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

 

Concerning the following divergent and convergent series, we see that

sum((-1)^n*(n^(1/n)-x), n = 1 .. infinity)=(1/2)*x-1/2+MRB*constant

and

sum((-1)^n *(n^((x)/(n))-x),n=1..infinity) -sum((-1)^n*(n^(x/n)-1), n = 1 .. infinity)= (1/2)*x-1/2

 

 

The MRB constant is  defined as follows. Consider the sequence of partial sums defined by 
S(n)= sum((-1)^n*n^(1/n),n=1..infinity) 
the sequence has two limit points at 0.187859... and 0.187859..-1. 
The upper limit point is sometimes known as the MRB constant after my initials. 
(See Sloane, N. J. A. Sequences A037077 in "The On-Line Encyclopedia of Integer Sequences.")

Here we will look at the family of divergent and convergent series related to the MRB constant; 
in particular  

g(x)=(∑) (-1)^n (n^((1)/(n))-x), f(x)=(∑) (-1)^n (n^((x)/(n))-1),
 and s(x)=(∑) (-1)^n (n^((x)/(n))-x).   
Interstingly, we will see  that perhaps g(x)=1/2*x-(1/2-MRB constant) and s(x)-f(x) = 1/2*x-1/2."
First we will graph g(x) and see what closed form best describes it.

g:=x-> evalf(sum((-1)^n*(n^(1/n)-x), n = 1 .. infinity));

f:=x->evalf(sum((-1)^n*(n^(x/n)-1), n = 1 .. infinity));

s:=x->evalf(sum((-1)^n*(n^(x/n)-x), n = 1 .. infinity));

Download sept042011c.mw for the rest of the document.

 

Accordingly,

sum((-1)^n*(n^(1/n)-x), n = 1 .. infinity)  minus sum((-1)^n *(n^((x)/(n))-x),n=1..infinity) plus sum((-1)^n*(n^(x/n)-1), n = 1 .. infinity) equals  sum((-1)^n*(n^(1/n)-1), n = 1 .. infinity).

As shown in Maple by

combine(sum((-1)^n*(n^(1/n)-x), n = 1 .. infinity)-(sum((-1)^n*(n^(x/n)-x), n = 1 .. infinity))+sum((-1)^n*(n^(x/n)-1), n = 1 .. infinity));

combine(sum((-1)^n*(n^(1/n)-x), n = 1 .. infinity)-(sum((-1)^n*(n^(x/n)-x), n = 1 .. infinity))+sum((-1)^n*(n^(x/n)-1), n = 1 .. infinity))

 

simplify((-n^(x/n)+x)*(-1)^n+(-1)^n*(n^(x/n)-1)+(-1)^n*(n^(1/n)-x));

simplify((-n^(x/n)+x)*(-1)^n+(-1)^n*(n^(x/n)-1)+(-1)^n*(n^(1/n)-x))

sum((-1)^n*(-1+n^(1/n)), n = 1 .. infinity)) - 0.1878596424620671202485179340542732300559030949;

evalf(sum((-1)^n*(-1+n^(1/n)), n = 1 .. infinity)) - 0.1878596424620671202485179340542732300559030949

 

 

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