Sum of Sequence of Harmonic Numbers/Proof 3

Proof
Let $\left \langle {a_n}\right \rangle$ be the sequence defined as:
 * $\forall n \in \N_{> 0}: a_n = H_n$

where $H_n$ denotes the $n$th harmonic number.

Let $G \left({z}\right)$ be the generating function for $\left \langle {a_n}\right \rangle$.

From Generating Function for Sequence of Harmonic Numbers:
 * $G \left({z}\right) = \dfrac 1 {1 - z} \ln \left({\dfrac 1 {1 - z} }\right)$

Then:

From Generating Function for Sequence of Partial Sums of Series, $\dfrac 1 {1 - z} G \left({z}\right)$ is the generating function for $\left \langle {b_n}\right \rangle$ where:
 * $b_n = \displaystyle \sum_{k \mathop = 0}^n H_k$

and so:
 * $\dfrac 1 {\left({1 - z}\right)^2} \ln \left({\dfrac 1 {1 - z} }\right) = \displaystyle \sum_{n \mathop \ge 0} \left({\sum_{k \mathop = 0}^n H_k}\right) z^n$

From Generating Function for Natural Numbers:
 * $\dfrac 1 {\left({1 - z}\right)^2} = \displaystyle \sum_{n \mathop \ge 0} \left({n + 1}\right) z^n$

That is:
 * $G' \left({z}\right) = \displaystyle \sum_{n \mathop \ge 0} \left({\sum_{k \mathop = 0}^n H_k + \left({n + 1}\right)}\right) z^n$

Now we have:

Equating coefficients of $z^n$ in these two expressions for $G' \left({z}\right)$:


 * $\displaystyle \sum_{k \mathop = 0}^n H_k + \left({n + 1}\right) = \left({n + 1}\right) H_{n + 1}$

The result follows.