How can we prove the following?
$-1+\frac1{12}{\pi }^{2}-\frac12\sum\limits_{n=2}^{\infty }\Gamma \left( n+1 \right) \sum\limits_{k=0}^{n+2} \,{\frac {\zeta \left( k \right) \left( - 1 \right) ^{-2+k} \left( {2}^{2-k}-2 \right) }{\Gamma \left( -1+k \right) \left( n+1-k \right) !}} =\sum\limits_{n=1}^{\infty }n \sum\limits_{j=2}^{\infty }{\frac { \left( -1 \right) ^{j-1}}{{j}^{ 2}} \left( 1-{j}^{-1} \right) ^{n-1}} =-\frac12$
We also have $\chi(n)=-\sum\limits_{k=0}^{n+2}\zeta \left( k \right) \left( k-n-2 \right) {n\choose k-2} \left( \left( -1 \right) ^{1+k}{2}^{1-k}+ \left( -1 \right) ^{k} \right) $ by replacing the $\Gamma$ factors with the corresponding binomial occeficient
Related question at References for $ \chi(n)=n\sum\limits_{j=2}^\infty\frac {(-1)^{j-1}}{j^2}\left(1-j^{-1}\right)^{n-1}$ in $\zeta$ expansion?
Digits := 37; flist(proc (n) options operator, arrow; evalf(coolchi(n)) end proc, 0 .. 75); Digits := 37 [ [0., -0.1775329665758867817637924166769874055, -0.158151287891164991627192075621149797, -0.100756475454096876860689273411921431, -0.052729122560581908477394880842614692, -0.022173447806981260740844120495648785, -0.005724758139923986695354623118140584, 0.001711181525440404872086057781767772, 0.004163504754435619487164038876686477, 0.004224146796392587120306729731994248, 0.003361436685455899377754627000698078, 0.00231507684250186191279422185478205, 0.00140312527387483007618661260684936, 0.00072272177218128957979732200349640, 0.00026840905319980399461855630763639, -0.00000386786514313326830151239928860, -0.00014529652172375174501064342971419, -0.00020079200685610596363763783920190, -0.00020486397550091241285507339640177, -0.00018177251376674866978557090651028, -0.00014732720568299891079533324844990, -0.00011103259250611536839455447354313, -0.00007801852557445637063802766476294, -0.00005057002459089850935720449502326, -0.00002924737982070438622370684257219, -0.0000136577182218588710540784871811, -0.0000029575122625340335255970335431, 0.0000038395199899767419714034429197, 0.0000076881193734840299247444522077, 0.0000094245398419219327153653943207, 0.0000097337529690318816412614692587, 0.0000091480274459270312096627770217, 0.0000080619145517535381759258176097, 0.0000067542476407336638941225910227, 0.000005411593443449269790648097261, 0.000004150125437131325581547336324, 0.000003034497471576312110001308212, 0.000002093267538135562252003012925, 0.000001330969825279401198332450463, -7 7.37209507955248062779620826 10 , -7 2.93263622398134958434524015 10 , -8 -2.3317072362848976102839973 10 , -7 -2.35650675570634484892471134 10 , -7 -3.65620346716670870284369471 10 , -7 -4.32803484845776951918534983 10 , -7 -4.53980382773092409894967670 10 , -7 -4.43028271969806636513667532 10 , -7 -4.11056601063391173974634568 10 , -7 -3.66679744724574133777868780 10 , -7 -3.16354929829636515323370167 10 , -7 -2.64737199886948816611138728 10 , -7 -2.15021028592691266141174466 10 , -7 -1.69250991341964200913477377 10 , -7 -1.28592821342563655928047463 10 , -8 -9.3562155186891501184884726 10 , -8 -6.4212079021714840683989162 10 , -8 -4.0282829593958497425360774 10 , -8 -2.1318155241615525408999560 10 , -9 -6.753256088468931634905523 10 , -9 4.020837075521115896921341 10 , -8 1.1611930078301706186481030 10 , -8 1.6596359914684719233773543 10 , -8 1.9500255425207775038798882 10 , -8 2.0789463805547333601557044 10 , -8 2.0865770984321044922048033 10 , -8 2.0067518073257429000271847 10 , -8 1.8673131375018625836228485 10 , -8 1.6906432963871255429917949 10 , -8 1.4942881292166557781340237 10 , -8 1.2916159703946332890495351 10 , -8 1.0924629207316840757383290 10 , -9 9.037454965393441382004053 10 , -9 7.300036840122634764357642 10 , -9 5.739399681529520904444054 10 , -9 -9 4.367571268906899802263293 10 , 3.185931532251771457815357 10 ] ]
listsum((20))= -0.4999999932944069400650356605249833795
Note: when evaluating this sum, at appears at least 28 digits of precision are required for accurate evaluation. If you use less than this (Maple defaults to 10) then the result will diverge quite wildly to an inaccurate answer.
