I need to calculate the following sum with generating functions:
$$\sum_{k=1}^{n-1}\frac{1}{k(n-k)}$$
I tried:
$$\sum_{n \geq 0}(\sum_{k=1}^{n-1}\frac{1}{k}\cdot\frac{1}{n-k})z^k=\sum_{n \geq 0}(\sum_{k=0}^{n}\frac{1}{k+1}\cdot\frac{1}{n-k+1})z^k$$
The inner sum is a Cauchy product, therefore: $$(\sum_{n \geq 0}\frac{z^n}{n+1})^2$$
Now I'm stuck. How can I calculate a closed form from this sum?
Method 1: Let $$S(x) = \sum_{n\geq 0} \frac{x^{n}}{n+1}$$ then \begin{align} D_{x}\left( x \, S(x) \right) = \sum_{n\geq 0} x^{n} = \frac{1}{1-x} \end{align} which leads to $$S(x) = - \frac{\ln(1-x)}{x}.$$ Now \begin{align} \sum_{n=1}^{\infty} a_{n} \, t^{n} &= \sum_{n=1}^{\infty} \sum_{k=1}^{n-1} \frac{1}{k(n-k)} \, t^{n} \\ &= \sum_{n=1}^{\infty} \sum_{n=1}^{\infty} \frac{t^{n+k}}{n \, k} \\ &= \left( \sum_{n=1}^{\infty} \frac{t^{n}}{n} \right)^{2} = \left( - \ln(1-t) \right)^{2} = \ln^{2}(1-t) \\ &= 2 \, \sum_{n=2}^{\infty} \frac{H_{n-1}}{n} \, t^{n} \end{align} where
$$\sum_{n=1}^{\infty} H_{n} \, t^{n} = - \frac{\ln(1-t)}{1-t}$$ was used. This yields $$a_{n} = \sum_{k=1}^{n-1} \frac{1}{k (n-k)} = \frac{2 \, H_{n-1}}{n}.$$
Method 2: Consider $$\frac{1}{k\, (n-k)} = \frac{1}{n} \, \left( \frac{1}{k} + \frac{1}{n-k} \right)$$ for which \begin{align} a_{n} &= \sum_{k=1}^{n-1}\frac{1}{k(n-k)} \\ &= \frac{1}{n} \, \left[ \sum_{k=1}^{n-1} \frac{1}{k} + \sum_{k=1}^{n-1} \frac{1}{n-k} \right] \\ &= \frac{1}{n} \, \left[ H_{n-1} + \sum_{k=1}^{n-1} \frac{1}{k} \right]\\ &= \frac{2 \, H_{n-1}}{n}, \end{align} where $H_{n}$ is the Harmonic number defined by $H_{n} = \sum_{k=1}^{n} \frac{1}{k}$.