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Variations to the MRB constant

As some of you know now, the MRB constant is computed by the convergent Sum[(-1)^n*(n^(1/n) - 1), {n, 1, Infinity}] .

I wondered what happens when you get the expression all mixed up and write Sum[(-1)^n ((1/n)^(n - 1)), {n, 1, Infinity}] ?

You get another convergent series! It is = -0.596965555578...

What can be said about that? `

POSTED BY: Marvin Ray Burns

Below we switch n^(1/n) with (1/n)^n.

While

Sum[(-1)^n (n^(1/n)), {n, 1, Infinity}] is divergent and Sum[(-1)^n (n^(1/n)-1), {n, 1, Infinity}] converges,

Sum[(-1)^n ((1/n)^n), {n, 1, Infinity}] converges and Sum[(-1)^n ((1/n)^n-1), {n, 1, Infinity}] is divergent.

I just noticed the additive inverse of Sum[(-1)^n ((1/n)^n), {n, 1, Infinity}] is in Steven R Finch's book Mathematical Constants P.449, one of two striking integrals just before the series for the MRB.constant.

Using Sum[(-1)^n ((1/n)^n), {n, 1, Infinity}] :

The following was of interest to me, where I use Mathematica's NSum for divergent series involving x: If

mm0=NSum[(-1)^n ((1/n)^n), {n, 1, Infinity}, WorkingPrecision -> 30, 
 Method -> "AlternatingSigns"] 

then NSum[(-1)^n ((1/n)^n - x), {n, 1, Infinity}, WorkingPrecision -> 30, Method -> "AlternatingSigns"] gives mm0+1/2*(x).

That is similar to what I earlier noted: If

 m0 = NSum[(-1)^n (n^(1/n) ), {n, 1, Infinity}, 
  WorkingPrecision -> 30, Method -> "AlternatingSigns"]

then NSum[(-1)^n (n^(1/n) - x), {n, 1, Infinity}, WorkingPrecision -> 30, Method -> "AlternatingSigns"] gives m0+1/2*(x).

POSTED BY: Marvin Ray Burns
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