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I am trying to show that the series $\sum({ -\ln({1-\frac{1}{2n+1})}})$ diverges, I am thinking of using the integral test, but is there a quicker way of doing it?

rubik
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continental
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    Limit comparison with $\sum \frac{1}{2n+1}$ also works and is probably a bit quicker. – mrf Sep 22 '15 at 12:13
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    Or remember that inside the $ln$ you can write it as $\frac{2n}{2n+1}$ which becomes $ln2n-ln(2n+1)$ What happens when you make a table based on a telescoping sum? – imranfat Sep 22 '15 at 12:20
  • Yes @mrf it is indeed quicker! – continental Sep 22 '15 at 12:21
  • @imranfat It doesn't really telescope (at least not without some additional mangling). – mrf Sep 22 '15 at 13:14
  • @mrf Yes, that is true, however it is reasonably easy to see that the infinite term does not go to zero, which is reason for suspicion that it doesn't converge, limit comparison however is probably quicker, I admit – imranfat Sep 22 '15 at 14:27

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Notice that $$-\log\left(1-\frac{1}{2n+1}\right)=-\log\left(\frac{2n}{2n+1}\right)=\log\left(\frac{2n+1}{2n}\right)=\log\left(1+\frac{1}{2n}\right)$$ And, using the bound $\log(x)>1-\frac{1}{x}$: $$\log\left(1+\frac{1}{2n}\right)\ge1-\frac{1}{1+\frac{1}{2n}}=1-\frac{2n}{2n+1}=\frac{1}{2n+1}$$ So your sum is larger than $$\sum_{n=1}^{\infty}\frac{1}{2n+1}$$ Which diverges.

preferred_anon
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$$\sum_{n=1}^{N}\log\left(1-\frac{1}{2n+1}\right)=\log\prod_{n=1}^{N}\frac{2n}{2n+1}=\log\frac{4^n n!^2}{(2n+1)!}=\log\left(\frac{\sqrt{\pi}\,\Gamma\left(N+1\right)}{2\,\Gamma\left(N+\frac{3}{2}\right)}\right)$$ hence the series is divergent by Gautschi's inequality, for instance.

Community
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Jack D'Aurizio
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