challenging alternating infinite series involving $\ln$ 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.
 A: Use $ \log\left(1 - \frac{1}{n(n-1)}\right) = \int_0^1 \frac{\mathrm{d} t}{n(1-n)+t}$.
Then
$$ \begin{eqnarray}
   \sum_{n=2}^\infty \frac{(-1)^n}{n - n^2 + t} &=& \sum_{n=2}^\infty \frac{2 (-1)^n }{\sqrt{4 t+1}} \left(\frac{1}{2 n+\sqrt{4 t+1}-1}-\frac{1}{2 n-\sqrt{4
   t+1}-1}\right) \\&=& 
\frac{1}{2 \sqrt{4 t+1}} \left( \psi ^{(0)}\left(-\frac{1}{4} \sqrt{4 t+1}-\frac{1}{4}\right)-\psi
   ^{(0)}\left(\frac{1}{4}-\frac{1}{4} \sqrt{4 t+1}\right) \right) \\ 
 &+&  \frac{1}{2 \sqrt{4 t+1}}  \left(\psi
   ^{(0)}\left(\frac{1}{4} \sqrt{4 t+1}+\frac{1}{4}\right)  -\psi
   ^{(0)}\left(\frac{1}{4} \sqrt{4 t+1}-\frac{1}{4}\right)\right)
\end{eqnarray}
$$
The latter comes about from $\sum_{n=1}^\infty \left(\frac{1}{n} -  \frac{1}{n+a}\right) = \gamma + \psi^{(0)}(a+1)$, and the summation above was split into summation over even and odd integers.
Integrating this expression out produces:
$$
  \text{log$\Gamma $}\left(\frac{1}{2}-\frac{\phi}{2} \right)-\text{log$\Gamma
   $}\left(-\frac{1}{2}+\frac{\phi }{2}\right)-\text{log$\Gamma $}\left(-\frac{\phi
   }{2}\right)+\text{log$\Gamma $}\left(\frac{\phi }{2}\right)+\log (2)
$$
where $\phi$ is Golden ratio.
Integration is trivial as $\frac{\mathrm{d} t}{\sqrt{4 t+1}} = \mathrm{d}\left(\frac{\sqrt{4 t+1}}{2}\right)$, and $\int \psi^{(0)}(u) \mathrm{d} u = \log\Gamma(u) + C$.

Numerical check in Mathematica:

In[85]:= N[
 Log[2] - LogGamma[-GoldenRatio/2] + 
  LogGamma[-(GoldenRatio/2) + 1/2] - LogGamma[GoldenRatio/2 - 1/2] + 
  LogGamma[GoldenRatio/2], 20]

Out[85]= -0.56655310975860303045 + 0.*10^-21 I

In[84]:= NSum[(-1)^n Log[1 - 1/(n (n - 1))], {n, 2, \[Infinity]}, 
 WorkingPrecision -> 20]

Out[84]= -0.566553109758603

A: 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
$$
\begin{align}
\frac{\Gamma(x+\alpha-k)}{x^{-k}\Gamma(x+\alpha)}
&=\frac{x^k}{(x+\alpha-k)(x+\alpha-k+1)(x+\alpha-k+2)\dots(x+\alpha-1)}\\
&\to1\text{ as }x\to\infty\tag{6}
\end{align}
$$
A: Put everything into an infinite product:
$$\log\prod_{n=1}^\infty \frac{1-\frac{1}{2n(2n-1)}}{1-\frac{1}{(2n+1)2n}}=\log\prod_{n=1}^\infty\frac{8n^3-4n-1}{8n^3-4n+1}$$
Factor the polynomials and look at partial products:
$$=\log\prod_{n=1}^{m-1}\frac{n+1/2}{n-1/2}\frac{n-\varphi/2}{n+\varphi/2}\frac{n-\Phi/2}{n+\Phi/2}$$
(NB $\varphi=\frac{1+\sqrt{5}}{2},\Phi=\frac{1-\sqrt{5}}{2}$.) Substitute using $\Gamma$'s properties:
$$\log\left((2m+1)\frac{\Gamma(m-\varphi/2)\Gamma(-\varphi/2)^{-1}}{\Gamma(m+\varphi/2)\Gamma(\varphi/2)^{-1}}\frac{\Gamma(m-\Phi/2)\Gamma(-\Phi/2)^{-1}}{\Gamma(m+\Phi/2)\Gamma(\Phi/2)^{-1}}\right)$$
Evaluate the limit as $m\to\infty$ using Stirling's Formula:
$$\log\left(\frac{2\Gamma(\varphi/2)\Gamma(\Phi/2)}{\Gamma(-\varphi/2)\Gamma(-\Phi/2)}\right).$$
