TITLE: challenging alternating infinite series involving $\ln$
QUESTION [15 upvotes]: I ran across an infinite series that is allegedly from a Chinese math contest.
Evaluate:
$\displaystyle\sum_{n=2}^{\infty}(-1)^{n}\ln\left(1-\frac{1}{n(n-1)}\right).$
I thought perhaps this telescoped in some fashion.  So, I wrote out
$\ln(1/2)-\ln(5/6)+\ln(11/12)-\ln(19/20)+\ln(29/30)-..............$
Separated the positive and negative using log properties:
$\ln(1/2)+\ln(11/12)+\ln(29/30)+......=\ln(\frac{1}{2}\cdot \frac{11}{12}\cdot \frac{29}{30}\cdot\cdot\cdot)$
$-(\ln(5/6)+\ln(19/20)+\ln(41/42)+......=-\ln(\frac{5}{6}\cdot \frac{19}{20}\cdot \frac{41}{42}\cdot\cdot\cdot) $
$\ln(\frac{1}{2}\cdot \frac{11}{12}\cdot \frac{29}{30}\cdot\cdot\cdot)-\ln(\frac{5}{6}\cdot \frac{19}{20}\cdot \frac{41}{42}\cdot\cdot\cdot)$
$=\displaystyle \ln\left(\frac{\frac{1}{2}\cdot \frac{11}{12}\cdot \frac{29}{30}\cdot\cdot\cdot}{\frac{5}{6}\cdot \frac{19}{20}\cdot \frac{41}{42}\cdot\cdot\cdot}\right)$
Maybe come up with a general term at the end of the partial sum? The terms in the numerator are $n=2,4,6,....$ and those in the denominator are $n=3,5,7,.....$
$\frac{N(N-1)-1}{N(N-1)}$.  But, I always end up with a limit of 1. This then gives $\ln(1)=0$.
The series does converge.  I managed to do some cancellations, but failed to wrap it up. 
I thought maybe I was onto something. I suppose I am and not seeing it.  What would be a good plan of attack for this one? Since it was in a contest, I assume it can be done. Any thoughts?
Thanks very much.

REPLY [12 votes]: Combining every two terms and substituting $n\mapsto2n$, we get
$$
\begin{align}
\sum_{n=2}^\infty(-1)^n\ln\left(1-\frac{1}{n(n-1)}\right)
&=\ln\left(\prod_{n=1}^\infty\frac{(2n+1)2n}{2n(2n-1)}\frac{(2n-\phi)(2n+\phi-1)}{(2n-\phi+1)(2n+\phi)}\right)\\
&=\ln\left(\prod_{n=1}^\infty\frac{n+1/2}{n-1/2}\frac{(n-\phi/2)(n+\phi/2-1/2)}{(n-\phi/2+1/2)(n+\phi/2)}\right)
\end{align}
$$
The partial product is
$$
\begin{align}
&\prod_{n=1}^N\frac{n+1/2}{n-1/2}\frac{(n-\phi/2)(n+\phi/2-1/2)}{(n-\phi/2+1/2)(n+\phi/2)}\\
&=(2N+1)\frac{\Gamma(N-\phi/2+1)}{\Gamma(1-\phi/2)}\frac{\Gamma(N+\phi/2+1/2)}{\Gamma(1/2+\phi/2)}\frac{\Gamma(3/2-\phi/2)}{\Gamma(N-\phi/2+3/2)}\frac{\Gamma(1+\phi/2)}{\Gamma(N+\phi/2+1)}
\end{align}
$$
Now we use the fact that $\displaystyle\lim_{x\to\infty}\frac{\Gamma(x+\alpha)}{x^\alpha\Gamma(x)}=1$ to get that the limit of the partial product is
$$
2\frac{\Gamma(3/2-\phi/2)\Gamma(1+\phi/2)}{\Gamma(1-\phi/2)\Gamma(1/2+\phi/2)}=2\frac{\Gamma(1/2-\phi/2)\Gamma(\phi/2)}{\Gamma(-\phi/2)\Gamma(\phi/2-1/2)}
$$
So the answer is
$$
\ln\left(2\frac{\Gamma(1/2-\phi/2)\Gamma(\phi/2)}{\Gamma(-\phi/2)\Gamma(\phi/2-1/2)}\right)
$$
Appendix:
Searching on the web, the the closest to a reference for
$$
\lim_{x\to\infty}\frac{\Gamma(x+\alpha)}{x^\alpha\Gamma(x)}=1
$$
I could find was Gautschi’s Inequality in the Digital Library of Mathematical Functions and no proof of this inequality is given.  Since I find it quite useful and it is a simple consequence of the log-convexity of $\Gamma$, I provide a proof here.
$\Gamma$ is the log-convex function for which $\Gamma(1)=1$ and $\Gamma(x+1)=x\Gamma(x)$. Thus, for $0\le\alpha\le1$,
$$
\begin{align}
\Gamma(x+\alpha)
&\le\Gamma(x)^{1-\alpha}\Gamma(x+1)^\alpha\\
&=\Gamma(x)x^\alpha\tag{1}
\end{align}
$$
and
$$
\begin{align}
\Gamma(x)x^\alpha
&=\Gamma(x+1)x^{\alpha-1}\\
&\le\Gamma(x+\alpha)^\alpha\Gamma(x+\alpha+1)^{1-\alpha}x^{\alpha-1}\\
&=\Gamma(x+\alpha)(x+\alpha)^{1-\alpha}x^{\alpha-1}\\
&=\Gamma(x+\alpha)(1+\alpha/x)^{1-\alpha}\tag{2}
\end{align}
$$
Combining $(1)$ and $(2)$, we get
$$
(1+\alpha/x)^{\alpha-1}\le\frac{\Gamma(x+\alpha)}{x^\alpha\Gamma(x)}\le1\tag{3}
$$
By the Sandwich Theorem, $(3)$ yields
$$
\lim_{x\to\infty}\frac{\Gamma(x+\alpha)}{x^\alpha\Gamma(x)}=1\tag{4}
$$
for $0\le\alpha\le1$. However, we can extend $(4)$ to any $\alpha$ using
$$
\begin{align}
\frac{\Gamma(x+\alpha+k)}{x^k\Gamma(x+\alpha)}
&=\frac{(x+\alpha)(x+\alpha+1)(x+\alpha+2)\dots(x+\alpha+k-1)}{x^k}\\
&\to1\text{ as }x\to\infty\tag{5}
\end{align}
$$
or
$$
\begin{align}
\frac{\Gamma(x+\alpha-k)}{x^{-k}\Gamma(x+\alpha)}
&=\frac{x^k}{(x+\alpha-k)(x+\alpha-k+1)(x+\alpha-k+2)\dots(x+\alpha-1)}\\
&\to1\text{ as }x\to\infty\tag{6}
\end{align}
$$