Re: Help on Infinite Series

*To*: mathgroup at smc.vnet.net*Subject*: [mg2451] Re: Help on Infinite Series*From*: Bob Hanlon <hanlon at pafosu2.hq.af.mil>*Date*: Wed, 8 Nov 1995 23:45:37 -0500

Using Mathematica: Needs["Algebra`SymbolicSum`"] PROBLEM: Find Sum[k^m, {k, 1, n}] f1[n_, m_] := Evaluate[SymbolicSum[k^m, {k, 1, n}]]; f1[n, m] Zeta[-m] - Zeta[-m, 1 + n] Table[f1[n, m], {m, 0, 4}] // TableForm n 1 -(--) - Zeta[-1, 1 + n] 12 -Zeta[-2, 1 + n] 1 --- - Zeta[-3, 1 + n] 120 -Zeta[-4, 1 + n] Except for the trivial case (m = 0), the results are not readily understandable since they are left in the form of generalized Zeta functions. However, an alternate -- indirect -- approach provides the results in a more useful form. Generalizing the sum to a power series in x: g1[n_, m_, x_] := Evaluate[SymbolicSum[k^m * x^k, {k, 1, n}]] g1[n, m, x] n x (LerchPhi[x, -m, 1] - x LerchPhi[x, -m, 1 + n]) g1[n, m, 1] Zeta[-m] - Zeta[-m, 1 + n, IncludeSingularTerm -> False] This is the same result; however, since, x * D[g1[n, m, x], x] = Sum[k^(m+1) * x^k, {k, 1, n}] = g1[n, m+1, x] then, g1 can be defined recursively. The starting point for the recursion is g1[n, 0, x] // Simplify n x (-1 + x ) ----------- -1 + x g2[n_, 0, x_] := x * (1 - x^n) / (1 - x); g2[n_, m_, x_] := g2[n, m, x] = Simplify[x * D[g2[n, m-1, x], x]]; mmax = 7; Table[g2[n, m, x], {m, 0, mmax}]; (* long output suppressed *) f2[n_, m_] := f2[n, m] = Limit[g2[n, m, x], x->1]; Table[f2[n, m] // Factor, {m, 0, mmax}] // TableForm n n (1 + n) --------- 2 n (1 + n) (1 + 2 n) ------------------- 6 2 2 n (1 + n) ----------- 4 2 (-1 - n) n (1 + 2 n) (1 - 3 n - 3 n ) ------------------------------------- 30 2 2 2 n (1 + n) (-1 + 2 n + 2 n ) ----------------------------- 12 3 4 n (1 + n) (1 + 2 n) (1 - 3 n + 6 n + 3 n ) ------------------------------------------- 42 2 2 2 3 4 n (1 + n) (2 - 4 n - n + 6 n + 3 n ) ---------------------------------------- 24 This agrees with published results (e.g., Hansen, Eldon R.; A Table of Series and Products, Prentice-Hall, New Jersey, 1975, Section 5.14) Checking: And @@ Flatten[ Table[f1[n, m] == f2[n, m], {n, 0, 7}, {m, 0, 7}]] True > From: mfischer at trincoll.edu To: mathgroup at smc.vnet.net > Newsgroups: sci.math.symbolic > Subject: Help on Infinite Series > Date: Mon, 6 Nov 1995 19:38:41 > > I know that the summation of n squared can be described as > > n(n+1)(2n+1) / 6 > > does anyone know how to prove this? > > I know that the difference in the differences in the series is > > 2n+1 but I don't know how to get the rest > > Also how would you describe n-cubed? > > the difference of the difference of the differences is > > 6(n+1) but I don't know how to encorporate that > > any help would be greatly appreciated > > please mail any help to me at fischer at micro5.cs.trincoll.edu > > thanks in advance > > matt