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