Sum of Sequence of Harmonic Numbers/Proof 3

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

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

Let $\map G z$ be the generating function for $\sequence {a_n}$.

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

Then:

From Generating Function for Sequence of Partial Sums of Series, $\dfrac 1 {1 - z} \map G z$ is the generating function for $\sequence {b_n}$ where:
 * $b_n = \ds \sum_{k \mathop = 0}^n H_k$

and so:
 * $\dfrac 1 {\paren {1 - z}^2} \map \ln {\dfrac 1 {1 - z} } = \ds \sum_{n \mathop \ge 0} \paren {\sum_{k \mathop = 0}^n H_k} z^n$

From Generating Function for Natural Numbers:
 * $\dfrac 1 {\paren {1 - z}^2} = \ds \sum_{n \mathop \ge 0} \paren {n + 1} z^n$

That is:
 * $\map {G'} z = \ds \sum_{n \mathop \ge 0} \paren {\sum_{k \mathop = 0}^n H_k + \paren {n + 1} } z^n$

Now we have:

Equating coefficients of $z^n$ in these two expressions for $\map {G'} z$:


 * $\ds \sum_{k \mathop = 0}^n H_k + \paren {n + 1} = \paren {n + 1} H_{n + 1}$

The result follows.