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# Variations to the MRB constant

Posted 9 years ago
 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? 
 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).