I don't know if anything good will come out of the following expansion; but since I don’t have anything better for "O," here it is anyway.



 

Let m = the MRB constant = = 0.1878596...

Proof that

 

To simplify the appearance of our work a little, below an infinite series of  is what is symbolized by .

Let m= the sum of the convergent infinite series of

(1.1)

In general we have the following:

Expanding

Collecting the constants we see that

By Grandi's series we know that

Collecting the infinite series we see that

(1.2)

Specifically, here where z = 1, by the lema

by one of the commutative laws the Cesaro sum of the divergent infinite series of 

Since

 (-1)^n = and  and the Cesaro sum of the divergent infinite series of

Also since and the Cesaro sum of the divergent infinite series of exp((I*(Pi*n^2-I*ln(n)))/n)+1/2=m.

Let t=, and then the infinite series of  = infinite series of exp(I*t) = .

Which means that 

Now we have t=, and since n>0

we can expand ln(n) into the series .

Thus the expanded version of t is t=.

 

Putting the expanded version of t into the formula  gives =m.

 

 

Distributing the "I" we get

 

Since  = , we have

 

Distributing the 1/n we get

 

=

 

Notice that "I" and the innermost series are both raised to the "k"th power, so we can combine like powers and get

 

 

Obviously -(I^2) =1; so we have

 

Moving the innermost series to the left gives us

 

Therefore distributing the 1/k! we get

 

QED

 

 



Download TripleMRBsum.mw

 

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