A closed form of $\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k!}\Gamma^2\left(\frac{k}{2}\right)$

I am looking for a closed form of the following series

\begin{equation} \mathcal{I}=\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k!}\Gamma^2\left(\frac{k}{2}\right) \end{equation}

I have no idea how to answer this question. Wolfram Alpha gives me result:

$$\mathcal{I}\approx2.7415567780803776$$

Could anyone here please help me to obtain the closed form of the series preferably (if possible) with elementary ways (high school methods)? Any help would be greatly appreciated. Thank you.

Solution 1:

You can use the identity given by the Euler Beta function $$\int_{0}^{1}x^{a-1} (1-x)^{b-1} \,dx=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$ to state: $$S=\sum_{k=1}^{+\infty}\frac{(-1)^{k+1}}{k!}\Gamma(k/2)^2=\sum_{k=1}^{+\infty}\frac{(-1)^{k-1}}{k}\int_{0}^{1}\left(x(1-x)\right)^{k/2-1}\,dx $$ and by switching the series and the integral: $$ S = \int_{0}^{1}\frac{\log(1+\sqrt{x(1-x)})}{x(1-x)}dx = 2\int_{0}^{1/2}\frac{\log(1+\sqrt{x(1-x)})}{x(1-x)}dx,$$ $$ S = 2\int_{0}^{1/2}\frac{\log(1+\sqrt{1/4-x^2})}{1/4-x^2}dx = 4\int_{0}^{1}\frac{\log(1+\frac{1}{2}\sqrt{1-x^2})}{1-x^2}dx,$$ $$ S = 4\int_{0}^{\pi/2}\frac{\log(1+\frac{1}{2}\sin\theta)}{\sin\theta}d\theta.$$ Now Mathematica gives me $\frac{5\pi^2}{18}$ as an explicit value for the last integral, but probably we are on the wrong path, and we only need to exploit the identity $$\sum_{k=1}^{+\infty}\frac{1}{k^2\binom{2k}{k}}=\frac{\pi^2}{18}$$ that follows from the Euler acceleration technique applied to the $\zeta(2)$-series. The other "piece" (the $U$-piece in the Marty Cohen's answer) is simply given by the Taylor series of $\arcsin(z)^2$. More details to come.

As a matter of fact, both approaches lead to an answer. The (Taylor) series approach, as Bhenni Benghorbal shows below, leads to the identity: $$\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k!}\Gamma^2\left(\frac{k}{2}\right)x^k= 2 \arcsin \left( x/2 \right) \left(\pi - \arcsin \left( x/2\right) \right),\tag{1}$$ while the integral approach, as Achille Hui pointed out in the comments, leads to: $$\begin{eqnarray*}\int_{0}^{\pi/2}\frac{\log(1+\frac{1}{2}\sin\theta)}{\sin\theta}\,d\theta&=&\int_{0}^{1}\log\left(1+\frac{t}{1+t^2}\right)\frac{dt}{t}\\&=&\int_{0}^{1}\frac{\log(1-t^3)-\log(1-t)-\log(1+t^2)}{t}\,dt\\&=&\int_{0}^{1}\frac{-\frac{2}{3}\log(1-t)-\frac{1}{2}\log(1+t)}{t}\,dt\\&=&\frac{1}{6}\sum_{k=1}^{+\infty}\frac{4+3(-1)^k}{k^2}=\frac{1}{6}\left(4-\frac{3}{2}\right)\zeta(2)=\frac{5\pi^2}{72}.\end{eqnarray*}\tag{2}$$

Thanks to both since now this answer may become a reference both for integral-log-ish problems like $(2)$ and for $\Gamma^2$-series like $(1)$.

Update 14-06-2016. I just discovered that this problem can also be solved by computing $$ \int_{-1}^{1} x^n\, P_n(x)\,dx, $$ where $P_n$ is a Legendre polynomial, through Bonnet's recursion formula or Rodrigues' formula. Really interesting.

