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.

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

