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# MRB constant (CMRB) Real-World, and beyond, Applications

Posted 1 year ago

The MRB constant record computations are given [here][1] in the Wolfram community. To help in the loading times of that extensive discussion, I've moved its applications to this separate discussion.

# CMRB and its applications

Definition 1 CMRB is defined at https://en.wikipedia.org/wiki/MRB_constant .

From Wikipedia:

References
Plouffe, Simon. "mrburns". Retrieved 12 January 2015.
Burns, Marvin R. (23 January 1999). "RC". math2.org. Retrieved 5 May 2009.
Plouffe, Simon (20 November 1999). "Tables of Constants" (PDF). Laboratoire de combinatoire et d'informatique mathématique. Retrieved 5 May 2009.
Weisstein, Eric W. "MRB Constant". MathWorld.
Mathar, Richard J. (2009). "Numerical Evaluation of the Oscillatory Integral Over exp(iπx) x^*1/x) Between 1 and Infinity". arXiv:0912.3844 [math.CA].
Crandall, Richard. "Unified algorithms for polylogarithm, L-series, and zeta variants" (PDF). PSI Press. Archived from the original (PDF) on April 30, 2013. Retrieved 16 January 2015.
(sequence A037077 in the OEIS)
(sequence A160755 in the OEIS)
(sequence A173273 in the OEIS)
Fiorentini, Mauro. "MRB (costante)". bitman.name (in Italian). Retrieved 14 January 2015.
Finch, Steven R. (2003). Mathematical Constants. Cambridge, England: Cambridge University Press. p. 450. ISBN 0-521-81805-2.




The following equation that was shown in the Wikipedia definition shows how closely the MRB constant is related to root two.


In[1]:= N[Sum[Sqrt[2]^(1/n)* Sqrt[n]^(1/n) - ((Sqrt[2]^y*Sqrt[2]^x)^(1/Sqrt[2]^x))^Sqrt[2]^(-y)/.
x -> 2*Log2[a^2 + b^2] /.
y -> 2*Log2[-ai^2 - bi^2] /.
a -> 1 - (2*n)^(1/4) /.
b -> 2^(5/8)*Sqrt[n^(1/4)] /.
ai -> 1 - I*(2*n)^(1/4) /.
bi -> 2^(5/8)*Sqrt[I*n^(1/4)], {n, 1, Infinity}], 7]

Out[1]= 0.1878596 + 0.*10^-8 I


The complex roots and powers above are found to be well-defined because we get all either "integer" and "rational" the first of the following lists only, also by working from the bottom to the top of the above list of equations.

Code:

In[349]:= Table[
Expand[(Sqrt[2])^-y/(Sqrt[2])^x] //.
x -> 2 (Log[1 + Sqrt[2] Sqrt[n]]/Log[2]) /.
y -> 2 (Log[-1 + Sqrt[2] Sqrt[n]]/Log[2])]], {n, 1, 10}]

Out[349]= {Integer, Rational, Rational, Rational, Rational, Rational, \
Rational, Rational, Rational, Rational}

In[369]:= Table[
Expand[(Sqrt[2])^-y/(Sqrt[2])^x] //.
x -> 2 (Log[1 + Sqrt[2] Sqrt[n]]/Log[3]) /.
y -> 2 (Log[-1 + Sqrt[2] Sqrt[n]]/Log[2])]], {n, 1, 10}]

Out[369]= {Times, Rational, Times, Times, Times, Times, Times, Times, \
Times, Times}


Definition 2 CMRB is defined at http://mathworld.wolfram.com/MRBConstant.html.

From MathWorld:

SEE ALSO:
Glaisher-Kinkelin Constant, Power Tower, Steiner's Problem
REFERENCES:
Burns, M. R. "An Alternating Series Involving n^(th) Roots." Unpublished note, 1999.

Burns, M. R. "Try to Beat These MRB Constant Records!" http://community.wolfram.com/groups/-/m/t/366628.

Crandall, R. E. "Unified Algorithms for Polylogarithm, L-Series, and Zeta Variants." 2012a.

Crandall, R. E. "The MRB Constant." §7.5 in Algorithmic Reflections: Selected Works. PSI Press, pp. 28-29, 2012b.

Finch, S. R. Mathematical Constants. Cambridge, England: Cambridge University Press, p. 450, 2003.

Plouffe, S. "MRB Constant." http://pi.lacim.uqam.ca/piDATA/mrburns.txt.

