The EulerGamma*Log[2] - Log[2]^2/2 and 5 (Log[Pi] - Log[2]) Problems

Questions
1

Considering the MRB constant which is approximated by

NSum[(-1)^n (n^(1/n) - 1), {n, 1, Infinity}, WorkingPrecision -> 70, 
 Method -> "AlternatingSigns"]

giving

0.18785964246206712024851793405427323005590309490013878617200468547342,

I wondered what happens when we change 1/n in n^(1/n) to 1 over other functions involving n?

Then I figured out that

Table[NSum[(-1)^n (1/n^(1/(n*10^x)) - 1), {n, 1, Infinity}, 
   WorkingPrecision -> 70, 
   Method -> 
    "AlternatingSigns"] + (EulerGamma*Log[2] - Log[2]^2/2)*10^-x, {x, 
  1, 30}]

and also

Table[NSum[(-1)^n (n^(1/(n*10^x)) - 1), {n, 1, Infinity}, 
   WorkingPrecision -> 70, 
   Method -> 
    "AlternatingSigns"] - (EulerGamma*Log[2] - Log[2]^2/2)*10^-x, {x, 
  1, 30}]

mantissa’s converge to the same constant even though one formula uses the reciprocal of what the other uses!

 {0.0002544931414649952541958458971418195836265633470373523204886572940\
 13, 2.5213764023532449457379170196206807316529005454922927744915334919\
 *10^-6, 2.\
 519038430344263802098258061692809581811767812023594203581344434*10^-8,
   2.51880480742043470745309496704890093429551288263349038480600225*10^\
 -10, 2.5187814468698119802633255716680901574100656515276061645434502*\
 10^-12, 2.\
 518779110832166299320533338340144497540312267653607154108212*10^-14, 
  2.51877887722857589613401551895739497023827511567917103396240*10^-16,
   2.5187788538682185974634313552273103698241994129327919890091*10^-18,
   2.518778851532182885012852605041013681772455184206979797782*10^-20, 
  2.51877885129857931394195952567425580667348404161289030118*10^-22, 
  2.5187788512752189568366118656930887417812470506544970303*10^-24, 
  2.518778851272882921126094516174526112522881012060672865*10^-26, 
  2.51877885127264931755504295538746538935935766591293015*10^-28, 
  2.5187788512726259571979378010504072724396184685687105*10^-30, 
  2.518778851272623621162227285634117940301609670334958*10^-32, 
  2.51877885127262338755865623409266317188334844194942*10^-34, 
  2.5187788512726233641982991289385194366894777279525*10^-36, 
  2.518778851272623361862263418423105080586570333924*10^-38, 
  2.51877885127262336162865984737156364515044562412*10^-40, 
  2.5187788512726233616052994902664095016085871417*10^-42, 
  2.518778851272623361602963454555894087254542116*10^-44, 
  2.51877885127262336160272985098484254582037185*10^-46, 
  2.5187788512726233616027064906277373916892954*10^-48, 
  2.518778851272623361602704154592026876399594*10^-50, 
  2.51877885127262336160270392098845582610468*10^-52, 
  2.5187788512726233616027038976280987334158*10^-54, 
  2.518778851272623361602703895292063147553*10^-56, 
  2.51877885127262336160270389505846082303*10^-58, 
  2.5187788512726233616027038950351129312*10^-60, 
  2.518778851272623361602703895032901548*10^-62}.

This is for the same reason that

NSum[(-1)^n (-1 + 1/n^(10^-x/n) + (10^-x Log[n])/n), {n, 1, Infinity},
   WorkingPrecision -> 70, Method -> "AlternatingSigns"] /. x -> 30

and

NSum[(-1)^n (-1 + n^(10^-x/n) - (10^-x Log[n])/n), {n, 1, Infinity}, 
  WorkingPrecision -> 70, Method -> "AlternatingSigns"] /. x -> 30

give

2.51877885127262336160270389503276304518288666816092953082281396924871\
*10^-62

(For version 3.1 you will have to use x->27 or less.)

(I wonder what closed form starts with 2.518778851272623361602703895?)

