Evaluating the series $\displaystyle{\sum_{n=1}^\infty\binom{n}{n/2}\frac{(-1)^{n}}{n(2a)^n}}$

Question

I encountered the following series while evaluating an integral

\begin{equation} \sum_{n=1}^\infty\binom{n}{n/2}\frac{(-1)^{n}}{n(2a)^n}, \quad a\in\mathbb{R} \end{equation}

and am looking for a closed form. I have looked up generating functions and found some close candidates such as,

\begin{equation} \sum_{n=1}^\infty\binom{2n}{n}\frac{x^n}{4^nn} = 2\log\left(\frac{2}{1+\sqrt{1-x}}\right) \end{equation}

but I can't figure out how to deal with the $\binom{n}{n/2}$ term. Looking at the power series for $(1+x)^n$, I assume finding such a function may be difficult, but I hope a closed form exists.

I will be happy to add my evaluation of the original integral up to the point where the series appears, if that may be necessary.

Thank you in advance.

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Edit:

Here is the integral I was evaluating, $$J(a)=\int_0^{\pi/2}\log\left(1+\frac{\sin x}{a}\right)\ dx,\quad a\in\mathbb{R}$$ which comes from this post, and the derivations of the series are in my answer there. I hope this helps @ClaudeLeibovici.

Answer 1

Let $x=-\frac 1 {2a}$ and consider $$S(x)=\sum_{n=1}^\infty \binom{n}{\frac{n}{2}}\frac{x^n}{n} $$ $$S'(x)=\sum_{n=1}^\infty \binom{n}{\frac{n}{2}}x^{n-1}=\frac{1}{x \sqrt{1-4 x^2}}-\frac{1}{x}+\frac{2 \sin ^{-1}(2 x)}{\pi x \sqrt{1-4 x^2}}$$ Integrating (do not forget the integration constant which is $\log(2)$) $$S(x)=\log \left(\frac{1-\sqrt{1-4 x^2}}{2 x^2}\right)+\frac 2 \pi \int \frac{\sin ^{-1}(2 x)}{ x \sqrt{1-4 x^2}}\,dx$$ Now, the problem is $$I=\int \frac{\sin ^{-1}(2 x)}{ x \sqrt{1-4 x^2}}\,dx$$ Integrating by parts $$I=-\sin ^{-1}(2 x) \tanh ^{-1}\left(\sqrt{1-4 x^2}\right)+2\int \frac{\tanh ^{-1}\left(\sqrt{1-4 x^2}\right)}{\sqrt{1-4 x^2}}\,dx$$ $$J=\int \frac{\tanh ^{-1}\left(\sqrt{1-4 x^2}\right)}{\sqrt{1-4 x^2}}\,dx$$ Using $t=\tanh ^{-1}\left(\sqrt{1-4 x^2}\right)$ $$J=-i\frac{x}{\sqrt{x^2}}\left(t\, \log \left(\frac{e^t-i}{e^t+i}\right)+\text{Li}_2\left(-i e^{-t}\right)-\text{Li}_2\left(i e^{-t}\right) \right)$$

Edit

From a formal point of view, the best expression is $$S(x)=\log \left(\frac{1-\sqrt{1-4 x^2}}{2 x^2}\right)+\frac {4x} \pi \,\, _3F_2\left(\frac{1}{2},1,1;\frac{3}{2},\frac{3}{2};4 x^2\right)$$ which has been obtained separating the odd and even terms.

So, the sum is title is $$\large\color{blue}{\log \left(2 a \left(a-\sqrt{a^2-1}\right)\right)-\frac{2}{a\pi }\,\, _3F_2\left(\frac{1}{2},1,1;\frac{3}{2},\frac{3}{2};\frac{1}{a^2}\right)}$$

Answer 2

Replacing the binomial coefficients by gamma functions, making the problem more general, at least for positive integer values of $k$, $$S_k(x)=\sum_{n=1}^\infty \binom{n}{\frac{n}{2}}\frac{x^n}{n^k}$$ write as the sum of two hypergeometric functions $$\color{blue}{S_2(x)=\frac{x^2}{2} \, _4F_3\left(1,1,1,\frac{3}{2};2,2,2;4 x^2\right)+}$$ $$\color{blue}{\frac {4x}\pi \, _4F_3\left(\frac{1}{2},\frac{1}{2},1,1;\frac{3}{2},\frac{3}{2 },\frac{3}{2};4 x^2\right)}$$ $$S_3(x)=\frac{x^2}{4} \, _5F_4\left(1,1,1,1,\frac{3}{2};2,2,2,2;4 x^2\right)+$$ $$\frac {4x}\pi \, _5F_4\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},1,1;\frac{3}{2 },\frac{3}{2},\frac{3}{2},\frac{3}{2};4 x^2\right)$$ $$\color{blue}{S_4(x)=\frac{x^2}{8} \, _6F_5\left(1,1,1,1,1,\frac{3}{2};2,2,2,2,2;4 x^2\right)+}$$ $$\color{blue}{\frac {4x}\pi \, _6F_5\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},1, 1;\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2} ;4 x^2\right)}$$ and we probably could conjecture what is $S_k(x)$ looking at the very simple patterns.