1
$\begingroup$

I am trying to make sense of a proof that the poles of $\Gamma(z)$ are at $z=-n$ and have residue $\frac{(-1)^n}{n}$. The proof reduces $\Gamma(z)$ to the sum of an (entire) incomplete gamma function and a meromorphic part given by $$\sum_{n=0}^\infty\frac{(-1)^n}{n!(z+n)}$$ My issue is the expansion $$\frac{1}{z}-\frac{1}{1+z}+\frac{1}{2(z+2)}- ...$$ does not resemble a Laurent Series since each term is centered around a different point. Can someone explain why it is still valid to take the coefficient of $(z-z_0)^{-1}$ in this form in order to find the residue at $z=z_0$?

$\endgroup$
  • $\begingroup$ Where did you find this proof? $\endgroup$ – José Carlos Santos Oct 26 '18 at 10:49
  • $\begingroup$ Complex Analysis lecture slides $\endgroup$ – P Collier Oct 26 '18 at 10:52
  • $\begingroup$ See en.wikipedia.org/wiki/Gamma_function#Residues $\endgroup$ – Nosrati Oct 26 '18 at 11:09
  • $\begingroup$ Oh yes I see now that you can simply recombine the terms without worrying about a Laurent series, thanks for pointing me in the right direction! $\endgroup$ – P Collier Oct 26 '18 at 11:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.