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MRB Constant X

Marvin Ray Burns 550 Maple
April 21 2012
0

 

The MRB constant is evaluated by

> 

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

.1878596425

 (1)

.

 

Now consider the following function

g := proc (x) options operator, arrow; sum((-1)^n*(n^(1/n)-1), n = 1 .. x) end proc

proc (x) options operator, arrow; sum((-1)^n*(n^(1/n)-1), n = 1 .. x) end proc

 (2)

.

 

g(1) is 1^(1/1)-1.

evalf(g(1))

0.

 (3)

g(2)*is*2^(1/2)-1+1^(1*1)-1 .

evalf(g(2))

.414213562

 (4)

We could evaluate g(3) and so on.

evalf(g(3))

-0.28036008e-1

 (5)

What happens if we sum g(1), g(2) and g(3) and then divide by 3?

(1/3)*evalf(sum(g(x), x = 1 .. 3))

.1287258513

 (6)

What if we stop at larger numbers?

 

(1/100)*evalf(sum(g(x), x = 1 .. 100))

.1872388060

 (7)

(1/1000)*evalf(sum(g(x), x = 1 .. 1000))

.1877875968

 (8)

``

(1/2000)*evalf(sum(g(x), x = 1 .. 2000))

.1878231516

 (9)

(1/3000)*evalf(sum(g(x), x = 1 .. 3000))

.1878352670

 (10)

(1/10000)*evalf(sum(g(x), x = 1 .. 10000))

.1878520839

 (11)

As the stopping number gets real large It looks like we arrive at the MRB constant again!

Can anyone help me explain why?

``

 

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