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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.

Sayan
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3 Answers3

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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.

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

Milly
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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.
Czylabson Asa
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