Proof of closed-form solution of the difference of two factorial series Context
I'm working on a problem tangentially related to the Kepler Equation 1. The details are very much in the weeds, and I'm not in a position to explain at this time exactly how I have arrived at Equation 1. Yet, I believe that the following holds true:
$$      \lim_{k\rightarrow \infty} \sum\limits_{s=0}^{ k- 1 } \,
        \dfrac{
        \left[  2^{ 2\,(k-  s)    } \left[
   ( k-  s)    !\right]^4
      - \pi\,2^{2(s-k) -1 } 
\,
 \left[   2\,(k - s)]!\right]^2    [ 2\,(k -s) ] 
     \right]
     }{[2(k-  s)]![2\,(k-  s) ]\,\left[  (  k - s)   !  \right]^2}
=
              \dfrac{ \pi   }{2 } - 1
 ~\text{Eq}.~1.$$
I've plotted Equation 1 for various $k$. My results seem to indicate that the expression above is plausibly true. Beyond $k=50$ I run into floating point issues in the numerical calculation, and series is not computable. 
I cannot figure out how to determine the veracity of the equation 1.  I've seen some closed-form solutions to factorial series (e.g., [2]).[] Yet, I have not seen such an expression elsewhere.
Questions


*

*Does anyone have one or more references to a book that has many factorial series?

*Can anyone prove Equation 1 true or false?

*Can anyone illustrate the results for $k >> 50$? 
Bibliography
1 Find the inverse of an equation reminiscent of Kepler's equation
[2] http://mathworld.wolfram.com/FactorialSums.html
 A: Here is the sum evaluated to 500 as well as a plot of the difference between successive terms, FYI.  
HTH


A: Rewriting,
$\begin{array}\\
f(k)
&=\sum\limits_{s=1}^{ k } \,
\dfrac{2^{2s}(s!)^4
- \pi\,2^{-2s-1} 
 (2s)!^2(2s)}
{2s(2s)!(s!)^2}\\
&=\sum\limits_{s=1}^{ k } \,
\dfrac{2^{2s}(s!)^4
}
{2s(2s)!(s!)^2}
-\pi\sum\limits_{s=1}^{ k } \,
\dfrac{2^{-2s-1} 
 (2s)!^2(2s)}
{2s(2s)!(s!)^2}\\
&=\sum\limits_{s=1}^{ k } \,
\dfrac{2^{2s-1}(s!)^2
}
{s(2s)!}
-\pi\sum\limits_{s=1}^{ k } \,
\dfrac{2^{-2s-1} 
 (2s)!}
{(s!)^2}\\
&=f_1(k)-\pi f_2(k)\\
\lim_{k \to \infty} f(k)
&=\dfrac{\pi}{2 } - 1
 ~\text{Eq}.~1.\\
\end{array}
$
According to Wolfy,
$f_1(k)
= \dfrac{(2^{2 k + 1} (2 k + 1) ((k + 1)!)^2)}{((k + 1) (2 (k + 1))!)} - 1
$
and
$f_2(k)
= \dfrac14\dfrac{(4^{-k} (k + 1) (2 (k + 1))!}{(k + 1)!^2} -\dfrac12
$.
Now,
take limits
(left to you).
A: Starting from @marty cohen's answer and fimplifying, we have
$$f_1(k)=\sqrt{\pi }\,\frac{ \Gamma (k+1)}{\Gamma \left(k+\frac{1}{2}\right)}-1$$
$$f_2(k)=\frac{\Gamma \left(k+\frac{3}{2}\right)}{\sqrt{\pi }\, \Gamma (k+1)}-\frac{1}{2}$$
$$f_1(k)-\pi f_2(k)=\frac \pi 2-1+\sqrt \pi\left(\frac{\Gamma (k+1)}{\Gamma \left(k+\frac{1}{2}\right)}-\frac{\Gamma
   \left(k+\frac{3}{2}\right)}{\Gamma (k+1)} \right)$$
Now, using Stirling approximation and continuing with Taylor series for large values of $k$
$$\log \left(\frac{\Gamma (k+1)}{\Gamma \left(k+\frac{1}{2}\right)}\right)=\frac{1}{2} \log \left({k}\right)+\frac{1}{8 k}-\frac{1}{192
   k^3}+O\left(\frac{1}{k^5}\right)$$
$$\frac{\Gamma (k+1)}{\Gamma \left(k+\frac{1}{2}\right)}=t+\frac{1}{8 t}+\frac{1}{128 t^3}-\frac{5}{1024
   t^5}+O\left(\frac{1}{t^7}\right)$$ where $\color{red}{t=\sqrt k}$.
Similarly
$$\log \left(\frac{\Gamma \left(k+\frac{3}{2}\right)}{\Gamma (k+1)}\right)=\frac{1}{2} \log \left({k}\right)+\frac{3}{8 k}-\frac{1}{8
   k^2}+\frac{3}{64 k^3}-\frac{1}{64k^4}+O\left(\frac{1}{k^5}\right)$$
$$\frac{\Gamma
   \left(k+\frac{3}{2}\right)}{\Gamma (k+1)}=t+\frac{3}{8 t}-\frac{7}{128 t^3}+\frac{9}{1024 t^5}+O\left(\frac{1}{t^7}\right)$$
$$f_1(k)-\pi f_2(k)=\frac \pi 2-1+\sqrt \pi\left(-\frac{1}{4 t}+\frac{1}{16 t^3}-\frac{7}{512t^5}+O\left(\frac{1}{t^7}\right)\right)$$
Computing for $k=10$, the exact value is
$$f_1(10)-\pi f_2(10)=\frac \pi 2-1+\frac{215955}{46189}-\frac{707825 \pi }{524288}\approx 0.4340968$$ while the above expansion gives
$$\frac{\pi }{2}-1-\frac{12487 }{51200}\sqrt{\frac{\pi }{10}}\approx 0.4340980$$
With regard to @user3113647's plots, we then have
$$S(k)-S(k-1)=\frac{1}{8} \sqrt{\pi }
   \left(\frac{1}{k}\right)^{3/2}+O\left(\frac{1}{k^{7/2}}\right)$$ and then a slope of $-\frac 32$ in the logarithmic scale.
