The final answer is a constant that is independent of the inital value; so this is not unique to the MRB constant.

Here I use 1/e in place of the MRB constant and arrive at the same final answer:

l=WolframAlpha["5000 digits of 1/e","Result"];

Table[c=Convergents[l,200];l=FromContinuedFraction[c];N[l,Floor[30]],{n,1,20}]

{0.733792275992314711414120159532,

0.575737042439875972092613628026,

0.553152533113037779991565734304,

0.557403692988915871648866020866,

0.555703065311148016537505263780,

0.555754325595562634464237138637,

0.555753142719770665959713854381,

0.555753104943645964882934314155,

0.555753104279444759120199111674,

0.555753104278046376035294972388,

0.555753104278045912459770901737,

0.555753104278045912445404105021,

0.555753104278045912445404118914,

0.555753104278045912445404118914,

0.555753104278045912445404118914,

0.555753104278045912445404118914,

0.555753104278045912445404118914,

0.555753104278045912445404118914,

0.555753104278045912445404118914,

0.555753104278045912445404118914}

marvinrayburns.com

I had to round down my final answers in the initial post.

Here is the work required to get the correct answers in Mathematica 8.0.1.0. I assume it takes a little more work with Maple, especially because Maple doesn't know the MRB constant.

l=WolframAlpha["5000 digits of the MRB Constant","Result"];

Table[c=Convergents[l,200];l=FromContinuedFraction[c];N[l,Floor[30]],{n,1,20}]

{0.900258255912555298965734800278,

0.607305254830648711246204213528,

0.547084962455398192802332932894,

0.557076724276875203120980274954,

0.555718946171123519667869271039,

0.555753940902258978980563984906,

0.555753133950772997064629963080,

0.555753105020833133379456068681,

0.555753104279310247183259953177,

0.555753104278046222593209949132,

0.555753104278045912442805167929,

0.555753104278045912445404117129,

0.555753104278045912445404118914,

0.555753104278045912445404118914,

0.555753104278045912445404118914,

0.555753104278045912445404118914,

0.555753104278045912445404118914,

0.555753104278045912445404118914,

0.555753104278045912445404118914,

0.555753104278045912445404118914}

ContinuedFraction[0.5557531042780459124454041189143]

{0,1,1,3,1,61,1,2,1,1,8,2,1,4,1,2,1,3,4,2,1,3,1,2,1,3,2,9,3,4,1,6}

The definition of a normal numbner is given at http://mathworld.wolfram.com/NormalNumber.html.

I had said that it is not too much of a stretch to conjecture that the digits of the MRB constant are also normally distributed. However, to say that the MRB constant is normal is to say that it is irrational with or without this theorem.

A rational number is not normal, because the last digits are all 0's or follow a repeating pattern; while an irrational number may be normal, (like sqrt(2) is conjectured to be), or  not normal like the Champernowne constant. So only irrational numbers are normal and if a number is normal it is irrational. So sadly my Algebraic normal conjecture 1 is useless!

Algebraic normal conjecture 1: The digits of the MRB constant are normally distributed, so it is irrational.

It has been conjectured that every irrational algebraic number is normal and the MRB constant is a sum of irrational algebraic numbers.  Thus it is not too much of a stretch to conjecture that the digits of the MRB constant are also normally distributed, so if you search far enough into that sting of digits, you will find any particular finite sequence you wish to find. That includes m consecutive 0’s and m digit patterns. However, according to the Archimedean property of real numbers, there exists an integer greater than m called m+1. Since there were only m digits in the pattern, the m+1 digit does not follow that pattern. Thus we have shown that if the MRB constant is normal then it is irrational.

 @Robert

I might never be able to prove the ir/rationality of the MRB constant, or I might only after 30 years of work and more education, or I just might get lucky and just happen to think of the right thing at the right time!

I am very scared at the thought of trying something so foolhardy; I might waste the rest of my life away.

It is a great unknown into which I now tread; the giants are too busy with things that really matter, and I feel that if I don't try it will never be done.

For inspiration I look back at what I said on Jan 11, 1999:

I have started a search for a new 
mathematical constant!
Does anyone want to help me?
Consider, 1^(1/1)-2^(1/2)+3^(1/3)...
This may be a contradiction but I call rc 
a dual constant.
The root constant for even numbers 
(n+1=even number and rc is negative.)
The root constant for odd numbers 
(n+1=odd number and rc is positive.)
rc=1^(1/1)-2^(1/2)+3^(1/3)+4^(1/4)...