While changing the sign of “1:”

NSum[(-1)^n (1 + n^(10^-x/n) - (10^-x Log[n])/n), {n, 1, Infinity}, 
  WorkingPrecision -> 70, Method -> "AlternatingSigns"] /. x -> 30

gives a “divergent sum” of

-1.0000000000000000000000000000000000000000000000000000000000038475945

And another constant, with formulas behaving differently, I noticed is from

Table[NSum[(-1)^n (n^(1/(n + 10^x)) - 1), {n, 1, Infinity}, 
   WorkingPrecision -> 70, 
   Method -> 
    "AlternatingSigns"] - (5 (Log[Pi] - Log[2]))*10^-(x + 1), {x, 1, 
  30}]

which gives

  {-0.002158265971057576296292446726899637434277905244619602754711500291\
  280, -0.00002326878830432348151050186189530613369186033070715442565971\
  5184159, -2.\
  344972498540413669051157226943914337661182682378121265593728211*10^-7,\
   -2.34679530411087431851660597513831629183877508983564770824474081*10^\
  -9, -2.3469777203157053575942148177302233368868838187419725158605417*\
  10^-11, -2.\
  346995963293941260610236486862825114739849987209310140146848*10^-13, \
  -2.34699778760534365532625414499456151604960848632944848056957*10^-15,\
   -2.3469979700366196841185707695593955974492155812208948712378*10^-17,\
   -2.346997988279748644892286166044070017626312946190793075881*10^-19, \
  -2.34699799010406155454860381961973372505412612579214679476*10^-21, \
  -2.3469979902864928456500250473931591201743603058719161031*10^-23, \
  -2.346997990304735974761525064795936837006618410461168616*10^-25, \
  -2.34699799030656028767268864548247023705012450346433171*10^-27, \
  -2.3469979903067427189638051393405861346146650120718647*10^-29, \
  -2.346997990306760962092916790084292349947997733009464*10^-31, \
  -2.34699799030676278640582795517224191773710290844785*10^-33, \
  -2.3469979903067629688371190716811726639785895468696*10^-35, \
  -2.346997990306762987080248183332067096497547977600*10^-37, \
  -2.34699799030676298890456109449715655333023197512*10^-39, \
  -2.3469979903067629890869923856136654991677088242*10^-41, \
  -2.346997990306762989105235514725316393937004272*10^-43, \
  -2.34699799030676298910705982763648148525584608*10^-45, \
  -2.3469979903067629891072422589275980128067172*10^-47, \
  -2.346997990306762989107260502056709849751672*10^-49, \
  -2.34699799030676298910726232636962287534485*10^-51, \
  -2.3469979903067629891072625088009325968910*10^-53, \
  -2.346997990306762989107262527044247758914*10^-55, \
  -2.34699799030676298910726252887042117380*10^-57, \
  -2.3469979903067629891072625290714575021*10^-59, \
  -2.346997990306762989107262529275751003*10^-61}

(Likewise I wonder what closed form starts with 2.34699799030676298910726?)

However here we have a different constant converged to by using the reciprocal

Table[NSum[(-1)^n (1/n^(1/(n + 10^x)) - 1), {n, 1, Infinity}, 
   WorkingPrecision -> 70, 
   Method -> 
    "AlternatingSigns"] + (5 (Log[Pi] - Log[2]))*10^-(x + 1), {x, 1, 
  30}]

