(This is partly recreational math, playing with this question originally looking for a useful and fast converging representation for $\pi$ ).
A bit aside of fiddling with series which approximate $\pi$ I came to that following variant
$$ s(x) = \sum_{k=0}^\infty {x^k\over \binom{2k+1}{k} \cdot \binom{2k+3}{2}} $$
which comes out to be related to $\pi$ (by 100 and more digits precision) in the following way:
$$ \begin{array} {}
s(1) &= \frac 41(1- { \pi \over 2 \sqrt 3}) \\
s(2) &=\frac 42(1- {\pi \over 4}) &=2 - { \pi \over 2} \\
s(3) &= \frac 43(1 - { \pi \over 3 \sqrt 3 } )\\
\end{array} $$
which looks somehow intriguing for me.
Background: I arrived at this series, when I invented a series for computation of $\pi$ by collecting partial expressions from the Leibniz-series, coming to $$ \pi=4 - 2( {0! \over 3!!}+{ 1!\over5!! }+{2! \over 7!! }+{ 3! \over 9!! } + ...) $$ (which is of course well known, accordingly to mathworld, even since Euler).
To have this series more smooth I converted the factorial and doublefactorial expressions into binomial-expressions, getting increasing powers of 2 in the numerators which gave me the idea to define this as $s(2)$ and to play with other $x$, giving a nice value for $s(3)$. (But I've nothing so far for, say $s(2.5)$ etc.)
For $s(4)$ the series converges very slowly, I summed up to 140 000 and to 140 200 terms and get the results
%3078 = 0.99763 1473716 \\ 140 000 terms
%3079 = 0.99763 3163689 \\ 140 200 terms
and I guess, that it might eventually run to 1.0 . Because it seems that there will be no index $k$ from where the rate of convergence improves, much likely an analytical approach is required. So my question:
Q: What would be an analytical expression for the sum? Or could we improve the serial summation with some convergence-acceleration? And of course: is the final value $s(4)=1$ ?
