5
$\begingroup$

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]
| cite | improve this question | |
$\endgroup$
  • 1
    $\begingroup$ A confirmation about the numeric evaluation being so slow would be nice though. $\endgroup$ – Diger Jul 15 '18 at 22:03
  • 1
    $\begingroup$ To be honest: I don't really know! I guess this is quite delicate as for when you would replace $(-1)^{j-1}$ by $1^{j-1}=1$ then the entire result diverges. Also the last sum with alternating $1$s and $-1$s is not really defined, but only in the sense of analytic continuation. Therefore I guess your sum above would have to be replaced by some function $$ f(x)=\sum _{n=1}^{\infty } \left( \sum _{j=2}^{\infty }{\frac { x^{j-1}n}{{j}^{ 2}} \left( 1-{j}^{-1} \right) ^{n-1}} \right) $$ with $|x|<1$, because then the series is absolutely and uniformly convergent and can you interchange $\endgroup$ – Diger Jul 16 '18 at 1:02
  • 1
    $\begingroup$ summation order. Then $f(x)$ is evaluated at $-1$ which amounts for, if your sum is the result of meromorphic function represented by this series, then the value of this function can be attached to it. $\endgroup$ – Diger Jul 16 '18 at 1:06
  • 1
    $\begingroup$ Mathematica code: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: $-1/2$ $\endgroup$ – Mariusz Iwaniuk Jul 18 '18 at 14:36
  • 1
    $\begingroup$ @crow Do you want a proof of the single identity in the title or of the double idendity below the title? Plus, what have you tried for proving them? $\endgroup$ – Saad Jul 18 '18 at 15:58
2
+50
$\begingroup$

In the following assume $|z|<1$ to begin with and I'll let $z \to 1$ to get the OP's left hand side formula: $$ S(z):=\sum_{n=1}^\infty z^n\,n \sum_{j=2}^\infty\, \frac{(-1)^{j-1}}{j^2}\big(1-\frac{1}{j}\big)^{n-1} .$$ $S(z)$ is solvable in closed-form in the following manner: Interchange summations and to eliminate the $n$ series use the derivative of the geometric series to obtain

$$S(z)= z\sum_{j=2}^\infty\, \frac{(-1)^{j-1}}{j^2(1-z(1-1/j))^2}=\frac{z}{(1-z)^2}\sum_{j=2}^\infty\, \frac{(-1)^{j-1}}{ (j+z/(1-z))^2 }.$$ This sum is solvable in the terms of the trigamma function, once a $j=1$ term is added and subtracted. Thus

$$S(z)= \frac{z/4}{(1-z)^2}\Big( \psi\,'\big(\tfrac{3}{2}+\tfrac{z}{2(1-z)}\big) - \psi\,' \big( 1+\tfrac{z}{2(1-z)} \big) \Big) .$$

To find the limit $z \to 1,$ it is seen that it is equivalent to

$$S(1) = \lim_{y \to 0} \frac{1}{4y^2} \Big( \psi\,'\big(\tfrac{3}{2}+\tfrac{1}{2y}) - \psi\,' \big(1+\tfrac{1}{2y} \big) \Big) = \lim_{x \to \infty} \frac{x^2}{4} \Big( \psi\,'\big(\tfrac{3}{2}+\tfrac{x}{2}\big) - \psi\,' \big( 1+\tfrac{x}{2} \big) \Big)$$ $$= \lim_{x \to \infty} \frac{x^2}{4} \Big(\frac{2}{x}-\frac{4}{x^2}+\mathit{O}(x^{-3}) - (\frac{2}{x}-\frac{2}{x^2}+{\mathit{O}}(x^{-3})) \Big)= -\frac{1}{2}$$ where the asymptotic formula for the trigamma function has been used.

| cite | improve this answer | |
$\endgroup$
  • 2
    $\begingroup$ Why is the interchange of a limit and a summation, licit? $\endgroup$ – Did Jul 18 '18 at 19:21
  • $\begingroup$ @skbmoore that's a pretty cool formula. I see that the series expansion of S has the n-th term in the expansion being $\chi(n)$ $\endgroup$ – crow Jul 19 '18 at 0:16
  • $\begingroup$ @Did. Assume z is 'small enough.' Do the sum and the final result tells us that there are no singularities until $|z|=1.$ Abel's theorem, I believe, lets us take the limit. $\endgroup$ – skbmoore Jul 19 '18 at 17:03
  • 2
    $\begingroup$ ?? "You think"? The reason that make you sure the step is legit should be written down explicitely in your answer. $\endgroup$ – Did Jul 19 '18 at 17:29
  • 1
    $\begingroup$ So, each inner series converges, right (and by much simpler arguments). But this is not the problem at all... $\endgroup$ – Did Jul 22 '18 at 20:38
4
$\begingroup$

Write $\chi(n) = n \sum _{j=2}^{\infty} \frac{(-1)^{j-1}}{j^2} \left( 1-\frac{1}{j} \right)^{n-1}$ using OP's notation. Its partial sum up to the $(N+1)$-th term can be simplified by interchanging the order of summation:

\begin{align*} \sum_{n=1}^{N+1} \chi(n) &= \sum _{j=2}^{\infty} \frac{(-1)^{j-1}}{j^2} \sum_{n=1}^{N+1} n \left( 1-\frac{1}{j} \right)^{n-1} \\ &= \sum_{j=2}^{\infty} (-1)^j \underbrace{ \left[ \left(1 + \frac{N+1}{j}\right)\left(1 - \frac{1}{j}\right)^{N+1} - 1 \right] }_{=: f_N(j)}, \tag{1} \end{align*}

Since $f_N(j) = \mathcal{O}(j^{-2})$ as $j\to\infty$ for each fixed $N$, $\text{(1)}$ is indeed a convergent series. Now grouping successive terms, $\text{(1)}$ simplifies to

\begin{align*} \sum_{n=1}^{N+1} \chi(n) &= - \sum_{k=1}^{\infty} (f_N(2k+1) - f_N(2k)) \\ &= - \sum_{k=1}^{\infty} \int_{2k}^{2k+1} f_N'(x) \, dx \\ &= - \sum_{k=1}^{\infty} (N+1)(N+2) \int_{2k}^{2k+1} \frac{\left(1 - \frac{1}{x}\right)^N}{x^3} \, dx \\ &= - \sum_{k=1}^{\infty} \frac{(N+1)(N+2)}{N^2} \int_{\frac{2k}{N}}^{\frac{2k+1}{N}} \frac{\left(1 - \frac{1}{Nu}\right)^N}{u^3} \, du \tag{$x=Nu$} \end{align*}

Then Fubini-Tonelli's theorem gives

$$ \sum_{n=1}^{N+1} \chi(n) = - \frac{(N+1)(N+2)}{N^2} \int_{0}^{\infty} \frac{\left(1 - \frac{1}{Nu}\right)^N}{u^3} \left( \sum_{k=1}^{\infty} \mathbf{1}_{\left[\frac{2k}{N}, \frac{2k+1}{N}\right]}(u) \right) \, du. $$

Since the integrand is bounded by the integrable dominating function $e^{-1/u}/u^3$, as $N\to\infty$ we have

$$ \lim_{N\to\infty} \sum_{n=1}^{N+1} \chi(n) = - \int_{0}^{\infty} \frac{e^{-1/u}}{u^3} \cdot \frac{1}{2} \, du = - \frac{1}{2}. $$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.