Calculating Harmonic Sums using Residues

[Amad27]

I posted the same question on Math Stackexchange: http://math.stackexchange.com/quest...g-harmonic-sums-with-residues/1085248#1085248

The answer there using complex analysis is great. I had questions, which Id like to get advice on here.

(1) How did he get the laurent series at negative integers? I can never understand how that works. Doesnt he need to find the coefficient a_n? Which is defined as a contour integral? How does the Laurent series work for digamma?

(2) How did he get the partial sum involving the q−1 in the upper index? And finally, I don't understand the step from: "which yields.. this implies that.." How did he make the transformation? It didnt change anything?

 

[jvicens]

Hello,

Thank you for sharing your questions about the answer on Math Stackexchange. I will do my best to explain the steps in the solution and address your concerns.

(1) In order to obtain the Laurent series at negative integers, the author used the fact that the digamma function can be expressed as a sum of residues of the logarithmic derivative of the gamma function. The coefficients of the Laurent series are then related to these residues through the residue theorem. The author did not explicitly calculate the coefficients, but instead used the fact that the residue at a pole of order n is given by the (n-1)th derivative of the function at that pole divided by (n-1)!. In this case, the pole at z=0 is of order n+1, so the coefficient a_n is given by the (n+1)th derivative of the digamma function at z=0 divided by n!.

(2) The author obtained the partial sum with the upper index q-1 by using the property of the digamma function that it can be expressed as a sum of logarithmic derivatives of the gamma function. This property allows us to rewrite the original sum as a telescoping series, where each term cancels out with the next term. The transformation from the original sum to the telescoping series did not change anything, it simply allowed the author to simplify the sum and obtain a closed form expression.

I hope this helps to clarify some of the steps in the solution. If you have any further questions, please don't hesitate to ask. Thank you for your interest in this topic and for reaching out for clarification.