Sloane, N. J. A. Sequences A037077 in "The On-Line Encyclopedia of Integer Sequences."

Referenced on Wolfram|Alpha: MRB Constant
CITE THIS AS:
Weisstein, Eric W. "MRB Constant." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/MRBConstant.html


How would we show that any of the series in the above MathWorld definition are convergent, or even absolutely convergent?

For "a"k=k1/k, given that the sequence is monotonically decreasing according to Steiner's Problem, next, we would like to show (5) is the alternating sum of a sequence that converges to 0 monotonically and use the Alternating series test to see that it is conditionally convergent

Here is proof that 1 is the limit of "a" as k goes to infinity:

Here are many other proofs that 1 is the limit of "a" as k goes to infinity.

Thus, (k1/k-1) is a monotonically decreasing and bounded below by 0 sequence.

If we want an absolutely convergent series, we can use (4).

Sk which, since the sum of the absolute values of the summands is finite, the sum converges absolutely!

There is no closed-form for CMRB in the MathWorld definition; this could be due to the following: in Mathematical Constants,( Finch, S. R. Mathematical Constants, Cambridge, England: Cambridge University Press, p. 450), Steven Finch wrote that it is difficult to find an "exact formula" (closed-form solution) for it.

# Real-World, and beyond, Applications

## CMRB as a Growth Model

Its factor ![a][5] models the interest rate to multiply an investment k times in k periods, as well as "other growth and decay functions involving the more general expression ![(1+k)^n][6], as in Plot 1A," because ![enter image description here][7]

r=(k^(1/k)-1);Animate[ListPlot[l=Accumulate[Table[(r+1)^n,{k,100}]], PlotStyle->Red,PlotRange->{0,150},PlotLegends->{"\!$$\*UnderscriptBox[\(\[Sum]$$, ]\)(r+1\!$$\*SuperscriptBox[\()$$, $$n$$]\)/.r->(\!$$\*SuperscriptBox[\(k$$, $$1/k$$]\)-1)/.n->"n},AxesOrigin->{0,0}],{n,0,5}]


Plot 1A ![enter image description here][8]

The discrete rates looks like the following.

r = (k^(1/k) - 1); me =
Animate[ListPlot[l = Table[(r + 1)^n, {k, 100}], PlotStyle -> Red,
PlotLegends -> {"(r+1)^n/.r->\!$$\*SuperscriptBox[\(k$$, \
$$1/k$$]\)=1/.n->", n}, AxesOrigin -> {0, 0},
PlotRange -> {0, 7}], {n, 1, 5}]


![enter image description here][9]

That factor ![enter image description here][10] models not only discretely compounded rates but continuous too, ie ![Pt=p0e^rt.][11]

By entering

Solve[P*E^(r*t) == P*(t^(1/t) - 1), r]


we see, for ![Pt=p0e^rt,][12] ![t>e][13]

gives an effect of continuous decay of ![enter image description here][14] Here Q1 means the first Quarter form 0 to -1.

The alternating sum of the principal of those continuous rates, i.e. P=(-1)t er t is the MRB constant (CMRB): ![enter image description here][15]

In[647]:= NSum[(-1)^t ( E^(r*t)) /. r -> Log[-1 + t^(1/t)]/t, {t, 1,
Infinity}, Method -> "AlternatingSigns", WorkingPrecision -> 30]

Out[647]= 0.18785964246206712024857897184


Its integral (MKB) is an analog to CMRB : ![enter image description here][16]

In[1]:= NIntegrate[(-1)^t (E^(r*t)) /. r -> Log[-1 + t^(1/t)]/t, {t,
1, Infinity I}, Method -> "Trapezoidal", WorkingPrecision -> 30] -
2 I/Pi

Out[1]= 0.0707760393115288035395280218303 -
0.6840003894379321291827444599927 I


So, integrating P yields about 1/2 greater of a total than summing:

In[663]:=
CMRB = NSum[(-1)^n ( Power[n, ( n)^-1] - 1), {n, 1, Infinity},
Method -> "AlternatingSigns", WorkingPrecision -> 30];

In[664]:=
MKB = Abs[
NIntegrate[(-1)^t ( E^(r*t)) /. r -> Log[-1 + t^(1/t)]/t, {t, 1,
Infinity I}, Method -> "Trapezoidal", WorkingPrecision -> 30] -
2 I/Pi];

In[667]:= MKB - CMRB

Out[667]= 0.49979272646562724956073343752


Next:

## CMRB from Geometric Series and Power Series

The MRB constant: ![enter image description here][17] is closely related to geometric series: ![enter image description here][18]

The inverse function of the "term" of the MRB constant, i.e. x^(1/x) within a certain domain is solved for in [this link,][19]

![enter image description here][20]

## ...

![enter image description here][21]

Now we have the following for the orientated area, from 0 to 1, between the graph of that term and the axis.

