Question of the value of the following double sum I hope to get the exact value of the following double series:
$$ \sum_{k \geq 0} \sum_{n \geq 0} \binom{2k}{ k} \frac{(-1)^n}{n! (2k+2n+1) 2^{4k+2n+1}}. $$
I am not sure it is possible or not.
I need your comments.

You can write your double series as:
$$\begin{eqnarray*}\color{red}{S}&=&\frac{1}{2}\int_{0}^{1}\sum_{k\geq 0}\sum_{n\geq 0}\binom{2k}{k}16^{-k}x^{2k}\frac{(-1)^n}{4^n\,n!}x^{2n}\,dx\\&=&\int_{0}^{1}\frac{e^{-x^2/4}}{\sqrt{4-x^2}}dx=\int_{0}^{1/2}\frac{dx}{e^{x^2}\sqrt{1-x^2}}=\color{red}{\int_{0}^{\pi/6}e^{-\sin^2\theta}d\theta},\end{eqnarray*}$$
but I would not bet that the last integral can be written in terms of elementary functions. 
Numerically, we have:
$$ S = 0.481561249\ldots $$
A fast converging series arises from considering the Fourier series of $e^{-\sin^2\theta}$:
$$e^{-\sin^2\theta}=\frac{1}{\sqrt{e}}I_0(1/2)+\frac{2}{\sqrt{e}}\sum_{m=1}^{+\infty}I_m(1/2)\cos(2m\theta),$$

$$ S = \frac{\pi}{6\sqrt{e}}I_0(1/2)+\frac{1}{\sqrt{e}}\sum_{m=1}^{+\infty}\frac{I_m(1/2)}{m}\sin\frac{\pi m}{3},$$

where $I_\alpha(\cdot)$ is the modified Bessel function of the first kind.
