How to express $\sum _{j=0}^{\infty } \frac{\binom{j-\frac{1}{2}}{j}}{2 j+2 k+1}$ as an explicit function of $k$ I'm trying to solve the period of a pendulum and right now I'm at
$$T=\frac{4}{\omega _0} \sum _{k=0}^{\infty } \large(\binom{-\frac{1}{2}}{k} \left(\sum _{j=0}^{\infty }
   \frac{\binom{j-\frac{1}{2}}{j}}{2 j+2 k+1}\right) u_0^{2k}\large)$$
The next step is to get rid of the inner infinite sum $$\sum _{j=0}^{\infty } \frac{\binom{j-\frac{1}{2}}{j}}{2 j+2 k+1}$$
The binomial coefficient can be broken into half factorials like $$\sum _{j=0}^{\infty } \frac{\frac{(j-\frac{1}{2})!}{j!(\frac{-1}{2})!}}{2 j+2 k+1}$$
which can be simplified into $$\sum _{j=0}^{\infty } \frac{(2j)!}{2^{2j}j!^2(2 j+2 k+1)}$$
I'm really not sure what to do next. Mathematica says it should be $\frac{(2k)!}{2^{2k+1}k!^2}\pi$, but I don't know how to get there.
 A: 
We obtain
  \begin{align*}
\color{blue}{\sum_{j=0}^\infty}&\color{blue}{\frac{1}{2j+2k+1}\binom{j-\frac{1}{2}}{j}}\\
&=\sum_{j=0}^\infty\binom{-\frac{1}{2}}{j}(-1)^j\int_{0}^{1}z^{2j+2k}\,dz\tag{1}\\
&=\int_0^1z^{2k}\sum_{j=0}^\infty\binom{-\frac{1}{2}}{j}(-1)^jz^{2j}\,dz\\
&=\int_0^1\frac{z^{2k}}{\sqrt{1-z^2}}\,dz\tag{2}\\
&=\int_0^{\frac{\pi}{2}}\sin^{2k}t\,dt\tag{3}\\
&=\frac{1}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\frac{e^{it}-e^{-it}}{2i}\right)^{2k}\,dt\tag{4}\\
&=\frac{(-1)^k}{2^{2k+1}}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sum_{j=0}^{2k}\binom{2k}{j}(-1)^je^{2i(j-k)}t\,dt\tag{5}\\
&\,\,\color{blue}{=\frac{\pi}{2^{2k+1}}\binom{2k}{k}}\tag{6}
\end{align*}

Comment:


*

*In (1)  we use an integral representation to get rid of the denominator. We also use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.

*In (2) we apply the binomial series expansion.

*In (3) we use the substitution $z=\sin t, dz = \cos t\, dt$.

*In (4) we represent $\sin  t$   using Euler's formula.
We  also  use the  symmetry  of the sine function and  integrate from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$.

*In (5) we use the binomial theorem.

*In (6) we note $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}e^{int}\,dt =0$  if $n\in\mathbb{Z}\setminus\{0\}$ and have therefore only to consider the term with $j=k$.

Recalling  the binomial  identity $\binom{-\frac{1}{2}}{k}=\frac{(-1)^k}{2^{2k}}\binom{2k}{k}$, we obtain from (6)
  \begin{align*}
T&=\frac{4}{\omega_{0}}\sum_{k=0}^\infty\binom{-\frac{1}{2}}{k}\frac{\pi}{2^{2k+1}}\binom{2k}{k}u_{0}^{2k}\\
&=\frac{2\pi}{\omega_{0}}\sum_{k=0}^\infty\binom{2k}{k}^2(-1)^k\left(\frac{u_0}{4}\right)^{2k}\tag{7}
\end{align*}
In (7) we see  the coefficients of the series are squared central binomial coefficients. This sequence is archived in OEIS as A002894. Since there is no closed form of (7) given, we rather don't expect it.

A: As $\dfrac{(2j)!}{2^{2j}j!^2}=\dfrac{(2j-1)!!}{(2j)!!}=(-1)^j\dbinom{-1/2}{j}$, the last sum is equal to $$\int_0^1 x^{2k}(1-x^2)^{-1/2}\ dx=\frac12\mathrm{B}\Big(k+\frac12,\frac12\Big)=\frac{(2k-1)!!}{(2k)!!}\frac{\pi}{2},$$
and your expression for $T$ leads to something like this (as one might expect).