![enter image description here][22]

In[344]:= f[x_] = x^(1/x);

In[346]:= CMRB =
NSum[(-1)^x (f[x] - 1), {x, 1, Infinity}, WorkingPrecision -> 20]

Out[346]= 0.18785964246207

In[350]:= (10 (CMRB + 3))/(3 (3 CMRB - 17)) -
NIntegrate[g = -x /. Solve[y == f[x], x], {y, 0, 1},
WorkingPrecision -> 20]

Out[350]= {1.5605*10^-11}


Consider the following about a slight generalization of that term.

![enter image description here][23]

CMRB can be written in geometric series form:

CMRB= ![enter image description here][24]

In[240]:= N[Quiet[(Sum[q^k, {x, 1, Infinity}] /.
k -> Log[-E^(I*Pi*x) + E^(x*(I*Pi + Log[x]/x^2))]/Log[q]) -
Sum[E^(I*Pi*x)*(-1 + x^(1/x)), {x, 1, Infinity}]]]

Out[240]= -4.163336342344337*^-16


Why would we express CMRB so? I'm not entirely sure, but we do have the following interestingly intricate graphs that go towards the value of the MRB constant and the MRB constant-1 as the input gets large. ![enter image description here][25] ![enter image description here][26] ![enter image description here][27] ![enter image description here][28]

![enter image description here][29] ![enter image description here][30]

see notebook [here.][31]

Next

## The Geometry of the MRB constant

In 1837 Pierre Wantzel proved that an nth root of a given length cannot be constructed if n is not a power of 2 (as mentioned [here][32] in Wikipedia). However, the following is a little different.

For ![,][33] on November 21, 2010, I coined a multiversal [analog][34] to, [Minkowski space][35] that plots their values from constructions arising from a peculiar non-euclidean geometry, below, and fully in [this vixra draft][36].

As in Diagram 2, we give each n-cube a hyperbolic volume (content) equal to its dimension,![enter image description here][37] Geometrically, as in Diagram3, on the y,z-plane line up an edge of each n-cube. The numeric values displayed in the diagram are the partial sums of S[x_] = Sum[(-1)^n*n^(1/n), {n, 1, 2*u}] where u is an positive integer. Then M is the MRB constant.

![enter image description here][38]

Join[ Table[N[S[x]], {u, 1, 4}], {"..."}, {NSum[(-1)^n*(n^(1/n) - 1), {n, 1, Infinity}]}]


Out[421]= {0.414214, 0.386178, 0.354454, 0.330824, "...", 0.18786}

Here are views of some regions of the plot of a definite integral equal to CMRB..

![enter image description here][39]

In[66]:= Csch[Pi t] Im[(1 + I t)^(1/(1 + I t))]

Out[66]= Csch[[Pi] t] Im[(1 + I t)^(1/(1 + I t))]

In[67]:= f[t_] = Csch[Pi t] Im[(1 + I t)^(1/(1 + I t))]

Out[67]= Csch[[Pi] t] Im[(1 + I t)^(1/(1 + I t))]

ReImPlot[Im[(1 + I t)^(1/(1 + I t))], {t, 0, 1}, PlotStyle -> Blue,
PlotLabels -> {Placed[(1 + I t)^(1/(1 + I t)), Above]}]


![enter image description here][40]

ReImPlot[Im[(1 + I t)^(1/(1 + I t))], {t, 0, 5}, PlotStyle -> Blue,
PlotLabels -> {Placed[(1 + I t)^(1/(1 + I t)), Above]}]


![enter image description here][41]

Show[ReImPlot[Csch[\[Pi] t], {t, 0, 1}, PlotStyle -> Yellow,
PlotLabels -> "Expressions"]]


![enter image description here][42]

Show[ReImPlot[Csch[\[Pi] t], {t, 0, 5}, PlotStyle -> Yellow,
PlotLabels -> "Expressions"]]


![enter image description here][43]

ReImPlot[f[t], {t, 0, 1},
PlotLabel -> NIntegrate[f[t], {t, 0, 1}, WorkingPrecision -> 20],
PlotStyle -> Green, PlotLabels -> "Expressions"]


![enter image description here][44]

ReImPlot[f[t], {t, 0, 5},
PlotLabel -> NIntegrate[f[t], {t, 0, 5}, WorkingPrecision -> 20],
PlotStyle -> Green, PlotLabels -> "Expressions"]


![enter image description here][45]

Next

# MeijerG Representation

From its integrated analog, I found a [MeijerG][46] representation for CMRB.

