# Computing $\sum _{k=1}^{\infty } \frac{\Gamma \left(\frac{k}{2}+1\right)}{k^2 \Gamma \left(\frac{k}{2}+\frac{3}{2}\right)}$ in closed form

What tools other than beta function you might like to use here?

$$\sum _{k=1}^{\infty } \frac{\displaystyle \Gamma \left(\frac{k}{2}+1\right)}{\displaystyle k^2 \Gamma \left(\frac{k}{2}+\frac{3}{2}\right)}\approx 1.27541$$

Supplementary question: calculating

$$\sum _{k=1}^{\infty } \frac{\displaystyle \Gamma \left(\frac{k}{2}+1\right)}{\displaystyle k^3 \Gamma \left(\frac{k}{2}+\frac{3}{2}\right)}\approx 1.02593$$ Is there a way to generalize it and get such a calculation?

$$\sum _{k=1}^{\infty } \frac{\displaystyle \Gamma \left(\frac{k}{2}+1\right)}{\displaystyle k^n \Gamma \left(\frac{k}{2}+\frac{3}{2}\right)}$$

• I would obviously start with the well-known identity $\displaystyle\bigg(\frac12\bigg){\large!}\cdot\sum_{k=\color{red}-1}^\infty \frac{\bigg(\dfrac k2\bigg)\large!}{\bigg(\dfrac{k+1}2\bigg)\large!}\cdot\sin^kx ~=~ \frac{\pi+2x}{\sin2x}$ – Lucian Jul 6 '15 at 21:33
• @Lucian thanks for the idea. – user 1357113 Jul 6 '15 at 21:42
• For $k$ even, let $k=2\ell$. Then, $$\frac{\Gamma(k/2+1)}{\Gamma(k/2+3/2)}=\frac{2^{\ell+1}\ell!}{(2\ell +1)!!\pi^{1/2}}$$ For $k$ odd, let $k=2m-1$. Then, $$\frac{\Gamma(k/2+1)}{\Gamma(k/2+3/2)}=\frac{(2m-1)!!\pi^{1/2}}{2^mm!}$$ I don't know if that helps at all. – Mark Viola Jul 6 '15 at 22:29

## 3 Answers

Writing $\dfrac{\Gamma\left(\frac{k}{2}+1\right)}{\Gamma\left(\frac{k}{2}+\frac{3}{2}\right)}$ as an integral and exchanging summation and integration, we get \begin{align} \sum^\infty_{k=1}\frac{\Gamma\left(\frac{k}{2}+1\right)}{k^2\Gamma\left(\frac{k}{2}+\frac{3}{2}\right)} &=\frac{2}{\sqrt{\pi}}\int^1_0\frac{x{\rm Li}_2(x)}{\sqrt{1-x^2}}\ {\rm d}x\tag1\\ &=-\frac{2}{\sqrt{\pi}}\int^1_0\frac{\sqrt{1-x^2}\ln(1-x)}{x}\ {\rm d}x\tag2\\ &=-\frac{1}{\sqrt{\pi}}\int^\pi_0\frac{\cos^2{x}\ln(1-\sin{x})}{\sin{x}}\ {\rm d}x\tag3\\ &=-\frac{1}{\sqrt{\pi}}\left[\sum^\infty_{n=1}\frac{(-1)^{n-1}}{n}\mathcal{A}_n+2\sum^\infty_{n=0}\frac{(-1)^{n-1}}{2n+1}\mathcal{B}_n\right]\tag4\\ \end{align} where \begin{align} \mathcal{A}_n =\int^\pi_0\frac{\cos^2{x}}{\sin{x}}(\cos(2nx)-1)\ {\rm d}x \ \ \ , \ \ \ \mathcal{B}_n =\int^\pi_0\frac{\cos^2{x}}{\sin{x}}\sin((2n+1)x)\ {\rm d}x\\ \end{align} Using simple trigonometric identities, it is not hard to see that, for $n\in\mathbb{N}$, \begin{align} \mathcal{A}_n-\mathcal{A}_{n-1}&=-\ \frac{1}{2n-3}-\frac{2}{2n-1}-\frac{1}{2n+1}\tag5\\ \mathcal{B}_n-\mathcal{B}_{n-1}&=0\tag6 \end{align} Thus we may obtain the closed forms for both sequences. \begin{align} \mathcal{A}_n=2H_n-4H_{2n}+\frac{1}{2n-1}-\frac{1}{2n+1}+2\ \ \ , \ \ \ \mathcal{B}_n=\pi-\frac{\pi}{2}\delta_{n0}\tag7 \end{align} Using the Taylor series for $\ln(1-x)$ and $\arctan{x}$, as well as the well-known identities \begin{align} \sum^\infty_{n=1}\frac{(-1)^{n-1}H_n}{n}&=\frac{\pi^2}{12}-\frac{1}{2}\ln^2{2}\tag8\\ \sum^\infty_{n=1}\frac{(-1)^{n-1}H_{2n}}{n}&=\frac{5\pi^2}{48}-\frac{1}{4}\ln^2{2}\tag9 \end{align} gives us \begin{align} \sum^\infty_{k=1}\frac{\Gamma\left(\frac{k}{2}+1\right)}{k^2\Gamma\left(\frac{k}{2}+\frac{3}{2}\right)} &=-\frac{1}{\sqrt{\pi}}\left[2\left(\frac{\pi^2}{12}-\frac{1}{2}\ln^2{2}\right)-4\left(\frac{5\pi^2}{48}-\frac{1}{4}\ln^2{2}\right)+2-2\pi\left(\frac{\pi}{4}\right)+2\left(\frac{\pi}{2}\right)\right]\\ &=\frac{3\pi^2-4\pi-8}{4\sqrt{\pi}}\tag{10} \end{align} as the closed form.

