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The question is simple. Does $$\prod _{n=2}^{\infty }\left(1+\frac{1}{n \, \ln(n)}\right)$$converge?

I know that it is pretty clear to show that

$$\prod _{n=1}^{\infty }\left(1+\frac{1}{n}\right)$$

diverges, while

$$\prod _{n=1}^{\infty }\left(1+\frac{1}{n^2}\right)$$

converges to $ \dfrac{\sinh(\pi)}{\pi} $.

Does it diverge in the way that $$\sum_{n=2}^{\infty} {\frac{1}{n \, \ln(n)}}$$ does?

How would you show this?

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    $\begingroup$ $\prod_{n=2}^N\left(1+\frac{1}{n\ln n}\right)>1+\sum_{n=2}^N\frac{1}{n\ln n}$. $\endgroup$ – Sergei Golovan Feb 16 '18 at 13:39
  • $\begingroup$ fr.m.wikipedia.org/wiki/Série_de_Bertrand this link can help you. You can also use $\ln(1+x) = x + o(x)$ $\endgroup$ – Youem Feb 16 '18 at 13:40
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    $\begingroup$ Is anyone going to adress the fact that these sums are all starting at $n=1$ yet $\ln(1)$ is in the denominator? $\endgroup$ – Tony S.F. Feb 16 '18 at 13:42
  • $\begingroup$ @TonyS.F. Well then we just have to show whether or not $$\prod_{n=2}^\infty\bigg(1+\frac{1}{n\ln n}\bigg)$$ converges. $\endgroup$ – Mr Pie Feb 16 '18 at 13:54
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No, it does not converge. First note that an infinite product $\prod\limits_{n=2}^\infty a_n$ converges iff $\sum\limits_{n=2}^\infty\log(a_n)$ converges. Then, by the limit comparison test for $x$ and $\log(1+x)$ we have,

$$\sum\limits_{n=2}^\infty \log\left(1+\frac{1}{n\log n}\right)\sim\sum\limits_{n=2}^\infty\frac{1}{n\log n},$$

where $\sim$ denotes that either both series converge or both diverge. You have already stated that the right hand side series diverges, so the left hand side must as well.

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