How do I prove the divergence of this series?

Question

How do I prove that $\displaystyle\sum_{n\geq 1}\frac {1}{\ln^2n}$ is a divergent series?

Answer 1

Can you prove $\log n\lt\sqrt n$? Then $1/(\log n)^2\gt1/n$, and off you go.

Answer 2

Short answer: Since $n > \log n$ for all $n \in \mathbb R^+$, we know that $\frac{1}{n (\log n)} \leq \frac{1}{(\log n)^2}$ for all $n \in \mathbb R^+$.

We know that $\sum_{n=1}^\infty \frac{1}{n (\log n)}$ diverges, so by the comparison test, so does $\sum_{n=1}^\infty \frac{1}{(\log n)^2}$.

Longer answer: Here's some information that explains the motivation of comparing the series with $\sum_{n=1}^\infty \frac{1}{n (\log n)}$. It is a commonly known fact that all the following series diverge:

• $\displaystyle \sum_{n=1}^\infty \frac{1}{n}$
• $\displaystyle \sum_{n=1}^\infty \frac{1}{n (\log n)}$
• $\displaystyle \sum_{n=1}^\infty \frac{1}{n (\log n) (\log \log n)}$
• $\displaystyle \sum_{n=1}^\infty \frac{1}{n (\log n) (\log \log n)(\log \log \log n)}$
• and so on

You can prove those by using the integral test (or the Cauchy condensation test, if you prefer). When you apply either of these tests, each series reduces to the series above it. We know the first one (which is the harmonic series) diverges, so by induction, all the other series diverge as well.

Answer 3

(Since I misinterpreted $\ln^2 n$ I rewrote the part specific to this question)

Suppose $\sum a_i$ and $\sum b_i$ are two infinite sums and $a_i\ge b_i\ge 0$ for all $i$.

We can you say if so that $\sum a_i\ge\sum b_i$, this is called a comparison test for convergence.

Suppose $\sum a_i$ is finite, then we have that $\sum b_i$ has to be finite as well. On the other hand, suppose $\sum b_i$ is infinite, in this case we have that $\sum a_i$ is infinite as well, simply by sandwiching: $$\lim_{n\to\infty}\sum_{i<n}b_i\le\lim_{n\to\infty}\sum_{i<n}a_i\le\infty$$

So in order to prove that $\sum\frac{1}{(\ln n)^2}$ is divergent, we need to find a sequence such that $a_n\ge(\ln n)^2$ and $\sum\frac{1}{a_n}$ diverges.

As Gerry suggests, $\ln n<\sqrt{n}$, which will lead to the required solution.