Series of logarithms $\sum\limits_{k=1}^\infty \ln(k)$ (Ramanujan summation?)

To address your last comment on evaluating infinite sums of powers of logarithms:

Letting

$$(-1)^n \zeta^{(n)}(s)=\sum_{k=1}^\infty \frac{(\log k)^n}{k^s}$$

the question can be recast as the evaluation of the quantity $(-1)^n \zeta^{(n)}(0)$ for various $n\in \mathbb N$.

Apostol and Choudhury had given a bunch of formulae for evaluating $\zeta^{(n)}(s)$; in particular, there is the formula

$$(-1)^n \zeta^{(n)}(0)=\frac{\Im((-\log\,2\pi-i\pi/2)^{n+1})}{\pi(n+1)}+\frac1{\pi}\sum_{j=1}^{n-1} a_j j!\binom{n}{j}\Im((-\log\,2\pi-i\pi/2)^{n-j})$$

where

$$a_j=c_{j+1}+\sum_{\ell=0}^j \frac{(-1)^\ell \gamma_\ell}{\ell!} c_{j-\ell}$$

$$c_k=-\frac{\gamma c_{k-1}}{k}-\frac1{k}\sum_{\ell=1}^{k-1} (-1)^\ell \zeta(\ell+1) c_{k-\ell-1},\qquad c_0=1$$

the $\gamma_n$ are the Stieltjes constants, and $\gamma=\gamma_0$ is the Euler-Mascheroni Constant.

Here are a few explicit evaluations of the formula:

$$\zeta^{\prime\prime}(0)=\frac{\gamma^2}{2}-\frac{\pi^2}{24}-\frac{(\log (2\pi))^2}{2}+\gamma_1$$

$$-\zeta^{\prime\prime\prime}(0)=-\gamma^3-\frac32 \gamma ^2 \log(2\pi)+\frac{\pi^2}{8}\log(2\pi)+\frac{(\log(2\pi))^3}{2}-3\gamma\gamma _1-3\gamma_1\log (2\pi)-\frac32\gamma_2+\zeta(3)$$

As can be surmised, the closed forms become rather unwieldy as $n$ gets large...

An extension of these results to noninteger powers would involve having to appropriately define the differintegral of the zeta function; going the series route would now involve terms containing the incomplete gamma function, but I've no knowledge of any closed forms for the resulting sum.


I think, thanks to the hint of J.M. I can answer Q1 myself now; only Q2 remains somehow vague - besides a simple empirical heuristic I did still not get the formally correct approach to the constant term/integral-definition in the Ramanujan summation - but this is now only a side problem here (hoewever it would be nice to get help also for this question).

With the help of the knowledge about the derivatives of the zeta at zero the relevant part of the problem could now satisfyingly be solved, so I put it here in an answer to my own question.

Final remark/conclusion: it is interesting, that the powerseries for the lngamma pops up here "automatically" - we need no other uniqueness criterion for the argument, that the (Eulerian) gamma-function gives "the correct" interpolation for the factorial problem.


Since the infinite series of like powers of logarithms can be expressed by the derivatives of $ \zeta(n) $ at $n=0$ we have expressions for $L_p$ and thus for $T_p(0)$ .
$$ L_p = T_p(0) = (-1)^p \zeta^{(p)}(0) = \ln(1)^p + \ln(2)^p + \ln(3)^p + \ldots $$

To actually compute that derivatives I use the powerseries representation of zeta using the Stieltjes-constants: $$ \zeta(z) = {1 \over z-1} + \sum_{k=0}^{\infty} (-1)^k \cdot \gamma_k \cdot { (z-1)^k \over k!} $$ where the $\gamma_k$ are the Stieltjes constants, beginning with $\gamma_0$ and the values $ \approx [0.5777...,-0.072..., -0.096...] $

The derivatives are then expressed by termwise differentation, giving $$ \begin{eqnarray} \zeta^{(p)}(z) &=& ({1 \over z-1})^{(p)} &+& \sum_{k=p}^{\infty} \binom kp \cdot p!\cdot (-1)^k \cdot \gamma_k \cdot { (z-1)^{k-p} \over k!} \\ &=& - (-1)^p {p! \over (z-1)^{p+1} }& +& (-1)^p \sum_{k=0}^{\infty} (-1)^k \cdot \gamma_{k+p} \cdot { (z-1)^k \over k!} \\ &=& {p! \over (1-z)^{p+1} }& +& (-1)^p \sum_{k=0}^{\infty} \gamma_{k+p} \cdot { (1-z)^k \over k!} \\ \end{eqnarray} $$
Thus $$ \begin{eqnarray} L_p = T_p(0) &=& (-1)^p*( {p! \over 1^{p+1} }& +& (-1)^p \sum_{k=0}^{\infty} \gamma_{k+p} \cdot { 1^k \over k!} ) \\ &=& (-1)^p*p! & +& \sum_{k=0}^{\infty} {\gamma_{k+p} \over k!} \end{eqnarray} $$

