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I dont see the transition from $(-z)^k$ in the fist sum to the transition to $(z+2)^k$ in the second sum?

How is that derived?

Amad27
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1 Answers1

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From $$\psi(1-z)=-\gamma-\sum_{k\geq 1}\zeta(k+1) z^{k} $$ and $\psi(2-z)=\psi(1-z)+\frac{1}{1-z}$ it follows that: $$\psi(2-z) = \frac{1}{1-z}-\gamma-\sum_{k\geq 1}\zeta(k+1) z^{k}=(1-\gamma)+\sum_{k\geq 1}(1-\zeta(k+1))\,z^k\tag{1}$$ now replace $z$ with $z+2$ to get the stated identity.

Jack D'Aurizio
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  • Okay then how can we use this to get the residue of $g(z)$ from here: http://math.stackexchange.com/questions/1102571/find-the-residue-at-z-2-for-gz-frac-psi-zz1z23/1102621?noredirect=1#comment2248081_1102621 at $z=-2$? – Amad27 Jan 13 '15 at 14:18
  • @Amad: OK, we can use this lemma and expand $\frac{1}{z+1}$ as $-\frac{1}{1-(z+2)}=-\sum_{n\geq 0}(z+2)^n$, so in that case you can compute the residue by multiplying two taylor series. But this is essentially the same as computing a couple of derivatives like I did. – Jack D'Aurizio Jan 13 '15 at 14:21
  • This relation seems to imply $\zeta^(1)=\gamma$ as the normalized value of the Riemann $\zeta$ function in $x=1$, which makes perfect sense when considering the extension $\zeta^(x)=\lim_{h\to0}\dfrac{\zeta(x+h)+\zeta(x-h)}2$ , and the fact that $\dfrac{\zeta\big(1^+\big)+\zeta\big(1^-\big)}2=\gamma$. – Lucian Jan 13 '15 at 17:24