The search for it began with the following:

On 10/10/2021, I found the following proper definite integral that leads to almost identical proper integrals from 0 to 1 for CMRB and its integrated analog.

![m vs m2 0 to 1][47]

Here is a [MeijerG][49] function for the integrated analog. See [(proof)][50] of discovery.

![enter image description here][51]

f(n)=![enter image description here][52].

In[135]:=f[n_]:=MeijerG[{{},Table[1,{n+1}]},{Prepend[Table[0,n+1],-n+1],{}},-\[ImaginaryI]\[Pi]];
In[337]:=M2=NIntegrate[E^(I Pi x)(SuperscriptBox["x", FractionBox["1", "x"]]-1),
{x,1,Infinity I},WorkingPrecision->100]

Out[337]=0.07077603931152880353952802183028200136575469620336302758317278816361845726438203658083188126617723821-0.04738061707035078610720940650260367857315289969317363933196100090256586758807049779050462314770913485 \[ImaginaryI]


![enter image description here][53]

I wonder if there is one for the MRB constant sum (CMRB)?

According to "Primary Proof 1" and "Primary Proof 3" shown below along with the section prefixed by the phrase "So far I came up with," it can be proven that for G being the Wolfram MeijerG function

and f(n)=![enter image description here][54], and ![enter image description here][55]

g[x_] = (-1)^x (1 - (x + 1)^(1/(x + 1)));

In[52]:= (1/2)*
NIntegrate[(g[-t] - g[t])/(Sin[Pi*t]*Cos[Pi*t]*I + Sin[Pi*t]^2), {t,
0, I*Infinity}, WorkingPrecision -> 100,

Out[52]= 0.\
1170836031505383167089899122239912286901483986967757585888318959258587\
7430027817712246477316693025869 +
0.0473806170703507861072094065026036785731528996931736393319610009025\
6586758807049779050462314770913485 I

In[57]:= Re[
NIntegrate[
g[-t]/(Sin[Pi*t]*Cos[Pi*t]*I + Sin[Pi*t]^2), {t, 0, I*Infinity},
WorkingPrecision -> 100,

Out[57]= 0.\
1878596424620671202485179340542732300559030949001387861720046840894772\
315646602137032966544331074969


# The Laplace transform analogy to the CMRB

![enter image description here][56] ![enter image description here][57] ![enter image description here][58] ![enter image description here][59] ![enter image description here][60] ![enter image description here][61] see [notebook][62]

Likewise, Wolfram Alpha [here][63] says

![enter image description here][64]

![enter image description here][65]

Interestingly,

![enter image description here][66]

That has the same argument, ![enter image description here][67], as the MeijerG transformation of CMRB. ![enter image description here][68]

# MRB constant formulas and identities

I developed this informal catalog of formulas for the MRB constant with over 20 years of research and ideas from users like you.

## 6/7/2022

CMRB

=![enter image description here][69]

=![enter image description here][70]

=![enter image description here][71]

=![enter image description here][72]

=![enter image description here][73]

=![enter image description here][74]

So, using induction, we have. ![enter image description here][75]

![enter image description here][76]

Sum[Sum[(-1)^(x + n), {n, 1, 5}] + (-1)^(x) x^(1/x), {x, 2, Infinity}]

## 3/25/2022

Formula (11) =

![enter image description here][77]

As Matheamatica says:

Assuming[Element[c, \[DoubleStruckCapitalZ]], FullSimplify[
E^(t*(r + I*Pi*(2*c + 1))) /. r -> Log[t^(1/t) - 1]/t]]


= E^(I (1 + 2 c) [Pi] t) (-1 + t^(1/t)) ![enter image description here][78]

Where for all integers c, (1+2c) is odd leading to ![enter image description here][79]

Expanding the E^log term gives

![enter image description here][80]

which is ![enter image description here][81],

That is exactly (2) in the above-quoted MathWorld definition: ![enter image description here][82]

