For what value of x is f(x)=x^(1/x) a maximum?

f'(x)=x^(1/x-2)(1-lnx)=0,         

x=e and gives a maximum of e^(1/e)=1.444667861...
Since         = 3, consider
.
Given that C MRB =
it follows that
where A is the Glaisher-Kinkelin constant.
user90369 (https://math.stackexchange.com/users/332823/user90369), High precision
evaluation of the series $\sum_{n=3}^\infty (-1)^n (1-n^{1/n})$, URL (version: 2019-07-16):
https://math.stackexchange.com/q/3294599
CMRB from Steiner's Problem
The MRB constant
(CMRB)
eta formula proof
.
These first equations were first published
in book form by
Finch, Steven R. (2003). Mathematical Constants.
Cambridge, England: Cambridge University Press. p. 450.
ISBN 0-521-81805-2
.
The following formulas were first published in book form by
Crandall, R. E. "The MRB Constant." §7.5 in Algorithmic Reflections:
Selected Works. PSI Press, pp. 28-29, 2012b.
They are an application of
Fubini’s theorem for double series.
The following identities are applications of the Abel-Plana formula,
first published  by
The Online Encyclopedia of Integer Sequences oeis.org/wiki/MRB_constant
Dark Malthorp
(https://math.stackexchange.com/users/532432/dark-malthorp), What
are some working models that are a fit the formula for the MRB
constant?, URL (version: 2020-01-12):
https://math.stackexchange.com/q/3505694
5,555,555 digit computation of CMRB
(MRB supercomputer output at 5:40:08 pm EDT  |  Monday, August 10, 2020)
Start time was Fri 19 Jun 2020 22:20:08.

Iterations required: 7369998

Will give 3599 time estimates, each more accurate than the previous.

Will stop at 7370752 iterations to ensure precision of around 5583332 decimal
places....
3166208 iter. done in 4.476*10^6 s. Ave. 0.70741iter./s. Should take 120.58 days
or 1.042*10^7s. Finish Sun 18 Oct 2020 12:18:31.

{42.9564,% done.}
MRB supercomputer
Processors: Intel Core i9-9900K (5.0 GHz Turbo) (16-Thread) (8-Core) 3.6
GHz. CPU Boost: Stage 2: Overclock CPU - Up to 5.1GHz on All CPU Cores at
3200 Mhz RAM Extreme Cooling: H20: Stage 2: Corsair H115i PRO - 280mm
Liquid CPU Cooler  with the following two exterior nodes: 4.7 GHz on all Intel
6 cores at 3000 MHz RAM and 4 cores of 3.6 GHz at 2400 MHz RAM!

{{u -> -3.205281240093347156628042917392342206845495258334899739232149151500845258217206440409918401887943074},

{u -> -1.975955817063408761652299553542124207955844914033641475935228264550492865978094627264109660184110877},
{u -> -1.028853359952178482391753039155168552490590931630510790413956782402957713372878008212396966920395899},
{u ->  0.02332059641642379960870201829771316178976827597634872153712935185694600973532346147587251063387838184},
{u -> 1.028851065679287940491239061962065335882572934488973325479147532561027453451485653303655759243419286},
{u -> 1.975930036556044011032057974393627818460615746707424999683678154612112980778643821981630632223625195},
{u -> 3.377688794565491686010340135121834601964847740798723565007400061225550945863584207850080547920439455},
{u -> 4.218664066279720330451890569753298644162048163522884025276106357736553673659421134883269758349165422}}
or
u=infinity


The MRB constant is defined at http://mathworld.wolfram.com/MRBConstant.html.
After a lot of looking I found a connection between the MRB constant and applied math:  

The MRB constant is ∑(−1)^k(k^(1/k)−1), and that k^(1/k)−1 is the interest rate
to multiply an investment k times in k periods -- as well as other growth models
involving the more general expression (1+k)^n --
since
((k^(1/k)−1)+1)^k|k∈Z+=k. and ((k^(1/n)−1)+1)^n|n∈Z+=k.

