Convergence of harmonic series multiplied by partial sums of harmonic series How can I show if the following series is convergent or divergent
$$\sum_{n=1}^\infty \frac1n \left ( 1 + \frac12 + \frac13 + \cdots + \frac1n \right) = \sum_{n = 1}^\infty\frac{1}{n}\sum_{k = 1}^n\frac{1}{k}$$
If it were a sequence I could have easily used Cauchy first theorem. 
 A: The series diverges. You have that $$1+1/2+\cdots \geq 1$$ and thus you only have to look at $$\sum_{n=1}^\infty n^{-1}$$ which is a lower bound to your series. The latter series is the harmonic series that is well-known to diverge.
A: Admitting that you can use harmonic number, let us consider $$S_p= \sum_{n = 1}^p\frac{1}{n}\sum_{k = 1}^n\frac{1}{k}=\sum_{n = 1}^p\frac{H_n}{n}=\frac{\left(H_p\right){}^2}{2}-\frac{H_p^{(2)}}{2}-\psi ^{(1)}(p+1)+\frac{\pi ^2}{6}$$ Considering the asymptotics $$H_p=\gamma +\log \left(p\right)+\frac{1}{2 p}-\frac{1}{12
   p^2}+O\left(\frac{1}{p^4}\right)$$ $$H_p^{(2)}=\frac{\pi ^2}{6}-\frac{1}{p}+\frac{1}{2 p^2}-\frac{1}{6
   p^3}+O\left(\frac{1}{p^4}\right)$$ $$\psi ^{(1)}(p+1)=\frac{1}{p}-\frac{1}{2 p^2}+\frac{1}{6 p^3}+O\left(\frac{1}{p^4}\right)$$ you should end with $$S_p=\frac{\pi ^2}{12}+\frac{\gamma ^2}{2}+\gamma  \log
   \left(p\right)+\frac{1}{2} \log ^2\left(p\right)+\frac{\log \left({p}\right)+\gamma -1}{2 p}+O\left(\frac{1}{p^2}\right)$$ which shows the result.
Computing exactly for $p=10$, we have $$S_{10}=\frac{32160403}{6350400}\approx 5.06431$$ while the above expression gives $ \approx 5.06308$.
A: Comparison:
$$
\sum_{n=1}^\infty \frac1n \left( 1 + \frac12 + \frac13 + \cdots + \frac1n \right) \ge \sum_{n=1}^\infty \frac 1 n = \infty.
$$
A: EDIT: I answered the question as it was originally stated, but I think you intended to ask a different question from the one you actually asked.
This is the mean of $S_n = \{1, \frac{1}{2}, \dots, \frac{1}{n}\}$. That's clearly bounded by $0$ and $1$, and also is clearly decreasing because to move from $S_n$ to $S_{n+1}$ we don't remove anything from $S_n$ but we add a smaller thing (namely $\frac{1}{n+1}$).
