# How to prove that this sequence converges? $\sum_{n=1}^{\infty} \frac{1}{n\ln^2(n)}$

I'm trying to prove this sequence converges: $\sum_{n=1}^{\infty} \frac{1}{n\ln^2(n)}$

I noticed that this is continuous function which its derivative is always less than $0$ for $x \gt 1$, so I tried to do the integral test but with no success...

Some help?

Thanks!

We can use the integral test for convergence. We have that $$\sum_{n=3}^\infty\frac1{n\log^2 n}\le\int_2^\infty\frac1{x\log^2 x}\mathrm dx=\biggl[-\frac1{\log x}\biggr]_2^\infty=\frac1{\log 2}.$$
• The derivative of $\log(x)$ is $\frac{1}{x}$, and the derivative of $\frac{1}{u}$ is $-\frac{u'}{u²}$. – Paul Picard Jun 17 '15 at 12:25
Since the sequence $\dfrac{1}{n\ln^2 n}$ is decreasing, you can apply Cauchy's condensation test, which says that if $a_n$ is decreasing, then $$\sum_{n=1}^\infty a_n <\infty \quad \text{if and only if}\quad \sum_{k=0}^\infty 2^k a_{2^k}<\infty.$$ Typically, this test works well with series with logarithms, you will see why once you apply it.