True of false: The sum of this infinite series. I'm fairly sure it is false, but I'm not quite sure about which test I should use to prove it.
$$\sum_{n=2}^\infty \ln\left(\frac{n-1}{n}\right) = -1 $$
I think using the integral test should work, but it may get kind of messy, so I'm looking for som advice. All I have to do is prove that it diverges, right? 
Thanks in advance.

If the series above did converged to -1, then
$$1=(-1)(-1)=-1\left(\sum^\infty_{n=2}ln\left(\frac{n-1}{n}\right)\right)=\sum^\infty_{n=2}-ln\left(\frac{n-1}{n}\right)=\sum^\infty_{n=2}ln\left(\frac{n}{n-1}\right)$$
would also be true. The series on the right was shown to be divergent elsewhere in this forum. 
Here: Why does the series $\sum_{n=1}^∞ \ln ({n \over n+1})$ diverges? And general tips about series and the logarithm
