Is there a way to re-write $\sum_{n=1}^{\infty}{\left(\sum_{k=1}^{n}{\frac{1}{k}}\right)}z^n$ 
Is there a way to re-write
  $\sum_{n=1}^{\infty}{\left(\sum_{k=1}^{n}{\frac{1}{k}}\right)}z^n$

I was thinking that I could use the cauchy product, but therefore I have to achieve this:$$\left(\sum_{n=0}^{\infty}{a_n}\right)\cdot \left(\sum_{n=0}^{\infty}{b_n}\right)$$ Any ideas?
 A: Consider the function $f(z)=\frac{-\ln(1-z)}{1-z}.$ By the Cauchy product we have
$$\begin{align*}
f(z) &= \left( \sum_{n=1}^\infty\frac{z^n}{n}\right )\left( \sum_{n=0}^\infty z^n\right )\\ 
 &= \sum_{n=1}^\infty\left(\sum_{k=1}^n\frac{1}{k} \right )z^n\\ 
 &= \sum_{n=1}^\infty H_n z^n.\\ 
\end{align*}$$
A: Well, the interior sum can be expressed as the $n^\text{th}$ Harmonic Number $H_n$ which thus leaves us with 
$$\sum_{n\geqslant1}H_n z^n$$
Or in other words: a generating function for the Harmonic Numbers. This, on the other hand, is a well-known one. 
Consider the MacLaurin Expansions of $-\log(1-z)$ and $\frac1{1-z}$ multiplied as direct discrete convolution, i.e. as Cauchy-Product. Doing so we obtain
$$-\frac{\log(1-z)}{1-z}=\left(\sum_{n\geqslant1}\frac{z^n}n\right)\left(\sum_{n\geqslant0}z^n\right)=\sum_{n\geqslant1}\left(\sum_{k=1}^n\frac1k\right)z^n=\sum_{n\geqslant1}H_n z^n$$

$$\therefore~\sum_{n\geqslant1}\left(\sum_{k=1}^n\frac1k\right)z^n~=~-\frac{\log(1-z)}{1-z}$$

A: Interchange order (valid inside radius of convergence) and get
$$
\sum_{n=1}^\infty\left(\sum_{k=1}^n\frac1k\right) z^n
=\sum_{k=1}^\infty\frac1k\sum_{n=k}^\infty z^n
$$
which allows you to calculate the sum for sufficiently small $\lvert z\rvert$.
To find the radius of convergence, recall $\sum_{k=1}^n\frac1k\sim\log n$ and use Cauchy-Hadamard or similar.
