# Kind of counter intuitive sum of log gamma

I came across an infinite series that appears to be rather counter intuitive.

Show that $\displaystyle \sum_{k=1}^{\infty}(-1)^{k+1}\ln(\Gamma(k+1))=\frac{-1}{4}\ln\left (\frac{\pi}{2}\right)$

At first glance, it obviously diverges. I ran this through Mathematica for a check and that is what it said, "sum does not converge".

But, I ran it through Maple as $\displaystyle \sum_{k=1}^{\infty}(-1)^{k+1}\ln(k!)$ and it actually returned the above result. Mathematica still would not.

If I entered it in using the Gamma function instead of its equivalent factorial, it would not return the result.

What is going on here?. I presume this has something to do with analytic continuation of some sort?.

Since $k!=\Gamma(k+1)$, why would Maple return the result for the factorial but would not for the Gamma even though they are essentially the same thing?.

It reminds me of $\zeta(0)=\frac{-1}{2}$.

If we just let $s=0$ in $\displaystyle\sum_{k=1}^{\infty}\frac{1}{k^{s}}$ we get an infinite string of 1's. But, using the functional equation, it can be shown to converge to -1/2.

How can the above sum be shown to equal $\frac{-1}{4}\ln\left (\frac{\pi}{2}\right)$?.

I searched all around for something on this, but could find nothing.

Thanks all. I hope you find this as interesting as I have.

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Ignore this as off-topic: Your question reads like a poem :) – Anon Jan 7 '13 at 20:45
What version of Maple you use? – Mhenni Benghorbal Jun 10 '13 at 0:08
I have an old version. 9.5 – Cody Jun 10 '13 at 12:57

## 5 Answers

This sum converges in Abel summation sense:

$$\sum_{k=1}^{\infty} (-1)^{k-1} \log (k!) = \lim_{x \uparrow 1} \sum_{k=1}^{\infty} (-1)^{k-1} x^{k} \log (k!).$$

Let us evaluate the sum inside the limit.

\begin{align*} \sum_{k=1}^{\infty} (-1)^{k-1} x^{k} \log (k!) &= \sum_{k=1}^{\infty} (-1)^{k-1} x^{k} \sum_{n=1}^{k} \log n = \sum_{n=1}^{\infty} \sum_{k=n}^{\infty} (-1)^{k-1} x^{k} \log n \\ &= \sum_{n=1}^{\infty} \frac{(-1)^{n-1} x^{n}}{1 + x} \log n \\ &= \frac{1}{x+1} \sum_{n=1}^{\infty} (-1)^{n-1} x^{n} \int_{0}^{\infty} \frac{e^{-t} - e^{-kt}}{t} \, dt \\ &= \frac{x}{x+1} \int_{0}^{\infty} \left( \frac{e^{-t}}{1 + x} - \frac{e^{-t}}{1 + x e^{-t}} \right) \, \frac{dt}{t} \\ &= - \frac{x^{2}}{(x+1)^{2}} \int_{0}^{\infty} \frac{e^{-t} (1 - e^{-t})}{1 + x e^{-t}} \, \frac{dt}{t} \\ & \xrightarrow{x \to 1^{-}} -\frac{1}{4} \int_{0}^{\infty} \frac{e^{-t}(1 - e^{-t})}{t(1 + e^{-t})} \, dt. \end{align*}

To evaluate the last integral, we introduce

$$I(s) = \int_{0}^{\infty} \frac{e^{-t}(1 - e^{-t})}{t(1 + e^{-t})} e^{-st} \, dt.$$

Differentiating,

\begin{align*} I'(s) &= - \int_{0}^{\infty} \frac{e^{-t}(1 - e^{-t})}{1 + e^{-t}} e^{-st} \, dt \\ &= - \int_{0}^{1} \frac{u^{s} - u^{s+1}}{1 + u} \, du \qquad (u = e^{-t}) \\ &= - \int_{0}^{1} \frac{u^{s} - 2u^{s+1} + u^{s+2}}{1 - u^{2}} \, du \\ &= - \frac{1}{2} \int_{0}^{1} \frac{v^{\frac{s-1}{2}} - 2v^{\frac{s}{2}} + v^{\frac{s+1}{2}} }{1 - v} \, dv \qquad (v = u^2) \\ &= \frac{1}{2} \left[ \psi_{0}\left( \frac{s+1}{2} \right) - 2 \psi_{0} \left( \frac{s}{2} + 1 \right) + \psi_{0} \left( \frac{s+3}{2} \right) \right]. \end{align*}

