# Evaluating another symbolic sum

Here I asked about a symbolic sum, and received three very insightful replies (from: მამუკა ჯიბლაძე, Carl Woll and Dr. Wolfgang Hintze) which did the trick. (Thank you again!)

Currently I am trying to solve a different symbolic sum, which in a general format is:

Sum[(-1 + p)^s p^t, {p, 2, m}, Assumptions -> (s | t) \[Element] Integers]


Each factor alone would lead to Harmonic numbers or Zeta functions, respectively. However, in this form, the sum returns unevaluated. I tried all of the three methods suggested previously (including integrating instead of differentiating, in one of the approaches), to no avail.

More specifically, the sum is:

Sum[(-1 + p)^(-1 + i - k) p^(-i + n), {p, 2, m}]


with m and n any positive integers, i values running from 1 to n-1 and k from 0 to i-1.

I also tried replacing the (p-1)^s with its own sum (with the binomial coefficient, etc.) and switching the order of the summations -- also to no avail.

Thank you for any direction, and for your time.

• Alternative: HoldForm[Sum[(-1 + p)^s*p^t, {p, 2, m}] == -HypergeometricPFQ[{-s}, {}, 1] + Sum[(-1)^j Binomial[s, j] HarmonicNumber[m, j - s - t], {j, 0, s}] == Integrate[-(( E^(-x - m x) (E^x - E^(m x)) x^(-1 - s - t) Hypergeometric1F1Regularized[-s, -s - t, x])/(-1 + E^x)), {x, 0, Infinity}](*integral for: Re[s+t]<0 and m>2 *)],but I can't solve :( – Mariusz Iwaniuk Nov 27 '19 at 18:24
• Mariusz Iwaniuk -- thank you for this attempt. I was able to get the first part, as I reported, writing out the (p-1)^s as a summation with the binomial coefficients, etc. How did you arrive at the integral? – Aharon Naiman Nov 27 '19 at 22:14
• Using very old trick:$$\sum _{n=2}^m f(n)=\int_0^{\infty } \left(\mathcal{L}_n^{-1}[f(n)](x)\right) \sum _{n=2}^m \exp (-n x) \, dx$$, then: Integrate[ InverseLaplaceTransform[(-1 + p)^s*p^t, p, x]* Sum[Exp[-p x], {p, 2, m}], {x, 0, Infinity}] – Mariusz Iwaniuk Nov 28 '19 at 16:05