And then a few days later I said of the constant and my search:

It has already gone through hundreds of test. 
It matches no other known constant or limited 
combination. I will be taking it apart and examining it "bit by bit". It is my hope to find connections to all kinds of arithmetical manipulations.I realize I am out in "no man's land" but, I work best there! If anyone else is foolhardy enough to come along an offer advice I welcome you. 

So here I go. I don’t know if anyone will care to help; but if I don’t do it, it won’t get done.

marvinrayburns.com

 p1/q1 =MRB constant. By definition then

p1/q1= sum((-1)^n*(n^(1/n)-1),n=1..infinity) and since sum((-1)^n*(n^(1/n)-1),n=1..infinity)-1 = sum((-1)^n*(n^(1/n)-1),n=1..infinity)-q1/q1= sum((-1)^n*(n^(1/n)),n=1..infinity), and p1/q1-q1/q1=(p1-q1)/q1 =p/q1=p/q. Thus q=q1 and

p/q= sum((-1)^n*(n^(1/n)),n=1..infinity). Multiplying by q we get

p= q*sum((-1)^n*(n^(1/n)),n=1..infinity). By an associate law,

p=sum((-1)^n*q*(n^(1/n)),n=1..infinity) Expressing n and q by the same exponent we get

p=sum((-1)^n*(n*q^n)^(1/n),n=1..infinity). When we expand the series we see that

p=-1^(1/1)*(q^1)^(1/1)+2^(1/2)*(q^2)^(1/2)-3^(1/3)*(q^3)^(1/3)+4^(1/4)*(q^4)^(1/4)-... . Since (A^B)^C=(A^C)^B when they are all positive:

p=-1^(1/1)*(q^(1/1))^1+2^(1/2)*(q^(1/2))^2-3^(1/3)*(q^(1/3))^3+4^(1/4)*(q^(1/4))^4... . Accordingly

p=-1^(1/1)*a1^1+2^(1/2)*a2^2-3^(1/3)*a3^3+4^(1/4)*a4^4-..., where ax=(q^(1/x))^x., and assuming q and x are positive  (q^(1/x))^x =q.

    Let x=q^(1/n) where n is the term of the power series

   So to determine if it was rational you would have to have integers,p1,q1,p,q such that
 
    p1/q1=sum((-1)^n*(n^(1/n)-1),n=1..infinity), which implies that

    p/q= sum((-1)^n*(n^(1/n)),n=1..infinity). Multiplying both sides by q we get 

    p= q*sum((-1)^n*(n^(1/n)),n=1..infinity). By an assosiate law,

    p=sum((-1)^n*q*(n^(1/n)),n=1..infinity) Expressing n and q by the same exponent we get

    p=sum((-1)^n*(n*q^n)^(1/n),n=1..infinity). When we expand the series we see that

_____________________________A change made here._____________________________________

    p=-q+sqrt(2)*sqrt(q^2)-3^(1/3)*(q^3)^(1/3)+4^(1/4)*(q^4)^(1/4)-5^(1/5)*(q^5)^(1/5)+6^(1/6)*(q^6)^(1/6)-...

    p+q=sqrt(2)*sqrt(q^2)-3^(1/3)*(q^3)^(1/3)+4^(1/4)*(q^4)^(1/4)-5^(1/5)*(q^5)^(1/5)+6^(1/6)*(q^6)^(1/6)-... since sqrt(q^2)=q,
   
    p+q=sqrt(2)*q-3^(1/3)*(q^3)^(1/3)+4^(1/4)*(q^4)^(1/4)-5^(1/5)*(q^5)^(1/5)+6^(1/6)*(q^6)^(1/6)-...

    p+q=q*(sqrt(2)-3^(1/3)+4^(1/4)-5^(1/5)+6^(1/6)-...). Diving by q!=0,

    (p+q)/q=(sqrt(2)-3^(1/3)+4^(1/4)-5^(1/5)+6^(1/6)-...). In series form we have

__________________________A new change made here.__________________________________

    (p+q)/q = sum((-1)^n*n^(1/n), n = 2 .. infinity), and

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

    (p+q)/q=MRB constant+1/2, But  p1/q1 =MRB constant, so by subsitution and reflextivity,

 (p+q)/q=p1/q1+1/2, Remembering that q=q1 we get

 (p+q)/q=p1/q+1/2. Since 1/2 =(1/2*q)/q,

 (p+q)/q=p1/q+(1/2*q)/q, Multiplying by q we get

 (p+q)=p1+1/2*q. Subtracting 1/2*q and by reflextivity

p1 =(p+q)-1/2*q. Simplifying gives an end result of

  .