Explanation:
$(1)$: Write $\displaystyle\frac{\Gamma\left(\frac{k}{2}+1\right)}{\Gamma\left(\frac{k}{2}+\frac{3}{2}\right)}=\frac{2}{\sqrt{\pi}}\int^1_0\frac{x^{k+1}}{\sqrt{1-x^2}}\ {\rm d}x$.
$(2)$: Integrated by parts.
$(3)$: Substitute $x\mapsto\sin{x}$ then $x\mapsto\pi-x$.
$(4)$: Use the fact that $\ln(2-2\sin{x})=2\mathrm{Re}\ln(1+ie^{ix})$ then expand the $\mathrm{RHS}$.
$(5)$: Write $\displaystyle\mathcal{A}_n-\mathcal{A}_{n-1}=-\int^\pi_0(1+\cos(2x))\sin((2n-1)x)\ {\rm d}x$.
$(6)$: Write $\displaystyle\mathcal{B}_n-\mathcal{B}_{n-1}=\int^\pi_0(1+\cos(2x))\cos(2nx)\ {\rm d}x$.
$(7)$: Sum up $(5)$ and $(6)$.
$(8),(9)$: Let $z=-1$, $z=i$ in $\displaystyle\sum^\infty_{n=1}\frac{H_n}{n}z^n={\rm Li}_2(z)+\frac{1}{2}\ln^2(1-z)$.
$(10)$: Apply $(7)$, $(8)$, $(9)$ to $(4)$.

• ..wow. That was cool. – DaveNine Jul 8 '15 at 3:47
• This is nice but it's missing some pretty crucial steps that would be really nice to see. What exactly is the integral representation you used for the ratio of the gamma functions? I'm guessing to get the second line from the first, you did an integration by parts? – Cameron Williams Jul 8 '15 at 4:10
• Good job! Nice! (+1) – user 1357113 Jul 8 '15 at 6:58
• @Lucian now that I see it written out, it's pretty clear. I had a feeling that was the case but there were a lot of moving parts in those first few steps. – Cameron Williams Jul 8 '15 at 8:37
• I was just wondering if this way also works for $$\sum _{k=1}^{\infty } \frac{\Gamma \left(k-\frac{1}{2}\right)}{k^5 \Gamma (k)}$$ $$=-8 \sqrt{\pi } \zeta (3)+4 \sqrt{\pi } \zeta (3) \log (4)+32 \sqrt{\pi }-\frac{4 \pi ^{5/2}}{3}-\frac{\pi ^{9/2}}{20}+\frac{1}{12} \sqrt{\pi } \log ^4(4)-\frac{2}{3} \sqrt{\pi } \log ^3(4)-\frac{1}{6} \pi ^{5/2} \log ^2(4)+4 \sqrt{\pi } \log ^2(4)+\frac{2}{3} \pi ^{5/2} \log (4)-8 \sqrt{\pi } \log (16)$$ – user 1357113 Jul 8 '15 at 9:55

