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How can I show that $\sum \limits_{n=2}^\infty\frac{1}{n\ln n}$ is divergent without using the integral test?

I tried using the comparison test but I could not come up with an inequality that helps me show the divergence of a series. I also tried using the limit comparison test but I was not successful.

Please do not give me solutions; just a hint so that I can figure it out myself.

  • This is just equal to \left(\frac{1}{n}\right)$ \left(\frac{1}{\text{ln} , n}\right)$. Do note that $\text{ln} , n \lt n$ therefore $\frac{1}{\text{ln} , n} \gt \frac{1}{n}$ thus it would be diverging faster than the harmonic series. Not that rigorous though, so you would have to do something else to prove it. – mopy Sep 03 '15 at 15:18
  • @Aldon: oh; I see this now. I was so stupid to not think of this. Thanks –  Sep 03 '15 at 15:21
  • Good thing you understood it even if I made a typo.

    P.S. Comparison tests are awesome.

     – mopy Sep 03 '15 at 15:30
  • See also: http://math.stackexchange.com/questions/574503/infinite-series-sum-n-2-infty-frac1n-log-n – Martin Sleziak Oct 31 '15 at 20:30

2 Answers2

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My usual way is to use Cauchy's condensation test and recall that the harmonic series is divergent (by the same reason, if you like it).

Jack D'Aurizio
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    In fact, this is such a spectacular method that you can use it to resolve $\sum_n 1/(n (\log n ) (\log \log n) \ldots (\log \log \ldots \log n)^p)$ for any $p$ and any number of iterations of the logarithm. – user2566092 Sep 03 '15 at 15:15
  • @user2566092 Spectacular is a bit overreaching, IMO, since the integral test works equally well for your example :). – Erick Wong Sep 03 '15 at 15:24
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Due to Chebyshev's estimates there exists constant $A$ and $n_0$ such that the $n$'th prime $p_n$ is bounded below by $An\log n$ for all $n>n_0$ Consequently, we have

$$ \sum_{n>n_0}^\infty{1\over n\log n}\ge\sum_{n>n_0}^\infty{A\over p_n} $$

and due to Euler the series on the RHS diverges, so the series $\sum{1\over n\log n}$ diverges.

TravorLZH
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