Prove that the circumference of an ellipse is given by : $$2\pi a\left[1-\sum_{n=1}^{\infty}\left(\frac{\left(2n-1\right)!!}{\left(2n\right)!!}\right)^{2}\frac{e^{2n}}{2n-1}\right]$$
The parametric of an ellipse is :
$$x=a\cos(\theta)$$
$$y=b\sin(\theta)$$ The Circumference of parametric curve can be computed via: $$\int_{\alpha}^{\beta}\ \sqrt{\left(\frac{dx}{dt}\right)^{2}+\left(\frac{dy}{dt}\right)^{2}}d\theta$$
Assuming the curve is not self-intersecting over the given interval.
Using this we see that the circumference of an ellipse is:
$$4\int_{0}^{\frac{\pi}{2}}\ \sqrt{a^{2}-\left(a^{2}-b^{2}\right)\cos^{2}\left(\theta\right)}d\theta$$
Assuming $a>b$,then $e=\frac{c}{a}=\frac{\sqrt{a^{2}-b^{2}}}{a}$ follows the integral is :
$$4a\int_{0}^{\frac{\pi}{2}}\sqrt{1-e^{2}\cos^{2}\left(\theta\right)}=4a\int_{0}^{\frac{\pi}{2}}\sqrt{1-e^{2}\sin^{2}\left(\theta\right)}$$
It's well-known that :$$\sqrt{1-x}=-\sum_{n=0}^{\infty}\binom{2n}{n}\frac{x^{n}}{4^{n}\left(2n-1\right)}$$
Which is convergent for $\left|x\right|<1$.
Since $0<e^{2}\sin^{2}\left(\theta\right)<1$,hence the integral maybe written as:
$$4a\left[\int_{0}^{\frac{\pi}{2}}d\theta+\int_{0}^{\frac{\pi}{2}}-\sum_{n=1}^{\infty}\binom{2n}{n}\frac{e^{2n}\sin^{2n}\left(\theta\right)}{4^{n}\left(2n-1\right)}d\theta\right]$$
Fubini/Tonelli theorems implies that the integral is indeed:
$$4a\left[\frac{\pi}{2}-\sum_{n=1}^{\infty}\binom{2n}{n}\frac{e^{2n}}{4^{n}\left(2n-1\right)}\int_{0}^{\frac{\pi}{2}}\sin^{2n}\left(\theta\right)d\theta\right]$$ From :
$$\int_{0}^{\frac{\pi}{2}}\sin^{2n}\left(\theta\right)d\theta=\frac{\pi}{2}\prod_{k=1}^{n}\frac{2k-1}{2k}\tag{$n \in \mathbb N^+$}$$ Then integral transforms to:
$$2\pi a\left[1-\sum_{n=1}^{\infty}\binom{2n}{n}\frac{e^{2n}}{4^{n}\left(2n-1\right)}\prod_{k=1}^{n}\frac{2k-1}{2k}\right]$$$$2\pi a\left[1-\sum_{n=1}^{\infty}\frac{\left(2n-1\right)!!}{\left(n!\right)^{2}4^{n}}\frac{e^{2n}}{2n-1}\right]$$
But how to finish ?
Note that
$$\begin{align} \binom{2n}{n}\frac{e^{2n}}{4^n(2n-1)}\prod_{k=1}^n\frac{(2k-1)}{2k}=\frac{(2n)!}{(n!)^2}\frac{e^{2n}}{4^n(2n-1)}\frac{(2n-1)!!}{(2n)!!}\tag1 \end{align}$$
We also have the identities
$$\begin{align} (2n)!!&=(2n)(2n-2)(2n-4)\cdots (2)\\\\ &=2^nn!\tag3 \end{align}$$
and
$$\begin{align} (2n-1)!!&=(2n-1)(2n-3)\cdot 1\\\\ &=\frac{(2n)(2n-1)(2n-2)(2n-3)\cdot 1}{(2n)(2n-2)\cdots (2)}\\\\ &=\frac{(2n)!}{2^n(n!)}\tag3 \end{align}$$
Using $(2)$ and $(3)$ in the right-hand side of $(1)$, we find that
$$\begin{align} \binom{2n}{n}\frac{e^{2n}}{4^n(2n-1)}\prod_{k=1}^n\frac{(2k-1)}{2k}=\left(\frac{(2n-1)!!}{(2n)!}\right)^2\frac{e^{2n}}{4^n(2n-1)} \end{align}$$
as was to be shown!