Let's write this as a sum of two hopefully simpler sums. $$\sum_{k=1}^\infty\frac{\Gamma(\frac k2+1)}{k^2\Gamma(\frac k2+\frac32)} =\sum_{k=1}^\infty\frac{\Gamma(k+\frac12)}{(2k-1)^2\Gamma(k+1)} +\sum_{k=1}^\infty\frac{\Gamma(k+1)}{4k^2\Gamma(k+\frac32)}\tag{1}$$ For integer $k$, we have $$\Gamma(k+\tfrac12)=\frac{\sqrt\pi}{4^k}\binom{2k}{k}k!\tag{2}$$ We will proceed by computing particular series.

The First Sum on the Right of $\boldsymbol{(1)}$

Identity $(2)$ says that $$\sum_{k=1}^\infty\frac{\Gamma(k+\frac12)}{(2k-1)^2\Gamma(k+1)} =\sqrt\pi\sum_{k=1}^\infty\frac{\binom{2k}{k}}{4^k(2k-1)^2}\tag{3}$$ Using the Extended Binomial Theorem, we get $$\sum_{k=0}^\infty\frac{\binom{2k}{k}}{4^k}x^{2k}=(1-x^2)^{-1/2}\tag{4}$$ Therefore, letting $x=\sin(u)$ and noting that $\int\frac{\mathrm{d}u}{1+\cos(u)}=\frac{\sin(u)}{1+\cos(u)}$, we get \begin{align} \sum_{k=1}^\infty\frac{\binom{2k}{k}}{4^k}\frac{x^{2k-1}}{2k-1} &=\int\frac{(1-x^2)^{-1/2}-1}{x^2}\,\mathrm{d}x\\ &=\int\frac{\mathrm{d}u}{1+\cos(u)}\\ &=\frac{\sin(u)}{1+\cos(u)}\\ &=\frac{x}{1+\sqrt{1-x^2}}\tag{5} \end{align} and \begin{align} \sum_{k=1}^\infty\frac{\binom{2k}{k}}{4^k}\frac{x^{2k-1}}{(2k-1)^2} &=\int\frac1{1+\sqrt{1-x^2}}\mathrm{d}x\\ &=\int\frac{\cos(u)}{1+\cos(u)}\,\mathrm{d}u\\ &=u-\frac{\sin(u)}{1+\cos(u)}\\ &=\sin^{-1}(x)-\frac{x}{1+\sqrt{1-x^2}}\tag{6} \end{align} Putting together $(3)$ and $(6)$ yields $$\sum_{k=1}^\infty\frac{\Gamma(k+\frac12)}{(2k-1)^2\Gamma(k+1)}=\sqrt\pi\left(\frac\pi2-1\right)\tag{7}$$

