# Series seems to converge extremely slow, but possibly towards 1. Does it possibly have a closed form?

(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 $s(4)=1$ ?

• Hint: This is merely another play on the lame old, same old $\displaystyle\sum_{n=1}^\infty\frac{(2x)^{2n}}{n^2~\displaystyle{2n\choose n}}=2\arcsin^2x$. Notice that $\displaystyle{2n+3\choose2}$ is a humble polynomial in n. – Lucian Sep 5 '15 at 21:06
• @Lucian 's comment tells pretty everything. – Jack D'Aurizio Sep 5 '15 at 21:08

$$s(x)=\frac{4}{x}-\frac{4\sqrt{4-x}}{x^{3/2}}\,\arcsin\left(\frac{\sqrt{x}}{2}\right) \tag{1}$$ hence $$\lim_{x\to 4^-} s(x) = \color{red}{1} \tag{2}$$ as conjectured.
• Ah, and after some simplification and substituting $y=4/x$ I get, feeding wolframalpha: the above series when expanded around $y=\infty$... I surely should get more experience with that "around infinity" thing... – Gottfried Helms Sep 5 '15 at 21:19