We can say, the result of summing, with alternating signs, the interest rate to multiply an investment k times in k
periods (or the equivalent growth model) could be the end "growth" rate resulting from growth, following decay,
following growth, ad infinitum.  
Real life application
Click for the Geometry
of the MRB constant
paper.
Geometry

The Geometry of the MRB Constant
Authors: Marvin Ray Burns
The MRB constant is the upper limit point of the sequence of
partial sums defined by S(x)=sum((- 1)^n*n^(1/n),n=1..x). The
goal of this paper is to show that the MRB constant is
geometrically quantifiable. To “measure” the MRB constant, we
will consider a set, sequence and alternating series of the nth
roots of n. Then we will compare the length of the edges of a
special set of hypercubes or ncubes which have a content of n.
(The two words hypercubes and n-cubes will be used
synonymously.) Finally, we will look at the value of the MRB
constant as a representation of that comparison, of the length of
the edges of a special set of hypercubes, in units of dimension 1/
(units of dimension 2 times units of dimension 3 times units of
dimension 4 times etc.). For an arbitrary example we will use
units of length/ (time*mass* …).
Comments: 8 Pages. This classic paper shows the utter
simplicity of the geometric description of the MRB constant (oeis.
org/A037077).
Download: PDF
Submission history
[v1] 2016-09-06 19:08:36
Inquiries?
bmmmburns@sbcglobal.net
Try to break these CMRB Computational
Records!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
(Click for the full amazing ride!)
copy to the right ->->
{{u ->1.333754341654332447320456098329979657122884399228753780402849117971679458574998581211071632367720673},
{u -> 2.451894470180356539050514838856255986670981536525622143367559422740744344429034215340055686290476534}}
or
u=infinity
Or let x=1 and
The MRB constant is related to the following oscillating divergent series
Its partial sums
are bounded within the closed interval
and where the MRB constant is defined as
since
The MRB constant can be explicitly defined by the following infinite sums
thus fulfilling the necessary and sufficient conditions for the
convergence of an alternating series.
where
converges. (Remember what this means in the case of infinite sums of positive
terms: it means that there is a number K such that every finite partial sum S is
bounded above by K; the least such upper bound will be the number that the
infinite sum converges to.) So take any such finite partial sum S, and rearrange
its terms so that the terms in the m=2 column come first, then the terms in the
m=3 column, and so on. An upper bound for the terms of S in the m=2 column is
ζ(2)(2)2!. Put that one aside.
So now we check the absolute convergence of the right-hand side, i.e., that
which certainly converges. So we have absolute convergence of the doubly infinite
sum.

Thus we are in a position to apply the Fubini theorem, which justifies the
rearrangement expressed in the first of the following equations
giving us what we wanted.
then
bounds together, an upper bound for the entire doubly infinite sum would be
For the m=3 column, an upper bound is
(we drop the n=1 term
which is 0). By calculus we have
for all n≥2, so this has upper
bound
by an integral test, which yields
as an upper
bound. Applying the same reasoning for the m column from m=4 on, an upper
bound for that column would be
Adding all those upper
Let x=25.65665403510586285599072933607445153794770546058072048626118194900973217186212880099440071247391598 and
I discovered the following via the Mathematica notebook to the right. ->->
Integrated analog of the CMRB series
The integrated analog of the series is a complex-valued integral of oscillatory character



Not convergent in the continuum limit at  ,         the limit of the sequence of integrals with an
integral difference in the upper limits 2n  exists.
Ultraviolet limit MI of the sequence of oscillatory integrals
The ultraviolet limit of the sequence of oscillatory integrals is defined as



and has been evaluated by
Richard J. Mathar.


The decimal expansion of the real part of MI is

The decimal expansion of the imaginary part of MI is


The MRB constant notebook to the right. ->->

m=the MRB constant. We looked at how n^m-m is similar to E^Pi-Pi (a near integer). One might think this is off the subject of breaking computational records of
the MRB constant, but it also could help show whether there exists a closed-form for computing and checking the digits of m from n^m-m=a near integer and n
is an integer.

So, I decided to make an extremely deep search of the n^m-m=a near integer, and n is an integer field. Here are the pearls I gleaned:

In[35]:= m =
 NSum[(-1)^n (n^(1/n) - 1), {n, 1, Infinity}, WorkingPrecision -> 100,
   Method -> "AlternatingSigns"];

