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[MUG] bug in sum?
| [MUG] bug in sum? |
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Author: Max_joern
Posted: Wed, 22 May 2002 10:33:19 +0200
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>> From: "max_joern"
Dear MUG,
I recently stumbled over this sum: (I'm using Maple 7)
Sum((k+1)*ee^k,k = 3 .. infinity). This should converge for
abs(ee)<1. We know that the sum from 0 to infinity equals in that
case 1/(1-ee)^2 which is checked by maple:
> sum((k+1)*(ee)^k,k=0..infinity);
1/((ee-1)^2)
(no convergence check b.t.w.)
If we plug in ee=-0.5 we obtain:
> subs(ee=-0.5,%);
0.4444444
Now Sum((k+1)*ee^k,k = 3 .. infinity) should be Sum((k+1)*ee^k,k = 0
.. infinity) - Sum((k+1)*ee^k,k = 0 .. 2), right? This we can
calculate for the case ee=-0.5 to be equal to 0.4444444-0.75=
-.3055555556. But here the surprise: if we use maple to calculate the
sum from 3 to infinity with arbitrary value of ee first and then
substitute ee=-0.5 we get:
> sum((k+1)*(ee)^k,k=3..infinity);
4*ee^3*(-9/ee^(1/2)/(-ee+1)^2*LegendreP(1,-3,(ee+1)/(-ee+1))-3/2*1/(ee-1)^5*(-ee+1)^2/ee+1/(ee-1)^5*(-ee+1)^2/ee^2-1/4*1/(ee-1)^5*(-ee+1)^2/ee^3-1/4*1/ee^3*(-ee+1))
> evalf(subs(ee=-0.5,%));
-.1666666667-0.*I
which is clearly different from the expected result. I assume that
already the expression with the associated Legendre functions is wrong
but I'm not sure. We get the expected result though if we take the
limit:
> sum((k+1)*(ee)^k,k=3..n);
ee^(n+1)*(-2+(n+1)*ee-n)/(ee-1)^2-ee^3*(-4+3*ee)/(ee-1)^2
>limit(subs(ee=-0.5,%),n=infinity);
-.3055555556
I suppose, this is a bug?
Thanks for a clarification,
Jorn Justiz
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