giving

{0.0027281794786312486595717428387460304770563641777457765720218456040\
40, 0.0000293550399240151127322722379565607620916546859019350100125647\
09573, 2.9551297193773754366885564592734651504693514196068487319926172\
96*10^-7, 
 2.95707365395153655826155461847169020429215068621347502129971599*10^-\
9, 2.9572678525605426063529814114458282642227732785054465491056943*10^\
-11, 2.957287270464422416597614464460409791592623556524877730647258*\
10^-13, 2.95728921223523164631517763738252605852113445125380329353929*\
10^-15, 2.9572894064121167732298398914267611418621574492035860745636*\
10^-17, 2.957289425829803327952191084785434676290469048265224666794*\
10^-19, 2.95728942777157196384472650970279641869142142361488168100*10^\
-21, 2.9572894279657488272381830467062488252286207798089293916*10^-23,
  2.957289427985166513575570730343167129427371191629855892*10^-25, 
 2.95728942798710828220928991900621614638391196404033743*10^-27, 
 2.9572894279873024590726616420755146114008339317553828*10^-29, 
 2.957289427987321876758998812424474393527194892652000*10^-31, 
 2.95728942798732381852763252943979067109607097234091*10^-33, 
 2.9572894279873240127044959011411265018465393721695*10^-35, 
 2.957289427987324032122182238311258126951706025741*10^-37, 
 2.95728942798732403406395087202827126988436394602*10^-39, 
 2.9572894279873240342581277353999725840002517184*10^-41, 
 2.957289427987324034277545421737142715594072394*10^-43, 
 2.95728942798732403427948719037085973059533356*10^-45, 
 2.9572894279873240342796813672342314505144463*10^-47, 
 2.957289427987324034279700784920568806696226*10^-49, 
 2.95728942798732403427970272668920438421309*10^-51, 
 2.9572894279873240342797029208660863609516*10^-53, 
 2.957289427987324034279702940283958748494*10^-55, 
 2.95728942798732403427970294222758788593*10^-57, 
 2.9572894279873240342797029424403697865*10^-59, 
 2.957289427987324034279702942645837845*10^-61}

I could also ask about the constant 2.957289427987324034279702942.

2

Unfortunately all the Inverse Symbolic Calculator web services seem to be down…

In Mathematica enter

=2.51877885127262336160270389503

followed by and when the result appears click the + at the right end of that and in a few seconds possible closed forms will be shown at the end of the results.

Likewise for

=2.34699799030676298910726253

I am not sure that any of those results will be exactly what you are looking for, but this method might lead you in the right direction.

If the Inverse Symbolic Calculator services come back and you increase the number of digits of precision then they might be able to find something interesting for you.

3

Considering the first new constant, 2.518778851272623361602703895,
by the function I used,

Table[NSum[(-1)^n (n^(1/(n*10^x)) - 1), {n, 1, Infinity}, 
   WorkingPrecision -> 70, 
   Method -> 
    "AlternatingSigns"] - (EulerGamma*Log[2] - Log[2]^2/2)*10^-x, {x, 
  1, 30}]

,
it is an approximation of the limit, as x goes to infinity, of Sum[(-1)^n (n^(1/(n10^x)) - 1]- (EulerGammaLog[2] - Log[2]^2/2)10^-x.
But we can go further find a beautifully concise pattern, for integer u, in Sum[(-1)^n (n^(u/(n10^x)) - 1], where x-> infinity.

Below we limit the magnitude of x to 60.

Enter

l = Table[
  NSum[(-1)^n (n^(u/(n*10^60)) - 1), {n, 1, Infinity}, 
    WorkingPrecision -> 140, Method -> "AlternatingSigns"] - 
   u (EulerGamma*Log[2] - Log[2]^2/2)*10^-60, {u, 1, 30}]

And you get a list of constants, the first of which is the one we are considering, that are related by the beautifully concise pattern.

The ratio of two consecutive elements of the list of constants, as x-> infinity is exactly equal to (u/(u - 1))^2. (I mean as shown, to a small degree (the magnitude of x used) in the next command:)

Table[l[[u]]/l[[u - 1]] - (u/(u - 1))^2, {u, 2, 30}]

,
giving

{4.121989880275861*10^-61, 2.318619307658961*10^-61, 
 1.8319955023485697*10^-61, 1.6101522969863107*10^-61, 
 1.4839163569027052*10^-61, 1.4026215564860330*10^-61, 
 1.3459558792769323*10^-61, 1.3042233605591426*10^-61, 
 1.2722190988536239*10^-61, 1.2469019387864491*10^-61, 
 1.2263771544652005*10^-61, 1.2094032808477541*10^-61, 
 1.1951331605562538*10^-61, 1.1829690345208172*10^-61, 
 1.1724771215035397*10^-61, 1.16333503457287374*10^-61, 
 1.15529820174103079*10^-61, 1.14817773671544167*10^-61, 
 1.14182545160274594*10^-61, 1.13612346075380612*10^-61, 
 1.13097681522584262*10^-61, 1.12630818526407794*10^-61, 
 1.12205395607027190*10^-61, 1.11816131735186967*10^-61, 
 1.11458606362931681*10^-61, 1.11129091077235512*10^-61, 
 1.10824419277921002*10^-61, 1.10541884225780500*10^-61, 
 1.10279158509433666*10^-61}

