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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.

Martin Sleziak
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MohammadReza Hosseini
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    Just as a sidenote: in general we have that if a sum $\sum a_n$ of positive terms diverges then so does $\sum \frac{a_n}{a_1+a_2+\ldots + a_n}$ (see e.g. https://math.stackexchange.com/questions/1135536/if-sum-n-0-infty-a-n-diverges-prove-that-sum-n-0-infty-fraca-n) – Winther Oct 18 '17 at 15:44
  • Compression test? What is that? – Did Oct 18 '17 at 15:44
  • @Did: the https://en.wikipedia.org/wiki/Cauchy_condensation_test is also known as compression test somewhere (like in Italy). Or maybe it is a miswording for comparison. – Jack D'Aurizio Oct 18 '17 at 15:53
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    @JackD'Aurizio Never seen the word used for that. I am ready to bet on a typo. – Did Oct 18 '17 at 16:00
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    Use $\ln\ln{n}\rightarrow+\infty$ for $n\rightarrow+\infty$ – Michael Rozenberg Oct 18 '17 at 15:31

2 Answers2

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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.

Anne Bauval
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Bernard
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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.

Jack D'Aurizio
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