Now if p is an integer and q is odd  then p1 is not an integer.

I'll have to look at this for a little bit. I doubt that I proved anything that easily.

After correcting all of my mistakes, I see that I haven't got anywhere!

p= q*sum((-1)^n*(n^(1/n)),n=1..infinity). By an associate law,

p=sum((-1)^n*q*(n^(1/n)),n=1..infinity) Expressing n and q by the same exponent we get

p=sum((-1)^n*(n*q^n)^(1/n),n=1..infinity). When we expand the series we see that

p=-1^(1/1)*(q^1)^(1/1)+2^(1/2)*(q^2)^(1/2)-3^(1/3)*(q^3)^(1/3)+4^(1/4)*(q^4)^(1/4)-... . Since (A^B)^C=(A^C)^B when they are all positive:

p=-1^(1/1)*(q^(1/1))^1+2^(1/2)*(q^(1/2))^2-3^(1/3)*(q^(1/3))^3+4^(1/4)*(q^(1/4))^4... . Accordingly

p=-1^(1/1)*a1^1+2^(1/2)*a2^2-3^(1/3)*a3^3+4^(1/4)*a4^4-..., where ax=(q^(1/x))^x., and assuming q and x are positive  (q^(1/x))^x =q.

    Let x=q^(1/n) where n is the term of the power series

   So to determine if it was rational you would have to have integers,p1,q1,p,q such that
 
    p1/q1=sum((-1)^n*(n^(1/n)-1),n=1..infinity), which implies that

    p/q= sum((-1)^n*(n^(1/n)),n=1..infinity). Multiplying both sides by q we get 

    p= q*sum((-1)^n*(n^(1/n)),n=1..infinity). By an assosiate law,

    p=sum((-1)^n*q*(n^(1/n)),n=1..infinity) Expressing n and q by the same exponent we get

    p=sum((-1)^n*(n*q^n)^(1/n),n=1..infinity). When we expand the series we see that

_____________________________A change made here._____________________________________

    p=-q+sqrt(2)*sqrt(q^2)-3^(1/3)*(q^3)^(1/3)+4^(1/4)*(q^4)^(1/4)-5^(1/5)*(q^5)^(1/5)+6^(1/6)*(q^6)^(1/6)-...

    p+q=sqrt(2)*sqrt(q^2)-3^(1/3)*(q^3)^(1/3)+4^(1/4)*(q^4)^(1/4)-5^(1/5)*(q^5)^(1/5)+6^(1/6)*(q^6)^(1/6)-... since sqrt(q^2)=q,
   
    p+q=sqrt(2)*q-3^(1/3)*(q^3)^(1/3)+4^(1/4)*(q^4)^(1/4)-5^(1/5)*(q^5)^(1/5)+6^(1/6)*(q^6)^(1/6)-...

    p+q=q*(sqrt(2)-3^(1/3)+4^(1/4)-5^(1/5)+6^(1/6)-...)

    (p+q)/q=(sqrt(2)-3^(1/3)+4^(1/4)-5^(1/5)+6^(1/6)-...). In series form we have

    (p+q)/q= sum((-1)^n*(n^(1/n)),n=1..infinity)+1. Thus

    (p+q)/q-1= sum((-1)^n*(n^(1/n)),n=1..infinity). But remember that

    p/q= sum((-1)^n*(n^(1/n)),n=1..infinity). Since q!=0,

    (p+q)/q-1 = p/q. Multiplying both sides by gives,

    (p+q)-q = p. By association,

    p+(q-q) = p. Since q-q=0,

    p=p.