Where, for a comparison of the magnitude of the result,

Table[l[[u]]/l[[u - 1]], {u, 2, 30}]

by itself gives

{4.0000000000000000000000000000000000000000000000000000000000004121989\
880275861, \
2.25000000000000000000000000000000000000000000000000000000000023186193\
07658961, \
1.77777777777777777777777777777777777777777777777777777777777796097732\
801263475, \
1.56250000000000000000000000000000000000000000000000000000000016101522\
969863107, \
1.44000000000000000000000000000000000000000000000000000000000014839163\
569027052, \
1.36111111111111111111111111111111111111111111111111111111111125137326\
675971441, \
1.30612244897959183673469387755102040816326530612244897959183686928946\
547871364, \
1.26562500000000000000000000000000000000000000000000000000000013042233\
605591426, \
1.23456790123456790123456790123456790123456790123456790123456802845647\
778659696, \
1.21000000000000000000000000000000000000000000000000000000000012469019\
387864491, \
1.19008264462809917355371900826446280991735537190082644628099185817490\
552916468, \
1.17361111111111111111111111111111111111111111111111111111111123205143\
919588652, \
1.15976331360946745562130177514792899408284023668639053254437881773816\
812663130, \
1.14795918367346938775510204081632653061224489795918367346938787339894\
426840825, \
1.13777777777777777777777777777777777777777777777777777777777789502548\
992813175, \
1.12890625000000000000000000000000000000000000000000000000000011633350\
3457287374, \
1.12110726643598615916955017301038062283737024221453287197231845463016\
6194864325, \
1.11419753086419753086419753086419753086419753086419753086419764568197\
1202408364, \
1.10803324099722991689750692520775623268698060941828254847645440781135\
4024540522, \
1.10250000000000000000000000000000000000000000000000000000000011361234\
6075380612, \
1.09750566893424036281179138321995464852607709750566893424036292488906\
4742538911, \
1.09297520661157024793388429752066115702479338842975206611570259196966\
1501614406, \
1.08884688090737240075614366729678638941398865784499054820415890237552\
7932168967, \
1.08506944444444444444444444444444444444444444444444444444444455626057\
6179631411, \
1.08160000000000000000000000000000000000000000000000000000000011145860\
6362931681, \
1.07840236686390532544378698224852071005917159763313609467455632414684\
2556525453, \
1.07544581618655692729766803840877914951989026063100137174211259367764\
2871885336, \
1.07270408163265306122448979591836734693877551020408163265306133503168\
0144147847, \
1.07015457788347205707491082045184304399524375743162901307966717329937\
2540349242}

.

4

I found the closed form for the first constant, 2.518778851272623361602703895.
It is simply

(1/24)*(Pi^2*Log[2]^2 - 
2*Pi^2*Log[2]*(EulerGamma + Log[2] - 12*Log[Glaisher] + 
Log[Pi]) - 6*Derivative[2][Zeta][2])*10^2

.
There is a series for that:

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

And there is a beautiful infinite pattern here. Watch!

We started with

NSum[(-1)^n (n^(1/(n*10^x)) - 1), {n, 1, Infinity}, 
 WorkingPrecision -> 100, Method -> "AlternatingSigns"]

and subtracted

(EulerGamma*Log[2] - Log[2]^2/2)*10^-x

.
There is a series for that:

Sum[(-1)^x ((Log[x]/x)^n/n!), {x, 1, Infinity}, {n, 1, 1}]*10^-x

.
So we have

Table[NSum[(-1)^n (n^(1/(n*10^x)) - 1), {n, 1, Infinity}, 
   WorkingPrecision -> 100, Method -> "AlternatingSigns"] - 
  Sum[(-1)^x ((Log[x]/x)^n/n!), {x, 1, Infinity}, {n, 1, 
     1}]*10^-x, {x, 1, 30}]

,
which gives the the first table in the original post.

