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!