The sum of the series

\frac{\pi}{2}=\sum_{k=0}^\infty\frac{k!}{(2k+1)!!}\tag{1}

can be derived by accelerating the Gregory Series

\frac{\pi}{4}=\sum_{k=0}^\infty\frac{(-1)^k}{2k+1}\tag{2}

using Euler’s Series Transformation. Mathematica is able to sum (1), so I assume there must be some method to sum the series in (1) directly; what might that method be?

**Answer**

First, (2k+1)!! = (2k+1)(2k-1) \cdots (1) = \frac{(2k+1)!}{(2k)(2(k-1)) \cdots 2(1)} = \frac{(2k+1)!}{2^k k!}.

So your sum can be rewritten as

\sum_{k=0}^\infty\frac{k! \, k! \, 2^k }{(2k+1)!} = \sum_{k=0}^\infty\frac{2^k}{(2k+1)\binom{2k}{k}}.

Variations of the sum of reciprocals of the central binomial coefficients have been well-studied. For example, this paper by Sprugnoli (see Theorem 2.4) gives the ordinary generating function of a_k = \frac{4^k}{(2k+1)}\binom{2k}{k}^{-1} to be

A(t) = \frac{1}{t} \sqrt{\frac{t}{1-t}} \arctan \sqrt{\frac{t}{1-t}}.

Subbing in t = 1/2 says that \sum_{k=0}^\infty\frac{2^k}{(2k+1)\binom{2k}{k}} = 2 \arctan(1) = \frac{\pi}{2}.

**Attribution***Source : Link , Question Author : robjohn , Answer Author : Mike Spivey*