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I am trying to learn about digamma function and it's uses. For example series, I found somewhere this solution: $$\begin{align*} \sum_{n=0}^{\infty} \frac{1}{(3n+2)\left ( 3n+3 \right )} &= \sum_{n=0}^{\infty} \left [ \frac{1}{3n+2} - \frac{1}{3n+3} \right ]\\ &=\frac{1}{3} \sum_{n=0}^{\infty} \left [ \frac{1}{n+ \frac{2}{3}} - \frac{1}{n+1} \right ]\\ &=\frac{1}{3} \left [ -\psi^{(0)} \left ( \frac{2}{3} \right ) + \psi^{(0)}(1) \right ]\\ &= \frac{\log 3}{2}- \frac{\pi}{6 \sqrt{3}} \end{align*}$$ I can prove the same result using a different method using integrals, but I am insterested about digamma function here, does this equality hold: $$-\psi(a) =\sum_{n=0}^{\infty} \frac{1} {n+a} ?$$ Now I know that the RHS diverges so this can't be true. My thought is that this hold if the series is not alone(we must have like in the solution from above two parts or more inside the series). Can you help me with a proof for this?

Also here:http://mathworld.wolfram.com/PolygammaFunction.html It's shown a general form for this, but of course this doesnt hold for digamma function right?

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  • $\begingroup$ 1. This series cannot represent the digamma function since it is divergent for every $a$. 2. For a representation of the digamma function as a series, please see the obvious. $\endgroup$ – Did Apr 22 '18 at 14:10
  • $\begingroup$ But isnt $\sum_{n=0}^{\infty} (\frac{1}{n+a}-\frac{1}{n+b})$ equal to $-\psi(a)+\psi(b)$? $\endgroup$ – Zacky Apr 22 '18 at 14:16
  • $\begingroup$ It is. And? (Hmmm... did you read the link I provided? No? Why?) $\endgroup$ – Did Apr 22 '18 at 14:17
  • $\begingroup$ so basically the first part is $-\psi(a) $ My question was how to prove that this holds if it is not alone. $\endgroup$ – Zacky Apr 22 '18 at 14:18
  • $\begingroup$ Maybe we have different definitions of "to read", then. $\endgroup$ – Did Apr 22 '18 at 14:23
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Weierstrass Formula: $$\Gamma(z)=\frac{e^{-\gamma z}}{z}\prod_{n=1}^\infty \frac{e^{\frac{z}{n}}}{1+\frac{z}{n}}$$

$$\log\Gamma(z)=-\gamma z-z+\sum_{n=1}^\infty\frac{z}{n}-\log(1+\frac{z}{n})$$

Differentiating:

$$\frac{\Gamma'(z)}{\Gamma(z)}=\psi(z)=-\gamma-1+\sum_{n=1}^\infty\frac{1}{n}-\frac{\frac{1}{n}}{1+\frac{z}{n}}$$ $$\psi(z)=-\gamma-1+\sum_{n=1}^\infty\frac{1}{n}-\frac{1}{z+n}$$

If $z\in\Bbb N^+$, this becomes a simple telescoping sum and $\psi(z)=-\gamma-1+H_z$ where $H_z$ is the $z$th harmonic number.

Furthermore, $\psi(z)\neq\sum_{n=0}^\infty\frac{1}{n+z}$ due to the divergence for any given $z$.

Just because $$\psi(z)-\psi(s)=(-\gamma-1+\sum_{n=1}^\infty\frac{1}{n}-\frac{1}{z+n})-(-\gamma-1+\sum_{n=1}^\infty\frac{1}{n}-\frac{1}{s+n})=\sum_{n=1}^\infty\frac{1}{s+n}-\frac{1}{z+n}$$

doesn't mean that $\psi(z)=\sum_{n=1}^\infty\frac{1}{z+n}$ and $\psi(s)=\sum_{n=1}^\infty\frac{1}{s+n}$.

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    $\begingroup$ I see, now I understand, basically when we substract gamma constant and the other part cancels out, thats why they are equal. Thank you! $\endgroup$ – Zacky Apr 22 '18 at 14:29
  • $\begingroup$ Glad I could help $\endgroup$ – aleden Apr 22 '18 at 14:31
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$$\sum_{n\geq 0}\frac{1}{n+a}$$ is blatantly divergent, the exploited identity was $$ \sum_{n\geq 0}\frac{1}{(n+a)(n+b)} = \frac{\psi(a)-\psi(b)}{a-b} $$ which comes from the Weierstrass product for the $\Gamma$ function.

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