## 2/21/2022

Directly from the formula of 12/29/2021 below, ![enter image description here][83] In

u = (-1)^t; N[
NSum[(t^(1/t) - 1) u, {t, 1, Infinity }, WorkingPrecision -> 24,
Method -> "AlternatingSigns"], 15]


Out[276]= 0.187859642462067

In

v = (-1)^-t - (-1)^t; 2 I N[
NIntegrate[Im[(t^(1/t) - 1) v^-1], {t, 1, Infinity I},
WorkingPrecision -> 24], 15]


Out[278]= 0.187859642462067

Likewise, ![enter image description here][84]

Expanding the exponents,

![enter image description here][85] This can be generalized to ![(x+log/][86]

Building upon that, we get a closed form for the inner integral in the following.

CMRB= ![enter image description here][87]

In[1]:=
CMRB = NSum[(-1)^n (n^(1/n) - 1), {n, 1, Infinity},
WorkingPrecision -> 1000, Method -> "AlternatingSigns"];

In[2]:= CMRB - {
Quiet[Im[NIntegrate[
Integrate[
E^(Log[t]/t + x)/(-E^((-I)*Pi*t + x) + E^(I*Pi*t + x)), {x,
I, -I}], {t, 1, Infinity I}, WorkingPrecision -> 200,
Method -> "Trapezoidal"]]];
Quiet[Im[NIntegrate[
Integrate[
Im[E^(Log[t]/t + x)/(-E^((-I)*Pi*t + x) + E^(I*Pi*t + x))], {x,
-t,  t }], {t, 1
, Infinity  I}, WorkingPrecision -> 2000,
Method -> "Trapezoidal"]]]}

Out[2]= {3.*10^-998, 3.*10^-998}


Which after a little analysis, can be shown convergent in the continuum limit at t → ∞ i.

## 12/29/2021

From "Primary Proof 1" worked below, it can be shown that ![enter image description here][88]

Mathematica knows that because

  m = N[NSum[-E^(I*Pi*t) + E^(I*Pi*t)*t^t^(-1), {t, 1, Infinity},
Method -> "AlternatingSigns", WorkingPrecision -> 27], 18];
Print[{m -
N[NIntegrate[
Im[(E^(Log[t]/t) + E^(Log[t]/t))/(E^(I \[Pi] t) -
E^(-I \[Pi] t))] I, {t, 1, -Infinity I},
WorkingPrecision -> 20], 18],
m - N[NIntegrate[
Im[(E^(Log[t]/t) + E^(Log[t]/t))/(E^(-I \[Pi] t) -
E^(I \[Pi] t))] I, {t, 1, Infinity I},
WorkingPrecision -> 20], 18],
m + 2 I*NIntegrate[
Im[(E^(I*Pi*t + Log[t]/t))/(-1 + E^((2*I)*Pi*t))], {t, 1,
Infinity I}, WorkingPrecision -> 20]}]


yields

  {0.*^-19,0.*^-19,0.*^-19}


Partial sums to an upper limit of (10^n i) give approximations for the MRB constant + the same approximation *10^-(n+1) i. Example:

-2 I*NIntegrate[
Im[(E^(I*Pi*t + Log[t]/t))/(-1 + E^((2*I)*Pi*t))], {t, 1, 10^7 I},
WorkingPrecision -> 20]


gives 0.18785602000738908694 + 1.878560200074*10^-8 I where CMRB ≈ 0.187856.

Notice it is special because if we integrate only the numerator, we have MKB=![enter image description here][89], which defines the "integrated analog of CMRB" (MKB) described by Richard Mathar in [https://arxiv.org/abs/0912.3844][90]. (He called it M1.)

Like how this:

NIntegrate[(E^(I*Pi*t + Log[t]/t)), {t, 1, Infinity I},
WorkingPrecision -> 20] - I/Pi


converges to

0.070776039311528802981 - 0.68400038943793212890 I.

(The upper limits " i infinity" and " infinity" produce the same result in this integral.)

## 11/14/2021

Here is a standard notation for the above mentioned

CMRB,![enter image description here][91] ![enter image description here][92].

