Generating function

Let $\mathcal{H}_n$ be the $n$ – th harmonic number. Prove that

$2 \sum_{n=1}^{\infty} \frac{\mathcal{H}_{n-1}}{n} x^n = \ln^2 (1-x) \quad \text{forall} \quad \left | x \right |< 1$

Solution

Recalling Cauchy’s product we have successively:

$\begin{align*} \ln^2 (1-x) &= \left ( \sum_{n=1}^{\infty} \frac{x^n}{n} \right ) \left ( \sum_{n=1}^{\infty} \frac{x^n}{n} \right ) \\ &= \sum_{n=1}^{\infty} \left ( \sum_{k=1}^{n-1} \frac{1}{k \left ( n-k \right )} \right )x^n \\ &= \sum_{n=1}^{\infty} \left ( \sum_{k=1}^{n-1} \frac{1}{n} \left ( \frac{1}{k} + \frac{1}{n-k} \right ) \right ) x^n \\ &= \sum_{n=1}^{\infty}\frac{1}{n} \left ( \sum_{k=1}^{n-1} \frac{1}{k} + \sum_{k=1}^{n-1} \frac{1}{n-k} \right ) x^n \\ &= \sum_{n=1}^{\infty} \frac{1}{n} \left ( \mathcal{H}_{n-1} + \mathcal{H}_{n-1} \right ) x^n \\ &= 2 \sum_{n=1}^{\infty} \frac{\mathcal{H}_{n-1}}{n} x^n \end{align*}$