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MRB constant R

Marvin Ray Burns 550 Maple
January 01 2012
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Consider*the*partial*sumsof S(N)=(∑)(-1)^(n)*(n^(1/(n))-1).

The MRB constant is the upper limit point of the partial sums of S according to http://en.wikipedia.org/wiki/MRB_constant .

According to Maple,

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sum((-1)^n*(n^(1/n)-1), n = 1 .. N) = (1/2)*(-1)^(N+1)+1/2+sum((-1)^n*n^(1/n), n = 1 .. N)

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Look*at*the*first*summand:

`$`(evalf(eval((1/2)*(-1)^(N+1)+1/2, N = x)), x = 1 .. 10) = 1.000000000, 0., 1.000000000, 0., 1.000000000, 0., 1.000000000, 0., 1.000000000, 0.

The first summand alternates between 0 and1. When N is even the first summand is 0.

 

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Is this true as n gets large?

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evalf(eval(sum((-1)^n*(n^(1/n)-1), n = 1 .. N), N = 1000)) = .191324039

evalf(eval(0+(sum((-1)^n*n^(1/n), n = 1 .. N)), N = 1000)) = .191324039

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What about at N = infinity?

 

evalf(sum((-1)^n*(n^(1/n)-1), n = 1 .. infinity)) = .1878596425

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evalf(0+(sum((-1)^n*n^(1/n), n = 1 .. infinity))) = -.3121403575NULL

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Interestingly the first summand is 1/2 at N =infninty.

evalf(1/2+sum((-1)^n*n^(1/n), n = 1 .. infinity)) = .1878596425``

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