So we have

$$I(s) = \log \left( \frac{\Gamma\left(\frac{s+1}{2}\right) \Gamma\left(\frac{s+3}{2}\right)}{\Gamma\left(\frac{s}{2}+1\right)^{2}} \right)$$

and the sum is equal to

$$\sum_{k=1}^{\infty} (-1)^{k-1} \log (k!) = -\frac{1}{4} I(0) = -\frac{1}{4} \log \left( \frac{\pi}{2} \right).$$

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Wonderful, SOS.:) – Cody Jun 11 '13 at 16:16

An idea: put

$$C:=\sum_{k=1}^\infty(-1)^{k+1}\log(k!)\Longrightarrow e^C=e^{\sum_{k=1}^\infty(-1)^{k+1}\log(k!)}=\prod_{k=1}^\infty (-1)^{k+1}k!$$

Try now to google (or directly in Wolfram) "Barnes G-Function", "Hyperfactorial function", etc. It is an old, but by no means trivial or even easy, result

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Continuing with this idea, zeta regularization would tell us that $C = -\hat{\zeta}'(0)$, where $\hat{\zeta}$ is the function defined by $$\hat{\zeta}(s) = \sum_{n=1}^{\infty} \left((k!)^{(-1)^{n+1}}\right)^{-s}.$$ No clue if this gets us anywhere. – Antonio Vargas Jan 7 '13 at 14:00
Thanks for the comments. I will look into it. If I find anything I will let you know. – Cody Jan 8 '13 at 11:35

I asked some others about this problem and they are adamant that the given closed form is wrong. The series obviously diverges.

I do not know where this problem originally came from nor why Maple gives its numerical equivalent. Perhaps a bug in Maple?.

Anyway, thanks all.

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If you use the Cesaro Sum or the Borel Formula this divergent series actually converges numerically to the value calculated in Maple, which is a little funny. So far I failed to calculate a closed form for this sum.

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Thanks a lot SOS. I always appreciate your input and clever solutions.

Anyway, sometime back, I combined consecutive terms to make it work.

The sum, as is, is certainly divergent. So, combine terms to try to make it convergent.

Let $\displaystyle S=\sum_{k=1}^{\infty}(-1)^{k+1}log(\Gamma(k+1))$

$=\displaystyle\frac{1}{2}\sum_{k=1}^{\infty}\left[(-1)^{k}log\Gamma(k)+(-1)^{k+1}log\Gamma(k+1)\right]$

$=\displaystyle\frac{1}{2}\sum_{k=1}^{\infty}(-1)^{k}log\frac{\Gamma(k)}{\Gamma(k+1)}$

$=\displaystyle\frac{1}{2}\sum_{k=1}^{\infty}(-1)^{k+1}log(k)$

This is still divergent, so combine terms again:

$S=\displaystyle\frac{1}{4}\sum_{k=1}^{\infty}\left[(-1)^{k+1}log(k)+(-1)^{k}log(k+1)\right]$

$=\displaystyle\frac{1}{4}\sum_{k=1}^{\infty}(-1)^{k}log\left(1+\frac{1}{k}\right)$

It now converges.

This is the log of the Wallis product formula $\displaystyle log(\frac{\pi}{2})=\sum_{k=1}^{\infty}(-1)^{k-1}log\left(1+\frac{1}{k}\right)$.

That comes from $\displaystyle\prod_{k=1}^{\infty}\frac{(2k)^{2}}{(2k-1)(2k+1)}=\frac{\pi}{2}$

This is a known sum that now equals $\displaystyle\frac{-1}{4}log(\frac{\pi}{2})$

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I also considered that technique, but I decided not to try it because I wanted to show that the sum is indeed summable in some sense. But once we know that the sum is summable in some summation sense that possesses regularity, linearity and stability as listed in this article, then your solution is justified, which is of course the case. – Sangchul Lee Jun 10 '13 at 17:29