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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?

Jack D'Aurizio
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jake walsh
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  • The correct formula is $\psi(s) = \frac{\Gamma'(s)}{\Gamma(s)} = -\gamma + \sum_{n=0}^\infty \frac{1}{n+1}-\frac{1}{n+s}$. The proof is that $\frac{\Gamma'(s)}{\Gamma(s)}-\sum_{n=0}^\infty \frac{1}{n+1}-\frac{1}{n+s}$ is a bounded entire function, so by Liouville's theorem it is constant. And the constant is by definition $-\gamma$. Showing it is the same $\gamma$ as in $\gamma = \lim_{n \to \infty} (\sum_{k=1}^n \frac{1}{k})-\ln n$ requires a little more work. – reuns Oct 16 '16 at 23:09
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    sorry but this doesn't answer my question how the -1/x goes into the sum – jake walsh Oct 17 '16 at 22:57
  • you didn't read carefully what I wrote – reuns Oct 18 '16 at 13:22

1 Answers1

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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.

Jack D'Aurizio
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