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

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. – Aldon Sep 3 '15 at 15:18
• @Aldon: oh; I see this now. I was so stupid to not think of this. Thanks – user265696 Sep 3 '15 at 15:21
• Good thing you understood it even if I made a typo. P.S. Comparison tests are awesome. – Aldon Sep 3 '15 at 15:30
• – Martin Sleziak Oct 31 '15 at 20:30

## 1 Answer

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).

• 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 3 '15 at 15:15
• @user2566092 Spectacular is a bit overreaching, IMO, since the integral test works equally well for your example :). – Erick Wong Sep 3 '15 at 15:24