The explicite sum in the last expression converges very well and can be computed to as many digits as the accuracy of the Stieltjes numbers allows. With this I get the logsums $L_0 = T_0(0) $ to $L_{23}=T_{23}(0) $ (read from left to right):

$ \qquad \small \begin{array} {rr} -0.500000000000 & 0.918938533205 \\ -2.00635645591 & 6.00471116686 \\ -23.9971031880 & 120.000232908 \\ -720.000936825 & 5039.99915018 \\ -40320.0002324 & 362880.000331 \\ -3628799.99946 & 39916800.0004 \\ -479001600.000 & 6227020800.00 \\ -87178291200.0 & 1.30767436800E12 \\ -2.09227898880E13 & 3.55687428096E14 \\ -6.40237370573E15 & 1.21645100409E17 \\ -2.43290200818E18 & 5.10909421717E19 \\ -1.12400072778E21 & 2.58520167389E22 \end{array} $

where the sequence of the deviation from the signed factorials is

$ \qquad \small \begin{array} {rr} 0.500000000000 & -0.0810614667953 \\ -0.00635645590858 & 0.00471116686225 \\ 0.00289681198629 & 0.000232907558455 \\ -0.000936825130051 & -0.000849823765002 \\ -0.000232431735512 & 0.000330589663612 \\ 0.000543234115780 & 0.000375493172907 \\ -0.0000196035362810 & -0.000407241232563 \\ -0.000570492013282 & -0.000393927078981 \\ 0.0000834588058255 & 0.000660943729629 \\ 0.00102622728654 & 0.000865575776779 \\ 0.0000192936717837 & -0.00135690605213 \\ -0.00269215645875 & -0.00305138562124 \\ -0.00142429184942 & 0.00270778921289 \\ 0.00860288096928 & 0.0135561620310 \\ 0.0127785133267 & 0.000576026175993 \\ -0.0264657041471 & -0.0641102553488 \\ -0.0940806022511 & -0.0801887691867 \\ 0.0263889614064 & 0.263594454732 \\ 0.608778480504 & 0.904597405553 \\ 0.786784764403 & -0.316856127028 \\ -2.99127389406 & -7.27888475259 \\ -11.6472353086 & -11.6164312194 \\ 1.20065709453 & 37.8116918598 \\ 105.106316717 & 189.660315159 \end{array} $

(I displayed more numbers here because the divergence of the sequence cannot be seen from the leading numbers only!)


A/the matrix $\Lambda$ contains then in each column the coefficients of the powerseries to compute $T_p(x)$ by the dot-product of a vandermonde rowvector of the type $V(z)=[1,z,z^2,z^3,...]$ with the column $p$ of $\Lambda$:

$ \qquad V(\ln(1+x))~ * \Lambda = \sum_{k=x}^\infty V(\ln(1+k))~ = [T_0(x),T_1(x),T_2(x),\ldots] $

and the sum of consecutive powers of the logarithms are taken by the difference of two $T_p(n)$-computations:

$ \qquad Sl_p(n) = T_p(0) - T_p(n) = \sum_{k=1}^n \ln(k) $


One of the reasons for that long derivations was, that the n are not required to be integer here, analoguosly to the solution for the sum of like powers by means of the Bernoulli polynomials.


Appendix:

Here is the top-left-segment of the matrix $\Lambda$:

$ \qquad \small \begin{bmatrix} -0.5000000 & 0.9189385 & -2.006356 & 6.004711 & -23.99710 & \ldots \\ -1.000000 & 0.5772157 & -0.1456317 & -0.02907109 & 0.008215338 & \ldots \\ -0.5000000 & -0.5338592 & 0.6345700 & -0.1858111 & -0.06050253 & \ldots \\ -0.1666667 & -0.3255788 & -0.3868589 & 0.6713043 & -0.2104926 & \ldots \\ -0.04166667 & -0.1252741 & -0.2407114 & -0.3127569 & 0.6958103 & \ldots \\ -0.008333333 & -0.03372565 & -0.09916203 & -0.1918968 & -0.2670350 & \ldots \\ -0.001388889 & -0.006859354 & -0.02847304 & -0.08146653 & -0.1603119 & \ldots \\ -0.0001984127 & -0.001172608 & -0.005923793 & -0.02452507 & -0.06891184 & \ldots \\ -0.00002480159 & -0.0001831833 & -0.0009884023 & -0.005280337 & -0.02143180 & \ldots \\ -0.000002755732 & -0.00002257650 & -0.0001620035 & -0.0008613236 & -0.004778981 & \ldots \\ \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \end{bmatrix} $