In[16]:= CMRB = 0.18785964246206712024851793405427323005590332204; \
CMRB - NSum[(Sum[
E^(I \[Pi] x) Log[x]^n/(n! x^n), {x, 1, Infinity}]), {n, 1, 20},
WorkingPrecision -> 50]

Out[16]= -5.8542798212228838*10^-30

In[8]:= c1 =
Activate[Limit[(-1)^m/m! Derivative[m][DirichletEta][x] /. m -> 1,
x -> 1]]

Out[8]= 1/2 Log[2] (-2 EulerGamma + Log[2])

In[14]:= CMRB -
N[-(c1 + Sum[(-1)^m/m! Derivative[m][DirichletEta][m], {m, 2, 20}]),
30]

Out[14]= -6.*10^-30


## 11/01/2021

: The catalog now appears complete, and can all be proven through Primary Proof 1, and the one with the eta function, Primary Proof 2, both found below.

a ≠b ![enter image description here][93] ![enter image description here][94]

g[x_] = x^(1/x); CMRB =
NSum[(-1)^k (g[k] - 1), {k, 1, Infinity}, WorkingPrecision -> 100,
Method -> "AlternatingSigns"]; a = -Infinity I; b = Infinity I;
g[x_] = x^(1/x); (v = t/(1 + t + t I);
Print[CMRB - (-I /2 NIntegrate[ Re[v^-v Csc[Pi/v]]/ (t^2), {t, a, b},
WorkingPrecision -> 100])]); Clear[a, b]
-9.3472*10^-94


Thus, we find

![enter image description here][95]

[here,][96] and ![enter image description here][97] next:

In[93]:= CMRB =
NSum[Cos[Pi n] (n^(1/n) - 1), {n, 1, Infinity},
Method -> "AlternatingSigns", WorkingPrecision -> 100]; Table[
CMRB - (1/2 +
NIntegrate[
Im[(t^(1/t) - t^(2 n))] (-Csc[\[Pi] t]), {t, 1, Infinity I},
WorkingPrecision -> 100, Method -> "Trapezoidal"]), {n, 1, 5}]

Out[93]= {-9.3472*10^-94, -9.3473*10^-94, -9.3474*10^-94, \
-9.3476*10^-94, -9.3477*10^-94}


CNT+F "The following is a way to compute the" for more evidence

For such n, ![enter image description here][98] converges to 1/2+0i.

(How I came across all of those and more example code follow in various replies.)

## On 10/18/2021

, I found the following triad of pairs of integrals summed from -complex infinity to +complex infinity.

![CMRB= -complex infinity to +complex infinity][99]

You can see it worked [ in this link here][100].

In[1]:= n = {1, 25.6566540351058628559907};

In[2]:= g[x_] = x^(n/x);
-1/2 Im[N[
NIntegrate[(g[(1 - t)])/(Sin[\[Pi] t]), {t, -Infinity I,
Infinity I}, WorkingPrecision -> 60], 20]]

Out[3]= {0.18785964246206712025, 0.18785964246206712025}

In[4]:= g[x_] = x^(n/x);
1/2 Im[N[NIntegrate[(g[(1 + t)])/(Sin[\[Pi] t]), {t, -Infinity I,
Infinity I}, WorkingPrecision -> 60], 20]]

Out[5]= {0.18785964246206712025, 0.18785964246206712025}

In[6]:= g[x_] = x^(n/x);
1/4 Im[N[NIntegrate[(g[(1 + t)] - (g[(1 - t)]))/(Sin[\[Pi] t]), {t, -Infinity I,
Infinity I}, WorkingPrecision -> 60], 20]]

Out[7]= {0.18785964246206712025, 0.18785964246206712025}


Therefore, bringing

![enter image description here][101]

back to mind, we joyfully find,

![CMRB n and 1][102]

In[1]:= n =
25.65665403510586285599072933607445153794770546058072048626118194900\
97321718621288009944007124739159792146480733342667100.;

g[x_] = {x^(1/x), x^(n/x)};

CMRB = NSum[(-1)^k (k^(1/k) - 1), {k, 1, Infinity},
WorkingPrecision -> 100, Method -> "AlternatingSigns"];

Print[CMRB -
NIntegrate[Im[g[(1 + I t)]/Sinh[\[Pi] t]], {t, 0, Infinity},
WorkingPrecision -> 100], u = (-1 + t); v = t/u;
CMRB - NIntegrate[Im[g[(1 + I v)]/(Sinh[\[Pi] v] u^2)], {t, 0, 1},
WorkingPrecision -> 100],
CMRB - NIntegrate[Im[g[(1 - I v)]/(Sinh[-\[Pi] v] u^2)], {t, 0, 1},
WorkingPrecision -> 100]]

During evaluation of In[1]:= {-9.3472*10^-94,-9.3472*10^-94}{-9.3472*10^-94,-9.3472*10^-94}{-9.3472*10^-94,-9.3472*10^-94}

