Why does the series $\sum_{n=1}^∞ \ln ({n \over n+1})$ diverges? And general tips about series and the logarithm

Why does the series $\sum_{n=1}^∞ \ln ({n \over n+1})$ diverges? I'm looking for an answer using the comparison test, I'm just not sure what I can compare it to.

And can I have some tips on what to look at when handling with series that have logarithms in the expression?

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We have $$\log\frac{n}{n+1}=\log n-\log (n+1)$$

Telescope, telescope, telescope.

Alternatively, $$\tag 1 \log\left(1+\frac{1}n\right)\sim\frac 1 n$$ as $n\to\infty$ cries out for the comparison test.

ADD Recall (or prove) that $$\lim_{x\to 0}\frac{\log(1+x)}x=1$$

This means $$\lim_{n\to \infty}\frac{\log\left(1+\frac{1}n\right)}{\frac1n}=1$$ which is what I write in $(1)$.

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Good one! Thanks. –  ohad Jul 15 '13 at 4:08
Why is $\ln (1+{1 \over n})$ ~ ${1 \over n}$? And what's the general principle? –  ohad Jul 15 '13 at 4:12
@ohad I added something. "the general principle"? –  Pedro Tamaroff Jul 15 '13 at 4:14
Nevermind. Thank you! –  ohad Jul 15 '13 at 4:18
The partial sum is telescopic $$\sum_{k=1}^n\log\left(\frac{k+1}{k}\right)=\sum_{k=1}^n(\log(k+1)-\log k)=\log(n+1)-\log 1=\log(n+1)\to\infty$$
$\sum_{n=1}^m \ln ({n \over n+1}) =\ln \prod_{n=1}^m ({n \over n+1}) = \ln \frac{1}{m+1}$
and since $\frac{1}{m+1} \to 0$ as $m \to =\infty$, $\ln \frac{1}{m+1} \to -\infty$ as $m \to \infty$.