Evaluation of tricky Gamma infinite sum

Question

I want to prove that: $$\sum_{n=0}^\infty\frac{n}{(n+1)^2}\frac{\Gamma\left(\frac{n}{2}\right)}{\Gamma\left(\frac{n+1}{2}\right)}=\frac{4G}{\sqrt{\pi}}+\sqrt{\pi}\log2$$ and $$\sum_{n=0}^\infty(-1)^n\frac{n}{(n+1)^2}\frac{\Gamma\left(\frac{n}{2}\right)}{\Gamma\left(\frac{n+1}{2}\right)}=\frac{4G}{\sqrt{\pi}}-\sqrt{\pi}\log2$$ where $G$ is Catalan's constant. I got these results, evaluating some random integrals in various ways and manipulating the results.

Now, my question is as follows:

is there a way to get the two results starting off from the series, without knowing the original integral, just using some algebraic manipulations and Gamma identities?

I tried doing this myself, because WolframAlpha was able to compute it, so I thought it could be done. Turns out this was not the case, as I got stuck before starting. My idea was turning Gammas into factorials, and then work with sums of factorials, but looking at the structure of the two Gammas in each sum, only one of them at a time can have an integer argument.

However, turns out $\Gamma(k+\frac12)$ has a nice structure as well, the problem is that in each sum, the integer argument in the gamma function is alternating between numerator and denominator, so things get ugly and messy splitting the sums in even and odd, and I doubt this is the right approach. I can't think of anything else though. Suggestions are appreciated.

I won't post the original integral, in order to prevent answers that start off from it. I'm looking for answers provided as we didn't even know the result. Any hints?

EDIT:

Splitting the first sum into even end odd terms, we simplify the problem to just proving this two results: $$\sum_{n=0}^\infty \frac{4^n}{(2n+1)^2}\frac{1}{{2n\choose n}}=2G$$ and $$\sum_{n=0}^\infty\frac{2n+1}{(n+1)^2}{2n\choose n}\frac{1}{4^n}=4\log2$$ and these would prove the full result. Still don't know how to do it though.

Answer 1

I should consider $$f(x)=\sum_{n=0}^\infty\frac{n}{(n+1)^2}\frac{\Gamma\left(\frac{n}{2}\right)}{\Gamma\left(\frac{n+1}{2}\right)}x^n=\sum_{n=0}^\infty a_n\,x^n$$ Expand and separate odd and even terms, we have $$S_1(x)=\sum_{n=0}^\infty a_{2n}\,x^{2 n}=\frac{2}{\sqrt{\pi }}+\frac{4 x^2 }{9 \sqrt{\pi }}\, _3F_2\left(1,\frac{3}{2},2;\frac{5}{2},\frac{5}{2};x^2 \right)$$ $$S_2(x)=\sum_{n=0}^\infty a_{2n+1}\,x^{2 n+1}=\frac{\sqrt{\pi }}{x}\left(\log (2)-\log \left(1+\sqrt{1-x^2}\right)\right)$$

Making $x=\pm 1$ to cover the two cases

$$S_1(\pm 1)=\frac{4 C}{\sqrt{\pi }}\qquad \text{and} \qquad S_2(\pm 1)=\pm \sqrt{\pi } \log(2)$$ and hence the results.

Edit (to clarify)

Write $$a_{2n}=\frac{2}{\sqrt{\pi }}\frac {b_n}{c_n}$$ The $b_n$ $$\{1,2,8,16,128,256,1024,2048,32768,65536\} $$ are the denominators of $$\frac 1 {4^n} \binom{2 n}{n}$$ and the $c_n$ $$\{1,9,75,245,2835,7623,39039,96525,1859715,4387955\}$$ are the denominators of the series expansion of $$\, _3F_2\left(\frac{1}{2},1,1;\frac{3}{2},\frac{3}{2};x \right)$$

Write $$a_{2n+1}=\frac{\sqrt{\pi }}4\frac {d_n}{e_n}$$

The $d_n$ $$\{1,3,5,35,63,77,429,6435,12155,46189\}$$ are the numerators of the series expansion of $$\cosh ^{-1}(x)-\log (2 x)$$ and the $e_n$ $$\{1,8,24,256,640,1024,7168,131072,294912,1310720\}$$ had to be worked !