Make the following pattern and continue accordingly:

Table[NSum[(-1)^n (n^(1/(n*10^x)) - 1), {n, 1, Infinity}, 
   WorkingPrecision -> 100, Method -> "AlternatingSigns"] - 
  Sum[(-1)^x ((Log[x]/x)^n/n!), {x, 1, Infinity}, {n, 1, 1}]*10^-x - 
  Sum[(-1)^x ((Log[x]/x)^n/n!), {x, 1, Infinity}, {n, 2, 
     2}]*10^-(2 x), {x, 1, 30}]

gives

{2.6152563377329180355755076385692350606137403807501201077512739577652\
30155711748968474297054286*10^-6, 
 2.5975510806215841352131245881772459934044789262707706824608260551575\
153529153078065860130552*10^-9, 
 2.5957907164044049555416666030609615227174545757219977214711548625409\
6270760022013399885265*10^-12, 
 2.5956147811345850391072016397448636016816067468380995558407178876124\
03640576904872093523*10^-15, 
 2.5955971886186606216766355866717505695849615841607205411291165263509\
470430179461102463*10^-18, 
 2.5955954293771782944330764101188081620108758515016065026457882853789\
8304596179515815*10^-21, 
 2.5955952534531311623924891484578779049113149028768314853181102937749\
91201171199883*10^-24, 
 2.5955952358607274601948068841647033463667699713497189640262365027022\
846590206722*10^-27, 
 2.5955952341014871000851019611295911764095765546980280700263928771072\
3874805853*10^-30, 
 2.5955952339255630641752321013987975046866912343260344114031870319184\
58716989*10^-33, 
 2.5955952339079706605852561217509840884611761674524087985158346970920\
299249*10^-36, 
 2.5955952339062114202262686338494549451235866036679862257139000755633\
2623*10^-39, 
 2.5955952339060354961903699861599345523114527609431373079264032591400\
54*10^-42, 
 2.5955952339060179037867801224019888382449944033014145447457240102639\
*10^-45, 2.\
59559523390601614454642113603630433009049565657904411607196536202*10^-\
48, 2.595595233906015968622385237399836979907567252336000585907314760*\
10^-51, 2.\
5955952339060159510299816475361912558955996266155269435028561*10^-54, 
 2.59559523390601594927074128854982669360446611619051742421930*10^-57,
  2.595595233906015949094817252651190237476453397669486757807*10^-60, 
 2.5955952339060159490772248490613265918646631321425891468*10^-63, 
 2.59559523390601594907546560870234022730349421565222672*10^-66, 
 2.595595233906015949075289684666441590847377425011342*10^-69, 
 2.5955952339060159490752720922628517272017657377101*10^-72, 
 2.59559523390601594907527033302249274083720364418*10^-75, 
 2.595595233906015949075270157098456842200654954*10^-78, 
 2.5955952339060159490752701395060532523277520*10^-81, 
 2.59559523390601594907527013774681289241565*10^-84, 
 2.595595233906015949075270137570888763943*10^-87, 
 2.5955952339060159490752701375532871030*10^-90, 
 2.59559523390601594907527013755060212*10^-93}

(2.59559523390601594907527013755060212…has a closed form of

1(1/24)*(Log[2]^3*Zeta[3] - 3*Log[2]^2*Derivative[1][Zeta][3] + 
Log[8]*Derivative[2][Zeta][3] + 3*Derivative[3][Zeta][3])*10^3

and a series of Sum[(-1)^x ((Log[x]/x)^n/n!), {x, 1, Infinity}, {n, 3, 3}]

So subtract that series and wallah!