Solution 2:

$($This is more of a comment than an answer, but $...)$

Consider the even and odd $k$ in separate sums:

Note: You probably do not want $\Gamma(0)$ in the sum, so I'll start at $k=1$.

$$\begin{align} \\S &=\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k!}\left[\Gamma\left(\frac{k}{2}\right)\right]^2 \\\\ &=\sum_{k=1}^\infty\frac{(-1)^{2k+1}}{(2k)!}\left[\Gamma\left(\frac{2k}{2}\right)\right]^2 ~+~\sum_{k=0}^\infty\frac{(-1)^{2k+1+1}}{(2k+1)!}\left[\Gamma\left(\frac{2k+1}{2}\right)\right]^2 \\\\ &=\sum_{k=1}^\infty\frac{-1}{(2k)!}\left[\Gamma\left(k\right)\right]^2 ~+~\sum_{k=0}^\infty\frac{1}{(2k+1)!}\left[\Gamma\left(k+\frac12\right)\right]^2 \\\\ &=-\sum_{k=1}^\infty\frac{((k-1)!)^2}{(2k)!} ~+~\sum_{k=0}^\infty\frac{1}{(2k+1)!}\left[\frac{\sqrt{\pi}(2k)!}{4^kk!}\right]^2 \\\\ &=-\sum_{k=1}^\infty\frac{((k-1)!)^2}{(2k)!} ~+~\pi\sum_{k=0}^\infty\frac{(2k)!}{(2k+1)16^k(k!)^2} \\\\ &=-T~+~\pi~U, \\\\ \end{align}$$

where $~T=\displaystyle\sum_{k=1}^\infty\frac{((k-1)!)^2}{(2k)!},~$ and $~U=\displaystyle\sum_{k=0}^\infty\frac{(2k)!}{(2k+1)16^k(k!)^2}.$

You can write these sums in terms of the central binomial coefficients $\displaystyle\binom{2k}{k}=\frac{(2k)!}{(k!)^2},$
and people who know more about these than I can possibly sum these.

I'll leave it at this.

Solution 3:

Here is a closed form

$$ \mathcal{I}=\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k!}\Gamma^2\left(\frac{k}{2}\right)x^k= 2 \arcsin \left( x/2 \right) \left(\pi - \arcsin \left( x/2\right) \right) .$$

Now just plug in $x=1$ and the result follows.

Solution 4:

This is more a comment that an answer

As Jack D'Aurizio commented, I suppose that the summation starts at $k=1$ and not $k=0$ as written in the post. Using a CAS, the result obtained is $$\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k!}\left[\Gamma\left(\frac{k}{2}\right)\right]^2=\frac{5 \pi ^2}{18}$$ which matches the value you obtained using Wolfram Alpha. Obtaining this result with elementary ways (such as high school methods) seems difficult (at least to me).

The only thing I have been able to obtain is that $$\sum_{k=1}^n\frac{(-1)^{k+1}}{k!}\left[\Gamma\left(\frac{k}{2}\right)\right]^2=\frac{(-1)^n \Gamma \left(\frac{n}{2}+1\right)^2 \, _3F_2\left(1,\frac{n}{2}+1,\frac{n}{2}+1;\frac{n}{2}+\frac{3}{2},\frac{n}{2}+2; \frac{1}{4}\right)}{\Gamma (n+3)}-\frac{(-1)^n \Gamma \left(\frac{n+1}{2}\right)^2 \, _3F_2\left(1,\frac{n}{2}+\frac{1}{2},\frac{n}{2}+\frac{1}{2};\frac{n}{2}+1,\frac{n }{2}+\frac{3}{2};\frac{1}{4}\right)}{\Gamma (n+2)}+\frac{5 \pi ^2}{18}$$ which is neither nice, neither easy to obtain.

One small thing I did was to compute the sum for odd and even terms. For odd values of $k$ the infinite sum is $\frac{\pi ^2}{3}$ while for even values of $k$ the infinite sum is $-\frac{\pi ^2}{18}$ and this makes the final result of $\frac{5\pi ^2}{18}$.

Added later

Looking at Marty Cohen's answer, and computing the terms he defined, I found that $T=\frac{\pi ^2}{18}$ and $U=\frac{\pi}{3}$ which again lead to the same final result.