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MathGroup Archive 2001

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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.


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