N[Table[NSum[(-1)^n (n^(1/(n*10^k)) - 1), {n, 1, Infinity}, 
    WorkingPrecision -> 200, Method -> "AlternatingSigns"] - 
   Sum[(-1)^x ((Log[x]/x)^n/n!), {x, 1, Infinity}, {n, 1, 1}]*10^-k - 
   Sum[(-1)^x ((Log[x]/x)^n/n!), {x, 1, Infinity}, {n, 2, 
      2}]*10^-(2 k) - 
   Sum[(-1)^x ((Log[x]/x)^n/n!), {x, 1, Infinity}, {n, 3, 
      3}]*10^-(3 k), {k, 1, 30}], 30]

gives

{1.9661103826902086500237501017910^-8,
1.9558467155681861378544506259010^-12,
1.9548249838900646639646550962010^-16,
1.9547228569090031931502193521910^-20,
1.9547126446726013654491204088310^-24,
1.9547116234535780627256746642010^-28,
1.9547115213317219010906536162610^-32,
1.9547115111195367466133616040110^-36,
1.9547115100983182357824943405710^-40,
1.9547115099961963847455762334410^-44,
1.9547115099859841996423461089210^-48,
1.9547115099849629811320277133310^-52,
1.9547115099848608592809959199410^-56,
1.9547115099848506470958927410610^-60,
1.9547115099848496258773824231810^-64,
1.9547115099848495237555313913910^-68,
1.9547115099848495135433462882110^-72,
1.9547115099848495125221277779010^-76,
1.9547115099848495124200059268710^-80,
1.9547115099848495124097937417610^-84,
1.9547115099848495124087725232510^-88,
1.9547115099848495124086704014010^-92,
1.9547115099848495124086601892210^-96,
1.9547115099848495124086591680010^-100,
1.9547115099848495124086590658810^-104,
1.9547115099848495124086590556610^-108,
1.9547115099848495124086590546410^-112,
1.9547115099848495124086590545410^-116,
1.9547115099848495124086590545310^-120,
1.9547115099848495124086590545310^-124}

Want to see that done two more times?

N[Table[NSum[(-1)^n (n^(1/(n*10^x)) - 1), {n, 1, Infinity}, 
    WorkingPrecision -> 200, Method -> "AlternatingSigns"] - 
   Sum[(-1)^x ((Log[x]/x)^n/n!), {x, 1, Infinity}, {n, 1, 1}]*10^-x - 
   Sum[(-1)^x ((Log[x]/x)^n/n!), {x, 1, Infinity}, {n, 2, 
      2}]*10^-(2 x) - 
   Sum[(-1)^x ((Log[x]/x)^n/n!), {x, 1, Infinity}, {n, 3, 
      3}]*10^-(3 x) - 
   Sum[(-1)^x ((Log[x]/x)^n/n!), {x, 1, Infinity}, {n, 4, 
      4}]*10^-(4 x), {x, 1, 30}], 30]

gives

{1.13988727053591376150910472612*10^-10, 
 1.13520558333662544579157137610*10^-15, 
 1.13473905215151555996041669650*10^-20, 
 1.13469241536807415602976577789*10^-25, 
 1.13468775185304046135429907865*10^-30, 
 1.13468728550317015609668432168*10^-35, 
 1.13468723886819945617277520989*10^-40, 
 1.13468723420470254948636257543*10^-45, 
 1.13468723373835286045078105450*10^-50, 
 1.13468723369171789156355349979*10^-55, 
 1.13468723368705439467499405030*10^-60, 
 1.13468723368658804498613973841*10^-65, 
 1.13468723368654141001725432355*10^-70, 
 1.13468723368653674652036578223*10^-75, 
 1.13468723368653628017067692809*10^-80, 
 1.13468723368653623353570804268*10^-85, 
 1.13468723368653622887221115414*10^-90, 
 1.13468723368653622840586146529*10^-95, 
 1.13468723368653622835922649640*10^-100, 
 1.13468723368653622835456299951*10^-105, 
 1.13468723368653622835409664982*10^-110, 
 1.13468723368653622835405001485*10^-115, 
 1.13468723368653622835404535136*10^-120, 
 1.13468723368653622835404488501*10^-125, 
 1.13468723368653622835404483837*10^-130, 
 1.13468723368653622835404483371*10^-135, 
 1.13468723368653622835404483324*10^-140, 
 1.13468723368653622835404483320*10^-145, 
 1.13468723368653622835404483319*10^-150, 
 1.13468723368653622835404483319*10^-155}

