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.

First, We need a lemma.
Lemma:
For $n>1$, 
$$1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2^n-1} < n$$
This could be checked easily with induction.

To prove the question we use Cauchy Condensation Test, and see that its equal to show $$\sum_{k=1}^{\infty}1+\frac{1}{2}+\frac{1}{3}+ ... +\frac{1}{2^k}$$ is divergent.
It can be checked easily that:
$$\frac{1}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2^k}} > \frac{1}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2^{k+1}-1}}$$
Now we apply our lemma for $k\geq 2$ and see:
$$\frac{1}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2^k}} > \frac{1}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2^k}+...+\frac{1}{2^{k+1}-1}} > k+1$$
hence:
$$\sum_{k=1}^{\infty}{\frac{1}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2^k}}} =  \frac{1}{1+\frac{1}{2}} + \sum_{k=2}^{\infty}{\frac{1}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2^k}}}  > \sum_{k=3}^{\infty}{k}$$
Since the left hand side of inequality is divergent, the other side is so. Finally by Cauchy Condensation Test we see that the we have proved the divergence.