p= q*sum((-1)^n*(n^(1/n)),n=1..infinity). By an associate law,

p=sum((-1)^n*q*(n^(1/n)),n=1..infinity) Expressing n and q by the same exponent we get

p=sum((-1)^n*(n*q^n)^(1/n),n=1..infinity). When we expand the series we see that

p=-1^(1/1)*(q^1)^(1/1)+2^(1/2)*(q^2)^(1/2)-3^(1/3)*(q^3)^(1/3)+4^(1/4)*(q^4)^(1/4)-... . Since (A^B)^C=(A^C)^B when they are all positive:

p=-1^(1/1)*(q^(1/1))^1+2^(1/2)*(q^(1/2))^2-3^(1/3)*(q^(1/3))^3+4^(1/4)*(q^(1/4))^4... . Accordingly

______________My mistake is still here.________________

p=-1^(1/1)*a1^1+2^(1/2)*a2^2-3^(1/3)*a3^3+4^(1/4)*a4^4-..., where ax=(q^(1/x))^x., and assuming q and x are positive  (q^(1/x))^x =q.

Let x=q^(1/n) where n is the term of the power series

 p=-1^(1/1)*x^1+2^(1/2)*x^2-3^(1/3)*x^3+4^(1/4)*x^4-… . Since 1^(1/1) =1,

 p=-1x^1+2^(1/2)*x^2-3^(1/3)*x^3+4^(1/4)*x^4-…  .  Factoring out a x we get

 p=x*(-1+2^(1/2)*x-3^(1/3)*x^2+4^(1/4)*x^3-…).  Dividing both sides by x!=0 we get 

 p/x  =  -1+2^(1/2)*x-3^(1/3)*x^2+4^(1/4)*x^3-… . Adding 1 to each side gives

 p/x+1  =  +2^(1/2)*x-3^(1/3)*x^2+4^(1/4)*x^3-… . Again, dividing both sides by x!=0 we get

 (p/x+1)/x  =  +2^(1/2)-3^(1/3)*x+4^(1/4)*x^2-… . Adding -2^(1/2) to each side gives

 (p/x+1)/x-2^(1/2)  =  -3^(1/3)*x+4^(1/4)*x^2-… . Repeating the previous two steps:

 ((p/x+1)/x-2^(1/2))/x  =  -3^(1/3)+4^(1/4)*x-… ., and

 ((p/x+1)/x-2^(1/2))/x+3^(1/3)  =  +4^(1/4)*x-5^(1/5)+... .

 ((((((p/x+1)/x-2^(1/2))/x+3^(1/3))/x)-4^(1/4))/x+5^(1/5))/…+…=0. In pretty print to see the pattern that is forming:

An inspection shows that, as n->infinity, the LHS becomes independent of p.

Now here do I dare say that p can be an integer even if q is?

Note to self:

Let x=q^(1/n) where n is the term of the power series.

marvinrayburns.com

The updated version of this message is found in Math Blog at http://math-blog.com/2010/11/21/the-geometry-of-the-mrb-constant/.

Download feb022011b.mw

There Maple 14 simply indicates that the sum is getting larger as j and k increase together.

Both 12 and 14 pretty much agree in their results up until the output gets to about -2.644*10^5.

It is likely that both of them are having troubles with accuracy; why else would 14 jump to  

from 10^164?

However it looks like we can agree with Maple 14, in that it indicates that the sum is getting larger as j and k increase together. Nonetheless, remember that the actual triple "infinite sum" is 0.1878596... .

I don't know where to go from here; whatever I do, it will probably be done in a new message. The formatting that I'm using takes way too long to load up on your computer -- that is if it is anything like mine.

Download feb02,2011.mw

marvinrayburns.com

@Robert Israel

You were correct. At first I thought you were saying the lower value  was -1. However, it was as you correctly said (or at least clearly meant) the value of the upper one minus1.

@Robert Israel

You were correct. At first I thought you were saying the lower value  was -1. However, it was as you correctly said (or at least clearly meant) the value of the upper one minus1.

 That is what the even-numbered partial sums were supposed to converge to. i.e. 0.1878... However, I thought the odd numbered partial sums were supposed to converge to 1 minus what the even-numbered converged to. i.e. -0.81214...

 EG. 

In Maple,

evalf(sum((cos(Pi*n)+I*sin(Pi*n))*(cosh(ln(n)/n)+sinh(ln(n)/n)), n = 1 .. 1001)) gives -.8156017982, and

evalf(sum((cos(Pi*n)+I*sin(Pi*n))*(cosh(ln(n)/n)+sinh(ln(n)/n)), n = 1 .. 3001)) gives -.8134758429.