And a little more general,

N[Table[NSum[(-1)^n (n^(1/(n*10^k)) - 1), {n, 1, Infinity}, 
    WorkingPrecision -> 500, Method -> "AlternatingSigns"] - 
   Sum[(-1)^x ((Log[x]/x)^n/n!), {x, 1, Infinity}, {n, 1, 1}]*10^-k - 
   Sum[Sum[(-1)^x ((Log[x]/x)^n/n!), {x, 1, Infinity}, {n, u, 
       u}]*10^-(k u), {u, 2, 12}], {k, 1, 30}], 30]

gives

{-5.45618018853714584642960759390*10^-30, \
-5.44024107910968662791592489621*10^-43, \
-5.43865173864928795594977880078*10^-56, \
-5.43849285018779289756765113776*10^-69, \
-5.43847696179736885286190437668*10^-82, \
-5.43847537296288358304480376121*10^-95, \
-5.43847521407948062728967349952*10^-108, \
-5.43847519819114078742630631670*10^-121, \
-5.43847519660230680799709093690*10^-134, \
-5.43847519644342341009974061218*10^-147, \
-5.43847519642753507031046129184*10^-160, \
-5.43847519642594623633153791693*10^-173, \
-5.43847519642578735293364562501*10^-186, \
-5.43847519642577146459385639627*10^-199, \
-5.43847519642576987575987747341*10^-212, \
-5.43847519642576971687647958112*10^-225, \
-5.43847519642576970098813979189*10^-238, \
-5.43847519642576969939930581297*10^-251, \
-5.43847519642576969924042241507*10^-264, \
-5.43847519642576969922453407529*10^-277, \
-5.43847519642576969922294524131*10^-290, \
-5.43847519642576969922278635791*10^-303, \
-5.43847519642576969922277046957*10^-316, \
-5.43847519642576969922276888073*10^-329, \
-5.43847519642576969922276872185*10^-342, \
-5.43847519642576969922276870596*10^-355, \
-5.43847519642576969922276870437*10^-368, \
-5.43847519642576969922276870422*10^-381, \
-5.43847519642576969922276870420*10^-394, \
-5.43847519642576969922276870420*10^-407}

If you don’t have half an hour to wait for all that presission you can enter,

  N[
 Table[NSum[(-1)^n (n^(1/(n*10^k)) - 1), {n, 1, Infinity}, 
    WorkingPrecision -> 400, Method -> "AlternatingSigns"] - 
   Sum[(-1)^x ((Log[x]/x)^n/n!), {x, 1, Infinity}, {n, 1, 1}]*10^-k - 
   Sum[Sum[(-1)^x ((Log[x]/x)^n/n!), {x, 1, Infinity}, {n, u, 
       u}]*10^-(k u), {u, 2, 12}], {k, 1, 30}], 30]

(*{-5.45618018853714584642960759390*10^-30, \
-5.44024107910968662791592489621*10^-43, \
-5.43865173864928795594977880078*10^-56, \
-5.43849285018779289756765113776*10^-69, \
-5.43847696179736885286190437668*10^-82, \
-5.43847537296288358304480376121*10^-95, \
-5.43847521407948062728967349952*10^-108, \
-5.43847519819114078742630631670*10^-121, \
-5.43847519660230680799709093690*10^-134, \
-5.43847519644342341009974061218*10^-147, \
-5.43847519642753507031046129184*10^-160, \
-5.43847519642594623633153791693*10^-173, \
-5.43847519642578735293364562501*10^-186, \
-5.43847519642577146459385639627*10^-199, \
-5.43847519642576987575987747341*10^-212, \
-5.43847519642576971687647958112*10^-225, \
-5.43847519642576970098813979189*10^-238, \
-5.43847519642576969939930581297*10^-251, \
-5.43847519642576969924042241507*10^-264, \
-5.43847519642576969922453407529*10^-277, \
-5.43847519642576969922294524131*10^-290, \
-5.43847519642576969922278635791*10^-303, \
-5.43847519642576969922277046957*10^-316, \
-5.43847519642576969922276888073*10^-329, \
-5.43847519642576969922276872185*10^-342, \
-5.43847519642576969922276870596*10^-355, \
-5.43847519642576969922276870437*10^-368, \
-5.43847519642576969922276870422*10^-381, \
-5.43847519642576969922276870420*10^-394, \
-5.438475196425769698337*10^-407}*)

The preceding is the result of three months of inspiration and half an hour of perspiration!

