TITLE: Convergence of harmonic series multiplied by partial sums of harmonic series
QUESTION [8 upvotes]: 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.

REPLY [6 votes]: 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$.