<s>[INST] Showing that $\sum_{n=1}^\infty n^{-1}\left(1+\frac{1}{2}+...\frac{1}{n}\right)^{-1}$ is divergent 
How to show that $\sum_{n=1}^\infty \frac{1}{n\left(1+\frac{1}{2}+...\frac{1}{n}\right)}$ diverges?

I used Ratio test for this problem and this is the result: $$\lim_{n\to\infty} \left(1+\frac{1}{n}\right)\left(1-\frac{1}{(n+1)+\frac{(n+1)}{2}+...+1}\right)= 1$$
Then I thought using abel or dirichlet test. But I couldn't solve it.
 [/INST] First note that
$$
\sum\limits_{k = 1}^n {\frac{1}{k}}  \le 1 + \sum\limits_{k = 2}^n {\int_{k - 1}^k {\frac{{dt}}{t} } }  = 1 + \int_1^n {\frac{{dt}}{t}}  = 1 + \log n,
$$
for all $n\geq 1$. Then for all $N\geq 3$,
\begin{align*}
& \sum\limits_{n = 1}^N {\frac{1}{{n\sum\nolimits_{k = 1}^n {\frac{1}{k}} }}}  > \sum\limits_{n = 3}^N {\frac{1}{{n\sum\nolimits_{k = 1}^n {\frac{1}{k}} }}}  \ge \sum\limits_{n = 3}^N {\frac{1}{{n(\log n + 1)}}} \\ & \ge \sum\limits_{n = 3}^N {\frac{1}{{n(\log n + \log n)}}}  = \frac{1}{2}\sum\limits_{n = 3}^N {\frac{1}{{n\log n}}} 
\\ &
 \ge \frac{1}{2}\sum\limits_{n = 3}^N {\int_{n}^{n+1} {\frac{{dt}}{{t\log t}}} }  = \frac{1}{2}\int_3^{N+1} {\frac{{dt}}{{t\log t}}} \\ & = \frac{{\log \log( N+1) - \log \log 3}}{2} .
\end{align*}
This shows that the series diverges.
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