5

I started the original post saying.

Considering the MRB constant which is approximated by

NSum[(-1)^n (n^(1/n) - 1), {n, 1, Infinity}, WorkingPrecision -> 70, 
 Method -> "AlternatingSigns"]

giving

0.18785964246206712024851793405427323005590309490013878617200468547342,

I wondered what happens when we change 1/n in n^(1/n) to 1 over
other functions involving n?

Then I found out an interesting pattern involving

Sum[(-1)^n (1/n^(1/(n*10^x)) - 1), {n, 1, Infinity}]

and

Sum[(-1)^n (n^(1/(n*10^x)) - 1), {n, 1, Infinity}]

.
But here are exact infinite sums of very partial sums of of those:

 FullSimplify[(Sum[(-1)^n (1/n^(1/(n*10^x)) - 1), {n, 1, 
      2}]) == (-E^(-10^(-1 - x) a) (-1 + E^(10^(-1 - x) a)) /. 
     a -> 5 Log[2]) == 
   Sum[(-1)^n (5 Log[2])^n/n!*10^(-(x + 1)*n), {n, 1, Infinity}]]

gives True.

And

FullSimplify[(Sum[(-1)^n (n^(1/(n*10^x)) - 1), {n, 1,
2}]) == (-E^(-10^(-1 - x) a) (-1 + E^(10^(-1 - x) a)) /.
a → 5 Log[1/2]) ==
Sum[(-1)^n (5 Log[1/2])^n/n!*10^(-(x + 1)*n), {n, 1, Infinity}]]

also gives True

.
**If any of my posts are useful, I hope that you will use those things in your research! And please acknowledge what my research did to help you, in particular the posts about the MRB constant like this one. I really desire for the world to find out more about the constant. (It’s my only legacy!) **!

6

The last reply is OK as far as

Sum[(-1)^n (1/n^(1/(n*10^x)) - 1), {n, 1, Infinity}]

and

Sum[(-1)^n (n^(1/(n*10^x)) - 1), {n, 1, Infinity}]

are concerned, but what about the MRB constant series,

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

?

Here is what we can say, which is prety much expected:

   FullSimplify[(f2 = (Sum[(-1)^n (n^(1/n) - 1), {n, 1,  2}])) 
== (E^(a) /.  a -> Log[f2])
== (Sum[(-1)^n (Log[1/f2])^n/n!, {n, 0, Infinity}])]


 (* True*)

  FullSimplify[(f3 = Sum[(-1)^n (n^(1/n) - 1), {n, 1, 3}]) 
== (E^(a) /.  a -> Log[f3]) 
== (Sum[(-1)^n (Log[1/f3])^n/n!, {n, 0, Infinity}])]

 (* True*)

  FullSimplify[(f4 = Sum[(-1)^n (n^(1/n) - 1), {n, 1, 4}]) 
== (E^(a) /.  a -> Log[f4]) 
== (Sum[(-1)^n (Log[1/f4])^n/n!, {n, 0, Infinity}])]

 (*True*)

  FFullSimplify[{(fI = N[Sum[(-1)^n (n^(1/n) - 1), {n, 1, Infinity}]]),
    fI - Re[E^(a) /. a -> Log[fI]],
    fI - Re[Sum[(-1)^n (Log[1/fI])^n/n!, {n, 0, Infinity}]], 
    N[fI - Re[
       Sum[(-1)^x (Log[x]/x)^n/n!, {x, 1, Infinity}, {n, 1, 
         Infinity}]]]}]

(* {0.18786, 0., 0., 3.33067*10^-16}*)