In[23]:= Quiet[
NIntegrate[
Im[g[(1 + I t)]/Sinh[\[Pi] t] -
g[(1 + I v)]/(Sinh[\[Pi] v] u^2)], {t, 1, Infinity},
WorkingPrecision -> 100]]

Out[23]= -3.\
9317890831820506378791034479406121284684487483182042179057328100219696\
20202464096600592983999731376*10^-55

In[21]:= Quiet[
NIntegrate[
Im[g[(1 + I t)]/Sinh[\[Pi] t] -
g[(1 - I v)]/(Sinh[-\[Pi] v] u^2)], {t, 1, Infinity},
WorkingPrecision -> 100]]

Out[21]= -3.\
9317890831820506378791034479406121284684487483182042179057381396998279\
83065832972052160228141179706*10^-55

In[25]:= Quiet[
NIntegrate[
Im[g[(1 + I t)]/Sinh[\[Pi] t] +
g[(1 + I v)]/(Sinh[-\[Pi] v] u^2)], {t, 1, Infinity},
WorkingPrecision -> 100]]

Out[25]= -3.\
9317890831820506378791034479406121284684487483182042179057328100219696\
20202464096600592983999731376*10^-55


## On 9/29/2021

I found the following equation for CMRB (great for integer arithmetic because

(1-1/n)^k=(n-1)^k/n^k. )

![CMRB integers 1][103]

So, using only integers, and sufficiently large ones in place of infinity, we can use

![CMRB integers 2][104]

See

In[1]:= Timing[m=NSum[(-1)^n (n^(1/n)-1),{n,1,Infinity},WorkingPrecision->200,Method->"AlternatingSigns"]][[1]]

Out[1]= 0.086374

In[2]:= Timing[m-NSum[(-1)^n/x! (Sum[((-1 + n)^k) /(k n^(1 + k)), {k, 1, Infinity}])^ x, {n, 2, Infinity}, {x, 1,100}, Method -> "AlternatingSigns",  WorkingPrecision -> 200, NSumTerms -> 100]]

Out[2]= {17.8915,-2.2*^-197}


It is very much slower, but it can give a rational approximation (p/q), like in the following.

In[3]:= mt=Sum[(-1)^n/x! (Sum[((-1 + n)^k) /(k n^(1 + k)), {k, 1,500}])^ x, {n, 2,500}, {x, 6}];

In[4]:= N[m-mt]

Out[4]= -0.00602661

Out[5]= Rational


Compared to the NSum formula for m, we see

In[6]:= Head[m]

Out[6]= Real


## On 9/19/2021

I found the following quality of CMRB.

![replace constants for CMRB][105]

## On 9/5/2021

I added the following MRB constant integral over an unusual range.

![strange][106]

See proof [in this link here][107].

## On Pi Day, 2021, 2:40 pm EST,

I added a new MRB constant integral.

![CMRB][108] ![=][109] ![integral to sum][110]

We see many more integrals for CMRB.

We can expand ![1/x][111] into the following.

![xx = 25.656654035][112]

xx = 25.65665403510586285599072933607445153794770546058072048626118194\
90097321718621288009944007124739159792146480733342667100.;

g[x_] = x^(xx/
x); I NIntegrate[(g[(-t I + 1)] - g[(t I + 1)])/(Exp[Pi t] -
Exp[-Pi t]), {t, 0, Infinity}, WorkingPrecision -> 100]

(*
0.18785964246206712024851793405427323005590309490013878617200468408947\
72315646602137032966544331074969.*)


Expanding upon the previously mentioned

![enMRB sinh][113]

we get the following set of formulas that all equal CMRB:

Let

x= 25.656654035105862855990729 ...

along with the following constants (approximate values given)

{u = -3.20528124009334715662802858},

{u = -1.975955817063408761652299},

{u = -1.028853359952178482391753},

{u = 0.0233205964164237996087020},

{u = 1.0288510656792879404912390},

{u = 1.9759300365560440110320579},

{u = 3.3776887945654916860102506},

{u = 4.2186640662797203304551583} or

$u = \infty .$

Another set follows.

let x = 1 and

along with the following {approximations}

{u = 2.451894470180356539050514},

{u = 1.333754341654332447320456} or

$u = \infty$

then

![enter image description here][114]

See [this notebook from the wolfram cloud][115] for justification.