This table is the partial sums of the terms, so I think the equality =-1/2 is correct
[0., -.1775329665758867817637924166769874055, -.3356842544670517733909844922981372025, -.4364407299211486502516737657100586335, -.4891698524817305587290686465526733255, -.5113433002887118194699127670483221105, -.5170680584286358061652673901664626945, -.5153568769031954012931813323846949225, -.5111933721487597818060172935080084455, -.5069692253523671946857105637760141975, -.5036077886669112953079559367753161195, -.5012927118244094333951617149205340695, -.4998895865505346033189751023136847095, -.4991668647783533137391777803101883095, -.4988984557251535097445592240025519195, -.4989023235902966430128607364018405195, -.4990476201120203947578713798315547095, -.4992484121188765007215090176707566095, -.4994532760943774131343640910671583795, -.4996350486081441618041496619736686595, -.4997823758138271607149449952221185595, -.4998934084063332760833395496956616895, -.4999714269319077324539775773604246295, -.5000219969564986309633347818554478895, -.5000512443363193353495584886980200795, -.5000649020545411942206125671852011795, -.5000678595668037282541381642187442795, -.5000640200468137515121667607758245795, -.5000563319274402674822420163236168795, -.5000469073875983455495266509292961795, -.5000371736346293136678853894600374795, -.5000280256071833866366757266830157795, -.5000199636926316330984998008654060795, -.5000132094449908994346056782743833795, -.5000077978515474501648150301771223795, -.5000036477261103188392334828407983795, -.5000006132286387425271234815325863795, -.4999985199611006069648714785196613795, -.4999971889912753275636731460691983795, -.4999964517817673723156103664483723795, -.4999961585181449741806519319243573795, -.4999961818352173370296280347643303795, -.4999964174858929076641129272354643795, -.4999967831062396243349832116049353795, -.4999972159097244701119351301399183795, -.4999976698901072432043450251075883795, -.4999981129183792130109815387751203795, -.4999985239749802764021555134096883795, -.4999988906547250009762892912784683795, -.4999992070096548306128046146486353795, -.4999994717468547175616212257873633795, -.4999996867678833102528873669618293795, -.4999998560188746522170882804392063795, -.4999999846116959947807442084866693795, -.5000000781738511816722453933713953795, -.5000001423859302033870860773605573795, -.5000001826687597973455835027213313795, -.5000002039869150389611089117208913795, -.5000002107401711274300405466264143795, -.5000002067193340519089246497050733795, -.5000001951074039736072184632240433795, -.5000001785110440589224992294505003795, -.5000001590107886337147241906516183795, -.5000001382213248281673905890945743795, -.5000001173555538438463456670465413795, -.5000000972880357705889166667746943795, -.5000000786149043955702908305462093795, -.5000000617084714316990354006282603795, -.5000000467655901395324776192880233795, -.5000000338494304355861447287926723795, -.5000000229248012282693039714093823795, -.5000000138873462628758625894053293795, -.5000000065873094227532278250476873795, -.5000000008479097412237069206036333795, -.4999999964803384723168071183403403795, -.4999999932944069400650356605249833795]
Sum[Sum[((-1)^(j - 1)*n)/j^2*(1 - j^-1)^(n - 1), {n, 1, Infinity}, Assumptions -> j > 1] // Simplify, {j, 2, Infinity}, Regularization -> "Dirichlet"]
gives: $-1/2$ $\endgroup$