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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.

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    $\begingroup$ Are you sure that the sum starts by $k=0$? The $\Gamma$ function has a pole in $0$. $\endgroup$ – Jack D'Aurizio Jul 26 '14 at 4:45
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    $\begingroup$ @JackD'Aurizio Sorry, it should be started from $k=1$. Edited! >_< $\endgroup$ – Anastasiya-Romanova 秀 Jul 26 '14 at 4:50
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    $\begingroup$ Multiplying and dividing each term by $\bigg(\dfrac k2\bigg)^2$, and using the fact that $x~\Gamma(x)=\Gamma(x+1)=x!$, we have $I=-4\cdot\displaystyle\sum_{k=1}^\infty\frac{(-1)^k}{\displaystyle{k\choose k/2}~k^2}$ . At the same time, we know that $\displaystyle\sum_{k=1}^\infty\frac{\big(2x\big)^{2k}}{\displaystyle{2k\choose k}~k^2}=2\arcsin^2x$. $\endgroup$ – Lucian Jul 26 '14 at 12:31
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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.

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    $\begingroup$ Nice use of the Beta function. I didn't think of that when I did my "answer". $\endgroup$ – marty cohen Jul 26 '14 at 5:06
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    $\begingroup$ The next question is: how to calculate the integral? Thanks anyway for your answer (>‿◠)✌ $\endgroup$ – Anastasiya-Romanova 秀 Jul 26 '14 at 5:25
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    $\begingroup$ @V-Moy Let $t = \tan\frac{\theta}{2}$, we have $$ 4\int_0^{\pi/2}\frac{\log(1+\frac12\sin\theta)}{\sin\theta}d\theta = 4\int_0^1\log(1+\frac{t}{1+t^2})\frac{dt}{t}\\ = 4\int_0^1( \log(1-t^3) - \log(1-t)-\log(1+t^2) )\frac{dt}{t}\\ = -\frac23\int_0^1 (4\log(1-t) + 3\log(1+t))\frac{dt}{t}\\ = \frac23 \int_0^1 \sum_{k=1}^\infty \left( \frac{4+3(-1)^k}{k}t^{k-1} \right) dt\\ = \frac23 \sum_{k=1}^\infty \frac{4+3(-1)^k}{k^2} = \frac23 \left( 4 - 3\left(1 - \frac{2}{2^2}\right)\right)\zeta(2) = \frac{5\pi^2}{18} $$ $\endgroup$ – achille hui Jul 26 '14 at 6:01
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    $\begingroup$ @achillehui I wish you were a 14 y.o. boy, we could be a best friend (or boyfriend) (★´−`)人(´▽`★) *kidding $\endgroup$ – Anastasiya-Romanova 秀 Jul 26 '14 at 6:22
  • $\begingroup$ @achillehui It seems Anna (eventually) has found the general formula of the integral you've calculated in her own post math.stackexchange.com/a/995449/146687 $\endgroup$ – Venus Nov 10 '14 at 17:08
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$($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.

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  • $\begingroup$ I did too split the series into odd and even part, but it's still hard for me to obtain the closed form. Thanks anyway for your answer (>‿◠)✌ $\endgroup$ – Anastasiya-Romanova 秀 Jul 26 '14 at 5:29
  • $\begingroup$ Your solution is nice and was very inspiring ! Cheers :) $\endgroup$ – Claude Leibovici Jul 26 '14 at 5:31
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    $\begingroup$ @Anastasiya-Romanova秀: We have $$T ~=~ \sum_{k=1}^\infty\frac1{\displaystyle{2k\choose k}k^2} ~=~ \frac{\pi^2}{18},$$ see my comment on the main post, and $$U ~=~ \sum_{k=0}^\infty{2k\choose k}\frac{16^{-k}}{2k+1} ~=~ \frac\pi3,$$ which follows from the well-known Taylor series of the arcsine function. $\endgroup$ – Lucian Sep 1 '17 at 3:55
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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.

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  • $\begingroup$ How to get it? Could you elaborate Mr. Mhenni Benghorbal? $\endgroup$ – Anastasiya-Romanova 秀 Jul 26 '14 at 6:37
  • $\begingroup$ @V-Moy: You can use Maple to get this cloded form. $\endgroup$ – Mhenni Benghorbal Jul 26 '14 at 6:46
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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.

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  • $\begingroup$ I remember to have seen similar series (the one that follows from the Euler acceleration techniques) in many papers that deal with the irrationality measure of $\pi^2$. The part corresponding to odd values of $k$ (central binomials in the numerator) depends on the Taylor series of the arcsine square, while the part corresponding to even values of $k$ is a rational multiple of $\zeta(2)$ by applying the right change of variables in the integral $\int_{[0,1]^2}\frac{dx\,dy}{1-xy}=\zeta(2)$. $\endgroup$ – Jack D'Aurizio Jul 26 '14 at 5:14
  • $\begingroup$ @JackD'Aurizio. This is a very interesting comment. Thanks ! $\endgroup$ – Claude Leibovici Jul 26 '14 at 5:22
  • $\begingroup$ Mr. Claude Leibovici: The term 'elementary ways' in my dictionary means without involving residues or hypergeometric function, but thank you for your answer. Cheers! (>‿◠)✌ $\endgroup$ – Anastasiya-Romanova 秀 Jul 26 '14 at 5:24
  • $\begingroup$ My knowledge (or lack of it) usually limits me to moderately elementary methods. I'm glad that I did not make any mistakes in my manipulations. $\endgroup$ – marty cohen Jul 26 '14 at 5:26
  • $\begingroup$ @V-Moy. I know and I wrote it in my answer ! If I wrote the partial sum, it was to show the complexity that I was unable to overcome. Fortunaly, Marty Cohen and Jack D'Aurizio provided interesting solutions. Cheers :) $\endgroup$ – Claude Leibovici Jul 26 '14 at 5:29

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