On the convergence (or divergence) of $\sum_{n\geq 1}\frac{1}{n H_n}$

Question

I have to study the series

$$\sum_{n=1}^{+\infty}\frac{1}{n\left(1+\frac{1}{2}+\ldots+\frac{1}{n}\right)}$$ Is it convergent? Is it divergent? I used the compression test with 1/n and 1/ln(n) but did not manage to get an answer. Raabe's test is beyond hardness and I would like some help.

Answer 1

You can use equivalents:

The harmonic series: $$H_n=1+\frac12+\dots+\frac1n\sim_\infty \log n,\enspace\text{ so }\quad \frac1{n(1+\frac12+\dots+\frac1n)}\sim_\infty\frac1{n\log n},$$ which is the general term of a divergent Bertrand's series.

Answer 2

By denoting the $n$-th harmonic number as usual, $$ H_n = 1+\frac{1}{2}+\ldots+\frac{1}{n},$$ we may notice that $$ H_n\leq \sum_{k=1}^{n} 2\,\text{arctanh}\left(\tfrac{1}{2k}\right)=\sum_{k=1}^{n}\log\left(\tfrac{2k+1}{2k-1}\right)=\log(2n+1) $$ hence $$\begin{eqnarray*} \sum_{n=1}^{N}\frac{1}{n H_n}\geq \sum_{n=1}^{N}\frac{1}{n\log(2n+1)}&\geq&\sum_{n=1}^{N}\frac{2}{(2n+1)\log(2n+1)}\\&\geq&\int_{1}^{N+1}\frac{2\,dx}{(2x+1)\log(2x+1)}\\[0.1cm]&=&\log\log(2N+3)-\log\log(3) \end{eqnarray*}$$ and the given series is (very slowly) divergent.