Answer 2

We can evaluate the first sum by

$$\begin{align*} & \sum_{n=0}^\infty \frac{n}{(n+1)^2}\frac{\Gamma\left(\frac{n}{2}\right)}{\Gamma\left(\frac{n+1}{2}\right)} x^n \\ &= \frac2{\sqrt\pi} + \sum_{n=1}^\infty \frac{2n}{(2n+1)^2}\frac{\Gamma(n)}{\Gamma\left(n+\frac12\right)} x^{2n} + \frac14 \sum_{n=0}^\infty \frac{2n+1}{(n+1)^2}\frac{\Gamma\left(n+\frac12\right)}{\Gamma(n+1)} x^{2n+1} \\ &= \frac2{\sqrt\pi} + \frac14 \lim_{x\to1^-} \left[4 \color{red}{F(x)} + \color{blue}{G(x)}\right] \\ &= \frac2{\sqrt\pi} \int_0^1 \frac{\arcsin y}{y\sqrt{1-y^2}} \, dy = \boxed{\frac{4G}{\sqrt\pi} + \sqrt\pi\,\log2} \end{align*}$$

and we get the alternating version for free since the only difference is the sign of the latter series in the second line.

We use the same ideas as in this answer to establish some preliminary results:

$$\begin{array}{rll} f(x) &= \displaystyle \sum_{n=1}^\infty \frac{\Gamma(n)}{\Gamma\left(n+\frac12\right)} x^{2n} &= \displaystyle \frac{2x \arcsin x}{\sqrt\pi \sqrt{1-x^2}} \\ x f'(x) &= \displaystyle \sum_{n=1}^\infty 2n \frac{\Gamma(n)}{\Gamma\left(n+\frac12\right)} x^{2n} &= \displaystyle \frac2{\sqrt\pi} \left(\frac{x^2}{1-x^2} + \frac{x \arcsin x}{\left(1-x^2\right)^{3/2}}\right) \\ \displaystyle \frac1x \int x f'(x) \, dx &= \displaystyle \sum_{n=1}^\infty \frac{2n}{2n+1} \frac{\Gamma(n)}{\Gamma\left(n+\frac12\right)} x^{2n} &= \displaystyle \frac2{\sqrt\pi} \left(\frac{\arcsin x}{x \sqrt{1-x^2}} - 1\right) \\ \displaystyle \int \frac1x \left[\int x f'(x) \, dx\right] \, dx &= \displaystyle \sum_{n=1}^\infty \frac{2n}{(2n+1)^2} \frac{\Gamma(n)}{\Gamma\left(n+\frac12\right)} x^{2n+1} &= \color{red}{\displaystyle \frac2{\sqrt\pi} \left(-x + \int_0^x \frac{\arcsin y}{y \sqrt{1-y^2}} \, dy\right) =: F(x)} \\[2ex] \hline g(x) &= \displaystyle \sum_{n=0}^\infty \frac{\Gamma\left(n+\frac12\right)}{\Gamma(n+1)} x^{2n+1} &= \displaystyle \frac{\sqrt\pi\,x}{\sqrt{1-x^2}} \\ g'(x) &= \displaystyle \sum_{n=0}^\infty (2n+1) \frac{\Gamma\left(n+\frac12\right)}{\Gamma(n+1)} x^{2n} &= \displaystyle \frac{\sqrt\pi}{\left(1-x^2\right)^{3/2}} \\ \displaystyle \frac1x \int g'\left(\sqrt x\right) \, dx &= \displaystyle \sum_{n=0}^\infty \frac{2n+1}{n+1} \frac{\Gamma\left(n+\frac12\right)}{\Gamma(n+1)} x^n &= \displaystyle \frac{2\sqrt\pi}{x\sqrt{1-x}} \\ \displaystyle \int \frac1x \left[\int g'\left(\sqrt x\right) \, dx\right] \, dx &= \displaystyle \sum_{n=0}^\infty \frac{2n+1}{(n+1)^2} \frac{\Gamma\left(n+\frac12\right)}{\Gamma(n+1)} x^{n+1} &= \color{blue}{\displaystyle 4\sqrt\pi \log\frac{2}{1+\sqrt{1-x}} =: G(x)} \end{array}$$