Generating Function $\sum_{k=1}^{\infty}\binom{2k}{k}^{-2}x^{k}$

Question

Closed Form For : $$S=\sum_{k=1}^{\infty}\binom{2k}{k}^{-2}x^{k}$$

Using the Series Expansion for $\arcsin^2(x)$ one can arrive at : $$\sum_{k=0}^{\infty}\binom{2k}{k}^{-1}x^{k}=\frac{4}{4-x}-4\arcsin\left(\frac{\sqrt{x}}{2}\right)\frac{1}{\sqrt{x\left(4-x\right)}}+16\arcsin\left(\frac{\sqrt{x}}{2}\right)\frac{1}{\left(4-x\right)\sqrt{x\left(4-x\right)}}\tag{1}$$

We need to add one more Binomial Term in the Denominator and I only know of two such Integrals (though there might be others?), The Beta Integral and the Odd Powers of $\sin$.
Here I will try to use the former.

$$\frac{r}{2}\int_{0}^{1}t^{r-1}\left(1-t\right)^{r-1}dt=\binom{2r}{r}^{-1}$$

Then replacing $x\to t(1-t)x$ and Integrating from $0$ to $1$ leads to (after some more algebraic manipulations) : $$S=\frac{x}{16-x}+16\frac{x\arctan\left(\sqrt{\frac{x}{16-x}}\right)}{\left(16-x\right)\sqrt{x\left(16-x\right)}}+2x\frac{d}{dx}\sqrt{x}\int_{0}^{1}\left(\frac{\arcsin\left(\frac{\sqrt{\left(t\right)\left(1-t\right)x}}{2}\right)}{\left(4-\left(t\right)\left(1-t\right)x\right)\sqrt{\left(t\right)\left(1-t\right)\left(4-\left(t\right)\left(1-t\right)x\right)}}\right)dt$$

Though I haven't had any luck in Calculating the Integral present above.
The variation of the Integral with only $t$ instead of $t(1-t)$ can be done :

$$\int_{0}^{1}\left(\frac{\arcsin\left(\frac{\sqrt{tx}}{2}\right)}{\left(4-tx\right)\sqrt{t\left(4-tx\right)}}\right)dt=\frac{\ln\left(1-\frac{x}{4}\right)}{4\sqrt{x}}+\frac{\arcsin\left(\frac{\sqrt{x}}{2}\right)}{2\sqrt{4-x}}$$

Wolfram Alpha says "no results found in terms of standard mathematical functions".
So is it not possible to solve this integral?

Although I believe this question might have already been asked before as its the Ordinary Generating Function for Inverse Of Central Binomial Coefficient Squared but was not able to find it.

Answer 1

This is not an answer.

You can simplify the first summation as $$\sum_{k=0}^{\infty}\binom{2k}{k}^{-1}x^{k}=\frac{x}{4-x}+\frac{4 \sqrt{x} }{(4-x)^{3/2}}\sin ^{-1}\left(\frac{\sqrt{x}}{2}\right)$$ But, as soon as you consider things such as $$\sum_{k=0}^{\infty}\binom{2k}{k}^{-n}x^{k}$$ you face (as usual) hypergeometric functions $$\sum_{k=1}^{\infty}\binom{2k}{k}^{-2}x^{k}=\frac{x}{4} \, \, _3F_2\left(1,2,2;\frac{3}{2},\frac{3}{2};\frac{x}{16}\right)$$

The integral is a nightmare !

Answer 2

I will provide the derivation of the hypergeometric form for the series, so you know the general method.

Let's call the common term $A_k$.

Then we have:

$$A_0 = 1$$

$$\frac{A_{k+1}}{A_k} = \frac{(k+1) \binom{2k}{k}^2}{\binom{2k+2}{k+1}^2} \frac{x}{k+1}$$

Using the definition in terms of factorials, we have:

$$\binom{2k}{k}= \frac{(2k)!}{k!^2}$$

Which gives us:

$$\frac{A_{k+1}}{A_k} = \frac{(k+1) (2k)!^2 (k+1)!^4}{(2k+2)!^2 k!^4} \frac{x}{k+1}$$

$$\frac{A_{k+1}}{A_k} = \frac{(k+1)^5}{(2k+2)^2 (2k+1)^2} \frac{x}{k+1}$$

$$\frac{A_{k+1}}{A_k} = \frac{(k+1)^5}{(k+1)^2 (k+1/2)^2} \frac{x/16}{k+1}$$

$$\frac{A_{k+1}}{A_k} = \frac{(k+1)^3}{(k+1/2)^2} \frac{x/16}{k+1}$$

By definition this means:

$$\sum_{k=0}^\infty \binom{2k}{k}^{-2} x^k = {_3 F_2} \left(1,1,1; \frac12, \frac12; \frac{x}{16} \right)$$

If we start with $k=1$, then:

$$\sum_{k=1}^\infty \binom{2k}{k}^{-2} x^k = {_3 F_2} \left(1,1,1; \frac12, \frac12; \frac{x}{16} \right) - 1$$