Re: An exact 1-D summation challenge - 3
From: Dana (delouis_at_bellsouth.net)
Date: 02/27/05
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Date: Sun, 27 Feb 2005 09:49:22 -0500
Hi. I couldn't get the program Mathematica to simplify your equation.
However, I did find an interesting "problem" with something.
See if you find this interesting...
Here's your equation:
equ=((HarmonicNumber[1/2+n]-2*HarmonicNumber[n])*(1+n)*Gamma[3/2+n])/(1+n)!^2;
I was trying to split it into two parts, to see if I could work with each
part by itself:
t=Apart[equ]
-((2*(1+n)*Gamma[3/2+n]*HarmonicNumber[n])/(1+n)!^2) +
((1+n)*Gamma[3/2+n]*HarmonicNumber[1/2+n])/(1+n)!^2
The first part did not simplify, but the second half did. It was the second
part that I had trouble with.
fx = ((1 + n)*Gamma[3/2 + n]*HarmonicNumber[1/2 + n])/(1 + n)!^2;
I wish to see how the number convergences as n gets larger...
{N[Sum[fx, {n, 0, 10}]],
N[Sum[fx, {n, 0, 100}]],
N[Sum[fx, {n, 0, 1000}]]
}
{2.0682716719330347, 2.0682716957999676, 2.0682716957999676}
Pretty safe to say that the answer is about 2.068...
If I take it out to infinity, I get the following equation, which has your
BesselI[0,1/2] in it...
k = Sum[((1 + n)*Gamma[3/2 + n]*HarmonicNumber[1/2 + n])/(1 + n)!^2, {n, 0,
Infinity}]
(1/2)*(EulerGamma*Sqrt[E*Pi]*BesselI[0, 1/2] +
EulerGamma*Sqrt[E*Pi]*BesselI[1, 1/2])
However, this number is 1.11444...
This doesn't seem to match the estimated numerical value, so I think this is
wrong.
N[k]
1.1144414478337052
So, I don't know what to think.
Good luck. :>)
-- Dana "Vladimir Bondarenko" <vb@cybertester.com> wrote in message news:1109442791.936042.151030@l41g2000cwc.googlegroups.com... > Thank you! > > Actually, I meant the exact symbolic solution. > > Something like > > 1/2*Pi^(1/2)*(2*exp(1/2)*BesselI(0,1/2)+... > > I do realize that Mathematica cannot sum it, just like Maple. > > I can sum it... but I am a human, not a CAS :) > > > Vladimir >
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