# Prove $\sum\limits_{n=2}^\infty \frac {1} {\ln{n}}$ is divergent.

Evaluate if the following series is convergent or divergent: $\sum\limits_{n=2}^\infty \frac {1} {\ln{n}}$.

I solved the problem using Weierstrass's comparison theorem once the integral would get messy:

$n+c\geqslant \ln(n)$, where $c$ is a constant. We can check that $$\frac{d\ln(x)}{dx}=\frac{1}{x}<1=\frac{dy}{dx},$$ for $y=x$ and $x\in\mathbb{N}$, which shows that $\ln(n)$ grows slower than $n$.

Then $$\sum\limits_{n=2}^\infty \frac {1} {{n}} <\sum\limits_{n=2}^\infty \frac {1} {{n}+c}<\sum\limits_{n=2}^\infty \frac {1} {\ln{n}},$$ proving the series diverge.

Question:

Is my answer right? If not why? What are other possibilities?

Thanks in advance!

• What "Weierstrass Comparison Theorem"?? – DonAntonio Apr 1 '18 at 11:01
• @DonAntonio It is a proof of the comparasion test. People call it Weierstrass's test in honour of the German mathematician that carried that surname. – Pedro Gomes Apr 1 '18 at 11:04
• The only Weierstrass Test I know is Weierstrass's $\;M\,-$ test, for power series. I'm not sure what you call Weierstrass Comparison theorem...perahsp you mean the usual comparison test for positive series? – DonAntonio Apr 1 '18 at 11:07
• @DonAntonio That is precisely what I mean. – Pedro Gomes Apr 1 '18 at 11:10
• Another alternative: $\frac{1}{\ln n} > \frac{1}{n \ln n}$ and the latter series diverges by the integral test $\int \frac{dx}{x\ln x} = \ln\ln(x) \to \infty$. – Winther Apr 1 '18 at 11:24

## 4 Answers

You can simply say that $(\forall n\in\mathbb{N}):\ln n<n$ and that therefore$$\sum_{n=2}^\infty\frac1{\ln n}\geqslant\sum_{n=2}^\infty\frac1n=+\infty.$$

Use that $$n>\log(n)$$ for $n>0$

Use the fact that $$n>\log(n)$$ So that we get $$\sum_{n=2}^\infty\frac{1}{\ln n}> \sum_{n=2}^\infty\frac{1}{n}=\infty$$

Use Cauchy's condensation test with $2^n > n$. (Easily proved by induction/binomial theorem.)

$$2^n a_{2^n} = \frac{2^n}{\ln 2^n} = \frac{2^n}{n \ln 2} > \frac{1}{\ln2}$$

So the series diverges.