A closed form expression of $\sum_{n \ge 0} \biggl( \sum_{k=1}^n \frac{1}{k} \biggr)z^n$ 
I am working on the following exercise: Use the identity $\frac{1}{1-z} = \sum_{n \ge 0} z^n$ and elementary operations on power series (addition, multiplication, integration, differentiation) to find closed form expressions of the following power series:

*

*$\sum_{n \ge 0} n^2z^n$

*$\sum_{n \ge 0} \frac{n}{n+1}z^n$

*$\sum_{n \ge 0} \biggl( \sum_{k=1}^n \frac{1}{k} \biggr)z^n.$

I have solved the first two points in the following way:

*

*Using differentiation on the identity $\frac{1}{1-z} = \sum_{n \ge 0} z^n$ we obtain

$$\frac{1}{(1-z)^2} = \sum_{n \ge 0} nz^{n-1}$$
and multiplying the above equality with $z$ yields $\frac{z}{(1-z)^2} = \sum_{n \ge 0} nz^{n}$ and again using differentiation and multplication with $z$ on this term yields $\frac{z^2+z}{(1-z)^3} = \sum_{n \ge 0} n^2z^n$ as desired.


*Integrating the term $\frac{z}{(1-z)^2} = \sum_{n \ge 0} nz^{n}$ from before yields

$$\ln(\vert 1-z \rvert)+\frac{1}{1-z} = \sum_{n \ge 0} \frac{n}{n+1}z^{n}$$
I realise that the third part is probably somehow related to the exercises before, but I do not see how. Could you please help me?

Where these series converge,
\begin{align}
&\Bigl( \frac11 \Bigr) \, z 
+ \Bigl( \frac11 + \frac12 \Bigr) \, z^2 
+ \Bigl( \frac11 + \frac12 + \frac13 \Bigr) \, z^3 
+ \cdots \\
&\qquad = \frac11 \Bigl( z + z^2 + z^3 + \cdots \Bigr) 
+ \frac12 \Bigl( z^2 + z^3 + \cdots \Bigr) 
+ \frac13 \Bigl( z^3 + \cdots \Bigr) \\ 
&\qquad = \frac11 z \Bigl(1 + z + z^2 + \cdots \Bigr) 
+ \frac12 z^2 \Bigl(1 + z + z^2 + \cdots \Bigr) 
+ \frac13 z^3 \Bigl(1 + z + z^2 + \cdots \Bigr) \\ 
&\qquad = \Bigl( \frac11 z + \frac12 z^2 + \frac13 z^3 + \cdots \Bigr) 
\Bigl(1 + z + z^2 + \cdots \Bigr) \\
&\qquad = -\frac{\ln(1 - z)}{1-z}.
\end{align}
