Digamma series Representation I was reading about where the series representation of digamma is proved and its states: $$-\frac{1}{x}-\gamma +\sum_{n=1}^{+\infty}\left(\frac{1}{n}-\frac{1}{n+x}\right) = -\gamma +\sum_{n=1}^{+\infty}\left (\frac{1}{n}-\frac{1}{n+x-1}\right) $$ I don't see where the $n+x-1$ comes from though any idea?

A good starting point is the Weierstrass product for the $\Gamma$ function: $$ \Gamma(x) = \frac{e^{-\gamma x}}{x}\prod_{n\geq 1}\left(1+\frac{x}{n}\right)^{-1}e^{x/n}\tag{1} $$ By considering $\frac{d}{dx}\log(\cdot)$ of both sides, $$ \psi(x) = -\frac{1}{x}-\gamma+\sum_{n\geq 1}\frac{x}{n(n+x)}=-\frac{1}{x}-\gamma+\sum_{n\geq 1}\left(\frac{1}{n}-\frac{1}{n+x}\right).\tag{2} $$ Since $x\,\Gamma(x)=\Gamma(x+1)$, that can be written also as: $$ \sum_{n\geq 1}\left(\frac{1}{n}-\frac{1}{n+x}\right) = \gamma+\psi(x+1).\tag{3}$$ Anyway, your concern is simply addressed by noticing that $$ \sum_{n\geq 1}\left(\frac{1}{n+x-1}-\frac{1}{n+x}\right)=\frac{1}{x}\tag{4}$$ since the LHS is a telescopic series.