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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.

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$\log(\frac{1}{1-a_n}) = -\log(1-a_n)$. 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.

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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 $a_n\ll 1$) $\log(1-a_n)\approx -a_n$...

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