In[63]:= 225897077238546^m - m

Out[63]= 496.99999999999999975304752932252481772179797865

In[62]:= 1668628852566227424415^m - m

Out[62]= 9700.9999999999999999994613109586919797992822178

In[61]:= 605975224495422946908^m - m

Out[61]= 8019.9999999999999999989515156294756517433387956

In[60]:= 3096774194444417292742^m - m

Out[60]= 10896.0000000000000000000000096284579090392932063

In[56]:= 69554400815329506140847^m - m

Out[56]= 19549.9999999999999999999999991932013520540825206

In[68]:= 470143509230719799597513239^m - m

Out[68]= 102479.000000000000000000000000002312496475978584

In[70]:= 902912955019451288364714851^m - m

Out[70]= 115844.999999999999999999999999998248770510754951

In[73]:= 2275854518412286318764672497^m - m

Out[73]= 137817.000000000000000000000000000064276966095482

In[146]:= 2610692005347922107262552615512^m - m

Out[146]= 517703.00000000000000000000000000000013473353420

In[120]:= 9917209087670224712258555601844^m - m

Out[120]= 665228.00000000000000000000000000000011062183643

In[149]:= 19891475641447607923182836942486^m - m

Out[149]= 758152.00000000000000000000000000000001559954712

In[152]:= 34600848595471336691446124576274^m - m

Out[152]= 841243.00000000000000000000000000000000146089062

In[157]:= 543136599664447978486581955093879^m - m

Out[157]= 1411134.0000000000000000000000000000000035813431

In[159]:= 748013345032523806560071259883046^m - m

Out[159]= 1498583.0000000000000000000000000000000031130944

In[162]:= 509030286753987571453322644036990^m - m

Out[162]= 1394045.9999999999999999999999999999999946679646


In[48]:= 952521560422188137227682543146686124^m - m

Out[48]=5740880.999999999999999999999999999999999890905129816474332198321490136628009367504752851478633240


In[26]:= 50355477632979244604729935214202210251^m - m

Out[26]=12097427.00000000000000000000000000000000000000293025439870097812782596113788024271834721860892874


In[27]:= 204559420776329588951078132857792732385^m - m

Out[27]=15741888.99999999999999999999999999999999999999988648448116819373537316944519114421631607853700001


In[46]:= 4074896822379126533656833098328699139141^m - m

Out[46]= 27614828.00000000000000000000000000000000000000001080626974885195966380280626150522220789167201350


In[8]:= 100148763332806310775465033613250050958363^m - m

Out[8]= 50392582.999999999999999999999999999999999999999998598093272973955371081598246


In[10]=  116388848574396158612596991763257135797979^m - m

Out[10]=51835516.000000000000000000000000000000000000000000564045501599584517036465406


In[12]:= 111821958790102917465216066365339190906247589^m - m

Out[12]= 188339125.99999999999999999999999999999999999999999999703503169989535000879619


In[33] := 8836529576862307317465438848849297054082798140^m - m

Out[33] = 42800817.00000000000000000000000000000000000000000000000321239755400298680819416095288742420653229


In[71] := 532482704820936890386684877802792716774739424328^m - m

Out[71] =924371800.999999999999999999999999999999999999999999999998143109316148796009581676875618489611792


In[21]:= 783358731736994512061663556662710815688853043638^m - m

Out[21]= 993899177.0000000000000000000000000000000000000000000000022361744841282020


In[24]:= 8175027604657819107163145989938052310049955219905^m - m

Out[24]= 1544126008.9999999999999999999999999999999999999999999999999786482891477\
944981


19779617801396329619089113017251584634275124610667^m - m
gives
1822929481.00000000000000000000000000000000000000000000000000187580971544991111083798248746369560.


130755944577487162248300532232643556078843337086375^m - m

gives

2599324665.999999999999999999999999999999999999999999999999999689854836245815499119071864529772632.
i.e.2, 599, 324, 665. 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 689

(51 consecutive 9 s)

322841040854905412176386060015189492405068903997802^m - m

gives

3080353548.000000000000000000000000000000000000000000000000000019866002281287395703598786588650156

i.e. 3, 080, 353, 548. 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000,019

(52 consecutive 0 s)


310711937250443758724050271875240528207815041296728160^m - m

gives

11195802709.99999999999999999999999999999999999999999999999999999960263763...
i.e. 11,195,802,709. 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 602, 637,63

(55 consecutive 9s)

1465528573348167959709563453947173222018952610559967812891154^ m - m  
gives
200799291330.9999999999999999999999999999999999999999999999999999999999999900450730197594520134278  
i. e 200, 799, 291, 330.999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 99

(62 consecutive 9 s).
I did a little experimenting and found that the existence of such precise near integers from n^m-m is not specific to the MRB constant due to any internal
quality, only due to the small value of m. You can find similarly precise approximations for n^x-x, where x is 1/E. 1/E(.37) has a value in the same neighborhood
of the MRB constant (.19).
MRB constant
CMRB=
1
OPEN QUESTION.
Which algorithm is faster?
OR
WHERE
From
https://www.sciencedirect.com/science/article/pii/S0022314X15001882
1

user43208 (https://math.stackexchange.com/users/43208/user43208),
Is there a more rigorous way to show these two sums are exactly
equal?, URL (version: 2016-02-28):
https://math.stackexchange.com/q/1675189
OR
WHERE
CMRB=