Show that $\sum a_n$ diverges if $\sum \log (\tfrac{1}{1-a_n})$ diverges Let $\{a_n\}$ be a sequence that satisfy $0\le a_n<1$ for all $n$. Given that the series
$\displaystyle \sum_{n=1}^{\infty} \log \left(\frac1{1-a_n}\right)$ diverges. Prove or disprove $\sum\limits_{n=1}^{\infty} a_n$ diverges.
I believe it diverges. But I couldn't really prove it. I was trying for the comparison test, but found that $\frac1{1-x} \ge e^{x}$. A hint or a counterexample would be appreciated.
 A: $\log(\frac{1}{1-a_n}) = -\log(1-a_n)$.  Now for $x > 0, \log(x) \leq x-1$, so $-\log(1-a_n)/a_n \geq -(1-(1-a_n))/a_n \to 1$ so the series diverges by the limit comparison test.
A: Hint: if $a_n\not\to 0$ then the series $\sum a_n$ diverges. If $a_n\to 0$, 
$$\sum \log\Big(\frac1{1-a_n}\Big)=-\sum \log (1-a_n),$$
and (for $a_n\ll 1$) $\log(1-a_n)\approx -a_n$...
A: Summary:
If $a_n \in [0,1)$ then
$$
\sum_n a_n  \ \text{ converges(diverges)}  \iff \sum_n \log\left( \frac{1}{1-a_n}\right) \ \text{ converges(diverges)} 
$$

*

*If $\sum_n a_n$  diverges:

*

*$a_n \le \log\left( \frac{1}{1-a_n}\right) = -\log( 1-a_n )$  by $\ e^{-x}\ge 1-x$

*Hence $\sum_n \log\left( \frac{1}{1-a_n}\right)$ diverges, by comparison.



*If $\sum_n a_n$ converges:

*

*$\lim_{x\to 0}\frac{-\log(1-x)}{x}=1$ by the L'Hospital/Bernoulli rule

*There is a $\delta>0$ for which $\frac{-\log(1-x)}{x} < 2$ if $0<x<\delta$, by definition of limit.

*Then $-\log(1-x) < 2x$ if $0\le x<\delta$ (note that here $x$ can be zero)

*There is a $n_{\delta}$ for which $a_n<\delta$ if $n>n_{\delta}$, therefore $-\log(1-a_n)<a_n$ for $n>n_{\delta}$, that is
$\sum_n \log\left( \frac{1}{1-a_n}\right)$ converges, by $a_n\to 0$ and comparison.