The Second Sum on the Right of $\boldsymbol{(1)}$

Identity $(2)$ says that $$\sum_{k=1}^\infty\frac{\Gamma(k+1)}{4k^2\Gamma(k+\frac32)} =\frac1{2\sqrt\pi}\sum_{k=1}^\infty\frac{4^k}{k^2(2k+1)\binom{2k}{k}}\tag{8}$$ Equation $(2)$ from this answer says that $$\sum_{k=0}^\infty\frac{4^kx^{2k}}{\binom{2k}{k}} =\frac1{1-x^2}\left[1+\sqrt{\frac{x^2}{1-x^2}}\sin^{-1}(x)\right]\tag{9}$$ Therefore, letting $x=\sin(u)$, \begin{align} \sum_{k=0}^\infty\frac{4^kx^{2k+1}}{(2k+1)\binom{2k}{k}} &=\int\frac1{1-x^2}\left[1+\sqrt{\frac{x^2}{1-x^2}}\sin^{-1}(x)\right]\,\mathrm{d}x\\ &=\int\sec(u)(1+\tan(u)u)\,\mathrm{d}u\\ &=u\sec(u)\\ &=\frac{\sin^{-1}(x)}{\sqrt{1-x^2}}\tag{10} \end{align} and \begin{align} \sum_{k=1}^\infty\frac{4^kx^{2k}}{2k(2k+1)\binom{2k}{k}} &=\int\frac{\sin^{-1}(x)-x\sqrt{1-x^2}}{x^2\sqrt{1-x^2}}\,\mathrm{d}x\\ &=\int\frac{u-\sin(u)\cos(u)}{\sin^2(u)}\,\mathrm{d}u\\ &=1-u\cot(u)\\ &=1-\frac{\sin^{-1}(x)\sqrt{1-x^2}}{x}\tag{11} \end{align} and \begin{align} \sum_{k=1}^\infty\frac{4^kx^{2k}}{4k^2(2k+1)\binom{2k}{k}} &=\int\frac{x-\sin^{-1}(x)\sqrt{1-x^2}}{x^2}\,\mathrm{d}x\\ &=\int\frac{\sin(u)\cos(u)-u\cos^2(u)}{\sin^2(u)}\,\mathrm{d}u\\ &=\frac12u^2+u\cot(u)-1\\ &=\frac12\sin^{-1}(x)^2+\frac{\sin^{-1}(x)\sqrt{1-x^2}}{x}-1\tag{12} \end{align} Combining $(8)$ and $(12)$ yields $$\sum_{k=1}^\infty\frac{\Gamma(k+1)}{4k^2\Gamma(k+\frac32)} =\frac2{\sqrt\pi}\left(\frac{\pi^2}8-1\right)\tag{13}$$

Putting together $(1)$, $(7)$, and $(13)$ gives $$\bbox[5px,border:2px solid #C0A000]{\sum_{k=1}^\infty\frac{\Gamma(\frac k2+1)}{k^2\Gamma(\frac k2+\frac32)} =\frac1{4\sqrt\pi}\left(3\pi^2-4\pi-8\right)}\tag{14}$$

Using some results from my other answer, we can compute the second sum requested.

Similar to my other answer, write $$\sum_{k=1}^\infty\frac{\Gamma(\frac k2+1)}{k^3\Gamma(\frac k2+\frac32)} =\sum_{k=1}^\infty\frac{\Gamma(k+\frac12)}{(2k-1)^3\Gamma(k+1)} +\sum_{k=1}^\infty\frac{\Gamma(k+1)}{8k^3\Gamma(k+\frac32)}\tag{1}$$

First Sum on the Right of $\boldsymbol{(1)}$

Using $(2)$ from my other answer, we get $$\sum_{k=1}^\infty\frac{\Gamma(k+\frac12)}{(2k-1)^3\Gamma(k+1)} =\sqrt\pi\sum_{k=1}^\infty\frac{\binom{2k}{k}}{4^k(2k-1)^3}\tag{2}$$

Starting from $(6)$ in my other answer, we get \begin{align} \sum_{k=1}^\infty\frac{\binom{2k}{k}}{4^k(2k-1)^3} &=\int_0^1\left(\frac{\sin^{-1}(x)}x-\frac1{1+\sqrt{1-x^2}}\right)\,\mathrm{d}x\\ &=\int_0^{\pi/2}\left(u\cot(u)-\frac{\cos(u)}{1+\cos(u)}\right)\,\mathrm{d}u\\ &=\left[u\log(\sin(u))\vphantom{\int}\right]_0^{\pi/2}-\int_0^{\pi/2}\log(\sin(u))\,\mathrm{d}u-\left[u-\frac{\sin(u)}{1+\cos(u)}\vphantom{\int}\right]_0^{\pi/2}\\ &=\frac\pi2\log(2)+1-\frac\pi2\tag{3} \end{align} Combining $(2)$ and $(3)$, $$\sum_{k=1}^\infty\frac{\Gamma(k+\frac12)}{(2k-1)^3\Gamma(k+1)} =\sqrt\pi\left(\frac\pi2\log(2)+1-\frac\pi2\right)\tag{4}$$

