Proving that the series $\sum_{k=2}^\infty \frac{1}{k \ln k}$ diverges?

Question

I don't know how to show this. The terms go to zero, and I can't really show that the terms dominate $\frac{1}{k}$ (a series with these terms diverges). Any other ideas?

Answer 1

Compare to the integral $\int \frac{dx}{x \log x}$ because the summand is a montone decreasing function. What do you get?

Answer 2

This is a Bertrand series.

Bertrand series are of the kind $\sum\limits_{k=2}^n\dfrac{1}{n^\alpha\ln(n)^\beta}$.

If $\alpha=1$ and $\beta\leq1$ then the series diverges.

Answer 3

Let $k\geq 2$ and let $x \in [k,k+1[$. Since $\displaystyle x \rightarrow \frac{1}{x\ln x}$ is a decreasing function, you may write $$ \frac{1}{x\ln x} \leq \frac{1}{k\ln k}, $$ integrating $$ \int_k^{k+1}\frac{1}{x\ln x}dx \leq \int_k^{k+1}\frac{1}{k\ln k} dx=\frac{1}{k\ln k} $$ then summing from $k=2$ to $N-1\geq2$, you get $$ \int_2^N \frac{1}{x\ln x} dx \leq \sum_{k=2}^{N}\frac{1}{k\ln k} $$ thus $$ \int_2^N \frac{1}{x\ln x} dx =\int_2^N \frac{(\ln x)'}{(\ln x)} dx = \log(\log(N))-\log( \log 2)\leq \sum_{k=2}^{N}\frac{1}{k\ln k} $$ and letting $N$ tend to $+\infty$ gives the divergence of the series.

Answer 4

If you refer Walter Rudin, Principles of Mathematical Analysis, there is a theorem:

If $ x_i $ is a non-negative monotone non-increasing sequence, then

$$\sum_{k=1}^{\infty}x_k \mbox{ converges} \iff \sum_{k=1}^{\infty}2^kx_{2^k} \mbox{ converges}$$