Determine whether the series converges or diverges. sum_{n=2}^inftyfrac{1}{nln n}

Given that:
The series ${\displaystyle \sum _{n=2}^{\mathrm{\infty }}\frac{1}{n\ln n}}$
By using,
Limit - Comparison test :
Suppose that we have two series ${\displaystyle \sum a_n}$ and ${\displaystyle \sum b_n}$ with ${\displaystyle a_n\ge 0,b_n>0}$ for all n
Define ${\displaystyle c=\underset{n\to \mathrm{\infty }}{lim}\frac{a_n}{b_n}}$
If c is positive (that is c>0) and is finite then either both series converges or both series diverge.
P - series test:
The series is of the form ${\displaystyle \sum \frac{1}{n^p}}$ is converges if ${\displaystyle p>1}$ and diverges if ${\displaystyle 0<p\le 1}$
Let ,
The original series is ${\displaystyle a_n=\frac{1}{n\ln n}}$
We need to find the series similar to the original series but simpler.
${\displaystyle \frac{1}{n\ln n}<\frac{1}{n}}$
Then,
The comparison series is ${\displaystyle b_n=\frac{1}{n}}$
By using the p- series test,
${\displaystyle \sum _{n=2}^{\mathrm{\infty }}\frac{1}{n}=\sum _{n=2}^{\mathrm{\infty }}\frac{1}{n^1}}$
Here p = 1
Then,
By the p - series test ${\displaystyle b_n}$ is divergent.
${\displaystyle c=\underset{n\to \mathrm{\infty }}{lim}\frac{a_n}{b_n}}$
${\displaystyle =\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n\ln n}\times \frac{n}{1}}$
${\displaystyle =\underset{n\to \mathrm{\infty }}{lim}\frac{1}{\ln n}}$
=diverge.
Then,
By the Limit - Comparison Test,
The series ${\displaystyle \sum _{n=2}^{\mathrm{\infty }}\frac{1}{n\ln n}}$ is diverges.
Therefore,
The series ${\displaystyle \sum _{n=2}^{\mathrm{\infty }}\frac{1}{n\ln n}}$ is diverges.