Second Sum on the Right of $\boldsymbol{(1)}$

Using $(2)$ from my other answer, we get $$\sum_{k=1}^\infty\frac{\Gamma(k+1)}{8k^3\Gamma(k+\frac32)} =\frac2{\sqrt\pi}\sum_{k=1}^\infty\frac{4^k}{8k^3(2k+1)\binom{2k}{k}}\tag{5}$$ Starting from $(12)$ in my other answer, we get \begin{align} \sum_{k=1}^\infty\frac{4^k}{8k^3(2k+1)\binom{2k}{k}} &=\int_0^1\left(\frac{\sin^{-1}(x)^2}{2x}-\frac{x-\sin^{-1}(x)\sqrt{1-x^2}}{x^2}\right)\,\mathrm{d}x\\ &=\int_0^{\pi/2}\left(\frac{u^2\cos(u)}{2\sin(u)}-\frac{\sin(u)\cos(u)-u\cos^2(u)}{\sin^2(u)}\right)\,\mathrm{d}u\\ &=\small\left[\frac{u^2}2\log(\sin(u))\right]_0^{\pi/2}-\int_0^{\pi/2}u\log(\sin(u))\,\mathrm{d}u-\left[\frac12u^2+u\cot(u)-1\right]_0^{\pi/2}\\ &=\frac{\pi^2}8\log(2)-\frac7{16}\zeta(3)-\frac{\pi^2}8+1\tag{6} \end{align} Combining $(5)$ and $(6)$, $$\sum_{k=1}^\infty\frac{\Gamma(k+1)}{8k^3\Gamma(k+\frac32)} =\frac1{8\sqrt\pi}\left(2\pi^2(\log(2)-1)-7\zeta(3)+16\right)\tag{7}$$

Putting together $(1)$, $(4)$, and $(7)$ yields $$\bbox[5px,border:2px solid #C0A000]{\sum_{k=1}^\infty\frac{\Gamma(\frac k2+1)}{k^3\Gamma(\frac k2+\frac32)} =\frac1{8\sqrt\pi}\left(6\pi^2(\log(2)-1)-7\zeta(3)+8\pi+16\right)}\tag{8}$$

Integrals Involving $\boldsymbol{\log(\sin(u))}$ Used Above

As shown in $(1)$ from this answer: $$\log(\sin(u))=-\log(2)+\sum_{k=1}^\infty(-1)^{k-1}\frac{\cos(2ku)}{k}\tag{9}$$ For $k\in\mathbb{Z}$ and $k\ne0$, we have $$\int_0^{\pi/2}\cos(2ku)\,\mathrm{d}u=0\tag{10}$$ and integration by parts gives \begin{align} \int_0^{\pi/2}u\cos(2ku)\,\mathrm{d}u &=-\frac1{2k}\int_0^{\pi/2}\sin(2ku)\,\mathrm{d}u\\ &=\frac1{4k^2}(1-\cos(k\pi))\\ &=\frac{[k\text{ is odd}]}{2k^2}\tag{11} \end{align} It follows that $$\int_0^{\pi/2}\log(\sin(u))\,\mathrm{d}u=-\frac\pi2\log(2)\tag{12}$$ and $$\int_0^{\pi/2}u\log(\sin(u))\,\mathrm{d}u=-\frac{\pi^2}8\log(2)+\frac7{16}\zeta(3)\tag{13}$$

• Clever as the previous solution (+1) – user 1357113 Jul 13 '15 at 16:30
• Impressive +1!) I really wasn't expecting this one to be solved because of how similar it is to this [long standing MSE question](math.stackexchange.com/questions/937912/… (one of my favorite problems of this entire site actually). Do you think your methods might shed any new light on that problem? – David H Jul 13 '15 at 16:48
• @DavidH: I have tried and made some progress, but I encountered integrals whose closed form involves $\mathrm{Li}_2$. I think that one may have a pretty complex answer, even using more esoteric special functions. – robjohn Jul 14 '15 at 5:12