty (-1)^k \dfrac{4^k(k!)^2}{(2k)!} a^{2k}\\ a^2S'(a) &= \sum_{k=0} ^\infty (-1)^k \dfrac{4^k(k!)^2}{(2k)!} a^{2k+2}\\ &= \sum_{k=1} ^\infty (-1)^{k-1} \dfrac{4^{k-1}((k-1)!)^2}{(2k-2)!} a^{2k}\\ (a^2+1)S'(a) &= \sum_{k=1} ^\infty (-1)^{k-1} \dfrac{4^{k-1}((k-1)!)^2}{(2k-2)!} a^{2k}+\sum_{k=0} ^\infty (-1)^k \dfrac{4^k(k!)^2}{(2k)!} a^{2k}\\ &= 1+\sum_{k=1} ^\infty \left((-1)^{k-1}\dfrac{4^{k-1}((k-1)!)^2}{(2k-2)!}+(-1)^k \dfrac{4^k(k!)^2}{(2k)!}\right) a^{2k}\\ &= 1+\sum_{k=1} ^\infty (-1)^{k-1} \dfrac{4^{k-1}((k-1)!)^2}{(2k-2)!}\left(1- \dfrac{4k^2}{(2k)(2k-1)}\right) a^{2k}\\ &= 1+\sum_{k=1} ^\infty (-1)^{k-1} \dfrac{4^{k-1}((k-1)!)^2}{(2k-2)!}\left(\dfrac{2k(2k-1)-4k^2}{(2k)(2k-1)}\right) a^{2k}\\ &= 1+\sum_{k=1} ^\infty (-1)^{k-1} \dfrac{4^{k-1}((k-1)!)^2}{(2k-2)!}\left(\dfrac{-2k}{(2k)(2k-1)}\right) a^{2k}\\ &= 1+\sum_{k=1} ^\infty (-1)^{k} \dfrac{4^{k-1}((k-1)!)^2}{(2k-1)!} a^{2k}\\ 1-aS(a) &= 1-a\sum_{k=0} ^\infty (-1)^k \dfrac{4^k(k!)^2}{(2k+1)!} a^{2k+1}\\ &= 1+\sum_{k=0} ^\infty (-1)^{k+1} \dfrac{4^k(k!)^2}{(2k+1)!} a^{2k+2}\\ &= 1+\sum_{k=1} ^\infty (-1)^{k} \dfrac{4^{k-1}((k-1)!)^2}{(2k-1)!} a^{2k}\\ \end{array} $

and they match.

Answer 2

Hint: write $a^2 S'$ and $a S$ as sums involving $a^{2k}$ rather than $a^{2k+2}$, so you can combine the terms.