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Exploring the Upper End of the Partial Sums of the MRB constant in Mma

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f[x_] := NSum[(-1)^n (n^(1/n) - 1), {n, x, Infinity}, Method -> "AlternatingSigns", WorkingPrecision -> 400]

As x grows without bound f[10^x] == 10^-x x Log[10]/2 leads to the following acceleration method for the partial sums of the MRB constant

enter image description here

Here is an example:

In[225]:= 
m = NSum[(-1)^n*(n^(1/n) - 1), {n, 1, Infinity}, 
   Method -> "AlternatingSigns", WorkingPrecision -> 400];

standard[x_] := (NSum[(-1)^n (n^(1/n) - 1), {n, 1, 10^x}, 
   Method -> "AlternatingSigns", WorkingPrecision -> 500])

plus[x_] := (standard[x] - 10^-x x Log[10]/2)

Table[N[{(m - standard[x]), (m - plus[x])}, 30], {x, 0, 100, 10}]

Out[228]= {{0.187859642462067120248517934054, 
  0.187859642462067120248517934054}, \
{-1.15129254776743274319429239444*10^-9, \
-1.27040990118529666709714903244*10^-18}, \
{-2.30258509299404568453691834027*10^-19, \
-5.18926885582869572712277056422*10^-38}, \
{-3.45387763949106852602698718214*10^-29, \
-1.17590768666018420974624000155*10^-57}, \
{-4.60517018598809136803598290937*10^-39, \
-2.09798339326141874738448664089*10^-77}, \
{-5.75646273248511421004497863671*10^-49, \
-3.28515400538657318555081659969*10^-97}, \
{-6.90775527898213705205397436405*10^-59, \
-4.73741952303564752424522987791*10^-117}, \
{-8.05904782547915989406297009140*10^-69, \
-6.45477994620864176346772647557*10^-137}, \
{-9.21034037197618273607196581874*10^-79, \
-8.43723527490555590321830639266*10^-157}, \
{-1.03616329184732055780809615461*10^-88, \
-1.06847855091263899434969696292*10^-176}, \
{-1.15129254649702284200899572734*10^-98, \
-1.31974306488711438843037161851*10^-196}}

That table looks like

enter image description here

We see,,for x>0, the simple 10^-x x Log[10]/2 increases the accuracy exponentialy!!

in the previous reply we saw something interesting about f[10^10^x], f[10^10^x] being isomorphic to 10^x Log[10]/2: as x grows without bound f[10^10^x]*10^10^x == 10^x Log[10]/2, but those arguments are so big and the results are so small, it is hard to conceptualize! I just now stumbled on to something interesting about the easier to imagine f[10^x] : As x grows without bound f[10^x] == 10^-x x Log[10]/2 . The former is just a special case of the later.. As always, your comments are welcome!

f[x_] := NSum[(-1)^n (n^(1/n) - 1), {n, x, Infinity}, 
      Method -> "AlternatingSigns", WorkingPrecision -> 400]

You can see as x grows without bound f[10^x] == 10^-x x Log[10]/2 in the following two pieces of code and results:

First we will compute some values of f:

In[126]:= Table[N[f[10^(x)], 30], {x, 0, 20}]

Out[126]= {0.187859642462067120248517934054, \
0.133553295001578355542577811018, -0.976341352833974415772015902634, \
0.00346732160172991953948973985134, \
0.000460749704412491717791977788008, \
0.0000575682039917382492456392183551, 
 6.90780620020678040526552496824*10^-6, 
 8.05905469826083516701897111654*10^-7, 
 9.21034126383164238279161762724*10^-8, 
 1.03616330307674596183280961166*10^-8, 
 1.15129254787756199809757665899*10^-9, 
 1.26642180131318965307268175465*10^-10, 
 1.38155105581618001913660951164*10^-11, 
 1.49668031044844208057889632681*10^-12, 
 1.61180956509609958086783296937*10^-13, 
 1.72693881974556492465977490514*10^-14, 
 1.84206807439524003003258758969*10^-15, 
 1.95719732904493922401341786377*10^-16, 
 2.07232658369464115957273029594*10^-17, 
 2.18745583834434340470892771857*10^-18, 
 2.30258509299404568455944419120*10^-19}

Then look the the difference between f[10^x] and x/2 Log[10] 10^-x. Notice the negative magnitude is doubled already, as x gets to 20.

In[133]:= Table[N[f[10^x] - x/2 Log[10] 10^-x, 30], {x, 0, 20}]

Out[133]= {0.187859642462067120248517934054, \
0.0184240403518760713416782382842, -0.999367203763914872612195817181, \
0.0000134439622388510134627526693130, 
 2.32685813682580988379497071386*10^-7, 
 3.57666688710714518943198794863*10^-9, 
 5.09212246433532115506041838758*10^-11, 
 6.87278167527295600102514565496*10^-13, 
 8.91855459646719651808498865426*10^-15, 
 1.12294254040247134570557355366*10^-16, 
 1.38053915608858093164841550814*10^-18, 
 1.66464526862786454571667564300*10^-20, 
 1.97526087258146388312728449837*10^-22, 
 2.31238596720188126042014063305*10^-24, 
 2.67602055238951095541466771904*10^-26, 
 3.06616462813141282526360313271*10^-28, 
 3.48281819442594105125308186257*10^-30, 
 3.92598125127289004134797998316*10^-32, 
 4.39565379867223450785790828180*10^-34, 
 4.89183583662397138196824776735*10^-36, 
 5.41452736512810029562830508680*10^-38}

Any comments as to why this is, would be welcome!

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