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

Determine whether the series converges or diverges.
$\sum _{n=2}^{\mathrm{\infty }}\frac{1}{n\mathrm{ln}n}$
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Ayesha Gomez

Given that:
The series $\sum _{n=2}^{\mathrm{\infty }}\frac{1}{n\mathrm{ln}n}$
By using,
Limit - Comparison test :
Suppose that we have two series $\sum {a}_{n}$ and $\sum {b}_{n}$ with ${a}_{n}\ge 0,{b}_{n}>0$ for all n
Define $c=\underset{n\to \mathrm{\infty }}{lim}\frac{{a}_{n}}{{b}_{n}}$
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 $\sum \frac{1}{{n}^{p}}$ is converges if $p>1$ and diverges if $0
Let ,
The original series is ${a}_{n}=\frac{1}{n\mathrm{ln}n}$
We need to find the series similar to the original series but simpler.
$\frac{1}{n\mathrm{ln}n}<\frac{1}{n}$
Then,
The comparison series is ${b}_{n}=\frac{1}{n}$
By using the p- series test,
$\sum _{n=2}^{\mathrm{\infty }}\frac{1}{n}=\sum _{n=2}^{\mathrm{\infty }}\frac{1}{{n}^{1}}$
Here p = 1
Then,
By the p - series test ${b}_{n}$ is divergent.
$c=\underset{n\to \mathrm{\infty }}{lim}\frac{{a}_{n}}{{b}_{n}}$
$=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n\mathrm{ln}n}×\frac{n}{1}$
$=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{\mathrm{ln}n}$
=diverge.
Then,
By the Limit - Comparison Test,
The series $\sum _{n=2}^{\mathrm{\infty }}\frac{1}{n\mathrm{ln}n}$ is diverges.
Therefore,
The series $\sum _{n=2}^{\mathrm{\infty }}\frac{1}{n\mathrm{ln}n}$ is diverges.