Re: Summation without Mathematica

*To*: mathgroup at smc.vnet.net*Subject*: [mg27440] Re: [mg27416] Summation without Mathematica*From*: BobHanlon at aol.com*Date*: Sun, 25 Feb 2001 20:55:44 -0500 (EST)*Sender*: owner-wri-mathgroup at wolfram.com

You can simplify the problem by generalizing the problem (make it harder). First let Mathematica solve the problem so we have an answer to check against. soln = Sum[x^3 * (7/8)^x, {x, 1, Infinity}] 18872 A more general form of the problem is genSoln = Sum[x^3 * y^x, {x, 1, Infinity}]//Simplify (y*(y^2 + 4*y + 1))/(y - 1)^4 Verifying this for the case at hand, (genSoln /. y -> 7/8) == soln True An easy way to derive the general result is by using the differential operator y*D[#, y]& y*D[y^x, y] == (y*D[#, y]& [y^x]) == x*y^x True Applying the operator three times, Nest[y*D[#, y]&, y^x, 3] == y*D[y*D[y*D[y^x, y], y], y] == x^3 * y^x True Substituting and pulling the operator outside the summation, Sum[y^x, {x, 1, Infinity}] -(y/(y - 1)) Application of the differential operator to this sum gives the desired result Simplify[Nest[y*D[#, y]&, %, 3]] == genSoln True Bob Hanlon In a message dated 2001/2/25 1:08:56 AM, mor3752 at alltel.net writes: >Can someone help me to evaluate the sum from x=1 to infinity for the >expression x^3(7/8)^x. I know the formulas that will evaluate either alone. >But the product is giving me difficulty.