<s>[INST] 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?
 [/INST] Expanding the sum: $$\sum_{n\ge 0}H(n)z^n=z+\left(1+\dfrac{1}{2}\right)z^2+\left(1+\dfrac{1}{2}+\dfrac{1}{3}\right)z^3+...=\frac{z}{1-z}+\frac{1}{2}z^2+\left(\dfrac{1}{2}+\dfrac{1}{3}\right)z^3+...=\frac{z}{1-z}+\frac{z^2}{2(1-z)}+...$$Where $H(n)$ is the $n$th harmonic number. Using induction, you could prove that: $$\sum_{n\ge 0}H(n)z^n=\sum_{n\ge 0}\frac{z^n}{n(1-z)}$$ Can you take it from here? (The series for $\ln (1-z)$ is the sum of $z^n/n$ from $n$ to $\infty$).
Note: I am assuming that the sums converge.
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