## 2020 and before:

Also, in terms of the Euler-Riemann zeta function,

CMRB =![enter image description here][116]

Furthermore, as ![enter image description here][117],

according to [user90369][118] at Stack Exchange, CMRB can be written as the sum of zeta derivatives similar to the eta derivatives discovered by Crandall. ![zeta hint ][119] Information about η(j)(k) please see e.g. [this link here][120], formulas (11)+(16)+(19).![credit][121]

In the light of the parts above, where

CMRB

= ![k^(1/k)-1][122]

= ![eta'(k)][123]

= ![sum from 0][124] ![enter image description here][125] as well as ![double equals RHS][126] an internet scholar going by the moniker "Dark Malthorp" wrote:

![eta *z^k][127]

## Primary Proof 1

CMRB=![enter image description here][128], based on

CMRB ![eta equals][129] ![enter image description here][130]

is proven below by an internet scholar going by the moniker "Dark Malthorp."

![Dark Marthorp's proof][131]

## Primary Proof 2

![eta sums][132] denoting the kth derivative of the Dirichlet eta function of k and 0 respectively, was first discovered in 2012 by Richard Crandall of Apple Computer.

The left half is proven below by Gottfried Helms and it is proven more rigorously![(][133]considering the conditionally convergent sum,![enter image description here][134]![)][135] below that. Then the right half is a Taylor expansion of eta(s) around s = 0.

![n^(1/n)-1][136]

At [https://math.stackexchange.com/questions/1673886/is-there-a-more-rigorous-way-to-show-these-two-sums-are-exactly-equal][137],

it has been noted that "even though one has cause to be a little bit wary around formal rearrangements of conditionally convergent sums (see the [Riemann series theorem][138]), it's not very difficult to validate the formal manipulation of Helms. The idea is to cordon off a big chunk of the infinite double summation (all the terms from the second column on) that we know is absolutely convergent, which we are then free to rearrange with impunity. (Most relevantly for our purposes here, see pages 80-85 of this [document][139], culminating with the Fubini theorem which is essentially the manipulation Helms is using.)"

![argument 1][140] ![argument 2][141]

## Primary Proof 3

Here is proof of a faster converging integral for its integrated analog (The MKB constant) by Ariel Gershon.

g(x)=x^(1/x), M1=![hypothesis][142]

Which is the same as

![enter image description here][143] because changing the upper limit to 2N + 1 increases MI by 2i/?.

MKB constant calculations have been moved to their discussion at [http://community.wolfram.com/groups/-/m/t/1323951?ppauth=W3TxvEwH][144] .

![Iimofg->1][145]

![Cauchy's Integral Theorem][146]

![Lim surface h gamma r=0][147]

![Lim surface h beta r=0][148]

![limit to 2n-1][149]

![limit to 2n-][150]

Plugging in equations [5] and [6] into equation [2] gives us:

![left][151]![right][152]

Now take the limit as N?? and apply equations [3] and [4] : ![QED][153] He went on to note that

![enter image description here][154]

I wondered about the relationship between CMRB and its integrated analog and asked the following. ![enter image description here][155] So far I came up with

Another relationship between the sum and integral that remains more unproven than I would like is

f[x_] = E^(I \[Pi] x) (1 - (1 + x)^(1/(1 + x)));
CMRB = NSum[f[n], {n, 0, Infinity}, WorkingPrecision -> 30,
Method -> "AlternatingSigns"];
M2 = NIntegrate[f[t], {t, 0, Infinity I}, WorkingPrecision -> 50];
part = NIntegrate[(Im[2 f[(-t)]] + (f[(-t)] - f[(t)]))/(-1 +
E^(-2 I \[Pi] t)), {t, 0, Infinity I}, WorkingPrecision -> 50];
CMRB (1 - I) - (M2 - part)
`

gives

6.10377910^-23 - 6.10377910^-23 I.

Where the integral does not converge, but Mathematica can give it a value:

# Update 2015

Here is my mini-cluster of the fastest 3 computers (the MRB constant supercomputer 0) mentioned below: The one to the left is my custom-built extreme edition 6 core and later with an 8 core 3.4 GHz Xeon processor with 64 GB 1666 MHz RAM.. The one in the center is my fast little 4-core Asus with 2400 MHz RAM. Then the one on the right is my fastest -- a Digital Storm 6 core overclocked to 4.7 GHz on all cores and with 3000 MHz RAM.

see notebook

Likewise, Wolfram Alpha here says