If $e_0 = \frac{1}{2} $and $\forall n\in\Bbb{N}:e_n=\frac{(2n-1)^2}{2n(2n+1)}e_{n-1}$, find $\sum_{n\geq 0} e_n$ Consider the following optimal stopping game:  The controller is presented with steps in a fair random walk (fair coin flips, $P(h)=P(t) = \frac{1}{2}$) and at each stage of the game, the controller can stop and take a payoff of $\frac{H}{H+T}$ where $H$ is the number of heads encountered up till that point, and $T$ is the number of tails.
A proposed stopping strategy is to stop the first time the payoff exceeds $\frac{1}{2}$. Note that this strategy does eventually stop, with probability 1. (This is probably in fact the optimal strategy, but that is not the point of this question.) What is the expectation value of the payoff if that strategy is used?
I have been able to calculate the probability $g_m$ of stopping at step $2m+1$ as
$$
g_m = \frac{1}{2^{2m+1}}\frac{1}{m+1} \binom{2m}{m}
$$
and the contribution to the expected payoff is from the possibility of stopping at step $2m+1$ is
$$
e_m = \frac{1}{2} g_m \left( 1+\frac{1}{2m+1} \right)
$$
It is then easy to show that $e_0 = \frac{1}{2}$ and for all integer $n>0$ 
$$
e_n=\frac{(2n-1)^2}{2n(2n+1)}e_{n-1}
$$
To find the expected payoff I need to find
$$\sum_{n=0}^\infty e_n
$$
Thus the question: 
If $e_0 = \frac{1}{2} $and $\forall n\in\Bbb{N}:e_n=\frac{(2n-1)^2}{2n(2n+1)}e_{n-1}$, find $\sum_{n\geq 0} e_n$
 A: $$\begin{align}e_n &= \frac{(2 n-1)^2}{(2 n+1) (2 n)} e_{n-1} \\ &=\frac{(2 n-1)^2}{(2 n+1) (2 n)} \frac{(2 n-3)^2}{(2 n-1) (2 n-2)}\cdots \frac{1^2}{3 \cdot 2} \cdot\frac12 \\ &= \frac{(2 n)!}{(2 n+1) n!^2} \frac1{2^{2 n+1}} \end{align} $$
Thus,
$$ \sum_{n=0}^{\infty} e_n = \sum_{n=0}^{\infty} \frac{(2 n)!}{(2 n+1) n!^2} \frac1{2^{2 n+1}} $$
This sum is straightforward.  Consider
$$f(x) = \sum_{n=0}^{\infty} \frac{(2 n)!}{(2 n+1) n!^2} x^{2 n+1} $$
Then
$$f'(x) = \sum_{n=0}^{\infty} \binom{2 n}{n} x^{2 n} = \left ( 1-4 x^2 \right )^{-1/2}$$
Then
$$f(x) = \frac12 \arcsin{2 x} $$
The sum is then
$$f \left ( \frac12 \right ) = \frac{\pi}{4} $$
A: Here is a slightly longer approach, but it might shed a little more light on how to get the answer without necessarily recognizing the series first.
Start with
$$
\begin{align}
e_n
&=\frac{(2n-1)^2}{2n(2n+1)}e_{n-1}\\
&=\frac12\frac{(2n-1)!!^2}{(2n)!!(2n+1)!!}\\
&=\frac1{2(2n+1)}\frac{(2n-1)!!}{(2n)!!}\\
&=\frac1{2(2n+1)}\frac{(n-\frac12)(n-\frac32)\cdots\frac12}{n!}\\
&=\frac1{2(2n+1)}\binom{n-\frac12}{n}\\
&=\frac{(-1)^n}{2(2n+1)}\binom{-\frac12}{n}\\
\end{align}
$$
Then,
$$
\begin{align}
\sum_{n=0}^\infty e_nx^{2n+1}
&=\frac12\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}\binom{-\frac12}{n}x^{2n+1}\\
&=\frac12\sum_{n=0}^\infty\int_0^x(-1)^n\binom{-\frac12}{n}t^{2n}\,\mathrm{d}t\\
&=\frac12\int_0^x\frac{\mathrm{d}t}{\sqrt{1-t^2}}\\
&=\frac12\sin^{-1}(x)
\end{align}
$$
Set $x=1$ to get
$$
\sum_{n=0}^\infty e_n=\frac\pi4
$$
