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MRB Constant I

Marvin Ray Burns 550
April 08 2010
0

Consider  S=sum((-1)^n*(n^(1/n)-a),n=1..infinity).
S=1/2*(a+2m-1), where m is the MRB constant.

Let m = MRB constant = sum((-1)^n*(n^(1/n)-1),n=1..infinity)

How to see this numerically:
a    sum((-1)^n*(n^(1/n)-a),n=1..infinity)
0.1 1/20*(20m-9) = m-9/20 = 1/2*(.1+2m-1)
0.2 1/20*(20m-8) = m-8/20 = 1/2*(.2+2m-1)
0.3 1/20*(20m-7) = m-7/20 = 1/2*(.3+2m-1)
0.4 1/20*(20m-6) = m-6/20 = 1/2*(.4+2m-1)
0.5 ...
0.6 ...
0.7 ...
0.8 1/20*(20m-...
0.9 1/20*(20m-1) = m-1/20 =1/2*(.9+2m-1)

Checking for known values for a:
a    sum((-1)^n*(n^(1/n)-a),n=1..infinity)
m                        = 1/2*(m+2m-1)=1/2(3m-1) TRUE!
2                        = 1/2*(2+2m-1) = 1/2(2m+1) TRUE!

According to Maple, evalf(sum((-1)^n*(n^(1/n)-2.),n=1..infinity)) gives 0.6878596425.

 

Maple uses the Levin's u-transform which can assign values to divergent series. Likewise, Euler found a value to the, divergent, first general infinite series, as seen in http://www.math.dartmouth.edu/~euler/docs/translations/E352.pdf .

 

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