Let $g(x) = \frac{1}{x} + \sum_{n=1}^{+\infty}\left(\frac{1}{x+n} + \frac{1}{x-n}\right)$. Show that for all $x \in \mathbb{R} \backslash \mathbb{Z}$, we have $$g\left(\frac{x}{2}\right) + g\left(\frac{1+x}{2}\right) = 2g(x)$$
Let $g(x) = \frac{1}{x} + \sum_{n=1}^{+\infty}\left(\frac{1}{x+n} + \frac{1}{x-n}\right)$. Show that for all $x \in \mathbb{R} \backslash \mathbb{Z}$, we have
$$g\left(\frac{x}{2}\right) + g\left(\frac{1+x}{2}\right) = 2g(x)$$