21
$\begingroup$

I had this question earlier, so to say as a "standalone" problem, but now it pops up in context of an analysis with the lngamma-function. As well as we can convert the question of sums of like powers $Su_p(n)=1^p+2^p+3^p+\cdots+n^p$ in terms of the Hurwitz-zeta $Su_p(n) = \zeta(p,1)-\zeta(p,n+1)$ (and solve using the Bernoulli-polynomials) I try to express this for sums of logarithms (and powers of logarithms) : $$ Sl_p(n) = \ln(1)^p+\ln(2)^p+\cdots+\ln(n)^p $$
I have seemingly proper coefficients for power series which allow that computations/approximations. The result is always found by the difference of $ Sl_p(n)-Sl_p(1) $ . Here the power series has zero as constant term.


However, to have the analogue to the Hurwitz-zeta I should have the constant term the "value" for the infinite sum $ Sl_p(\infty) $ instead. (This cancels properly if I formulate the sum of (powers of) logarithms always as that difference $ Sl_p(n)-Sl_p(1) $) And also, I'd like that this agreed conceptually more to the notion: $$ \begin{array} {rll} Sl_p(n)-Sl_p(1) &=& T_p(0)-T_p(n) \\\ & =& (\ln(1)+\ln(2)+\ln(3)+\cdots) \\\ & & - (\ln(n+1)+\ln(n+2)+\ln(n+2)+\cdots) \end{array} $$
as this agrees more with the idea of the Hurwitz-zeta-difference.

But this requires, that $T_p(0)$ represents the (infinite) sum of the $p$'th powers of the logarithms, and the constant in the power series of $T_p(0)$ must contain just such a value. Let's talk about the first powers of the logs $p=1 $ first and denote the assumed sum as $L_1$ : $ L_1 = \lim_{n\to \infty} T_1(0) $


[update 2]
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 (however 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, see that answer below....

Final remark/conclusion: it is interesting, that the power series 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. It is just the result of the construction of an operator for the "indefinite summation" (which was the comceptual goal from where the problem/question arose initially)

In the following I keep the rest of the original question although the approach to the Ramanujan summation contains an error
[end Update 2]


I tried to give sense to that divergent series by replacing the powers of $x$ in the Mercator series for $ \ln(1+x) $ by appropriate zetas at negative integer arguments. If I understand things correctly then this is similar to the method of Ramanujan summation, where the Bernoulli numbers are just replaced by the according zeta-values (appropriately scaled). With this I got then an approximation of $$L_1 \approx -0.0810614667953 $$
which seems a rather "random" value...
By searching in other online sources I got the suggestion (OEIS), that this is also $$ \int_1^2 \ln(\Gamma(t)) \; dt \approx -0.08106146679532725821967026 $$

With this my power series for $T_1(x)$ begins like

$ \qquad \small \begin{array} {l} - & 0.0810614667953 \\ - & 0.577215664902x \\ + & 0.533859200973x^2 \\ + & 0.325578788221x^3 \\ + & 0.125274140308x^4 \\ + & 0.0337256506589x^5 \\ + & 0.00685935357296x^6 \\ + & 0.00117260810356x^7 \\ + & O(x^8) \end{array} $
The other coefficients occur also in the power series for $f(x)=\ln(\Gamma(\exp(x)) $

Also, $L_1$ seem to satisfy the following expression: $$ \exp(L_1) = {\sqrt{(2\pi)} \over e} \approx 0.9221370088957891168791517 $$

The questions are: $\qquad$ Q1: Is that value $-0.0801\ldots$ a meaningful (or even correct) representation for the infinite sum of logarithms?
$\qquad$ Q2: Did I reproduce the Ramanujan summation correctly here?


[update 1]: J.M. mentions the relation to $ y= - \zeta'(0) \approx 0.918938533205 $ where the representation of the derivative of zeta at 0 equals formally just the sum of logarithms. Now my $L_1$ and the $y$ are related by $L_1 = y - 1$. So I expect some error in my derivation which leads just to that unit difference...

$\endgroup$
9
  • $\begingroup$ Wait, I am confused. How'd you obtain $T_p$ from $Sl_p$? $\endgroup$ May 16 '11 at 11:25
  • 1
    $\begingroup$ @J.M.: by the idea $T_p(n) = L_p - S_p(n)$ so if $S_p(4)=\ln(1)^p+\ln(2)^p+\ln(3)^p+\ln(4)^p$ and $L_p=\ln(1)^p+\ln(2)^p+\ln(3)^p+\ln(4)^p+\ldots $ then $T_p(n)=\ln(n+1)^p+\ln(n+2)^p+\ln(n+3)^p+\ln(n+4)^p+\ldots $. As the difference is only the constant $L_p$ the powerseries are the same except of the sign and the constant. (Well, this all has little to do with consideration of convergence... ) $\endgroup$ May 16 '11 at 11:47
  • $\begingroup$ Hmm, so $L_1$ is effectively equivalent to $-\zeta^\prime(0)$? I seem to be getting something different from $-0.08106\dots$ maybe I'm missing a step in your derivation. $\endgroup$ May 16 '11 at 12:33
  • $\begingroup$ Hmm, due to your hint to the derivative of the zeta I found formula 43 at mathworld.wolfram.com/RiemannZetaFunction.html .For s=0 this denotes the sums of the logarithms. On the other hand, for this (in terms of the derivative of zeta at zero) mathematica gives -0.9189385332046727.... Hmm... $\endgroup$ May 16 '11 at 13:01
  • 2
    $\begingroup$ @J.M.: formula 38: should have seen this before. The mentioned paper: very complex; I'll look at it another day. Now, even if I seem to have now the answer for the case p=1 I'm interested whether the cases for higher p come out to be solvable too... $\endgroup$ May 16 '11 at 15:01
15
$\begingroup$

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.

$\endgroup$
4
  • 1
    $\begingroup$ M.:Yes, thank you; just got at least the derivative-of-zeta connection yesterday in the evening. For small values I got values using Pari/GP by doing ${\zeta(0+h/2)-\zeta(0-h/2)\over h}$ with small h (and analoguously for higher derivatives). I think, what I'd missed in my question above was the introduction of the leading integral of the function $\ln(n)$ according to the wikipedia article on Ramanujan-summation. With that my Ramanujan-sums and the derivatives of the zetas agree now. (I'll include that corrections in my question later) $\endgroup$ May 17 '11 at 3:58
  • $\begingroup$ @Gottfried: You probably will be interested in this MO answer where I detailed various good algorithms for numerical derivatives. $\endgroup$ May 17 '11 at 4:43
  • $\begingroup$ (correction of previous comment):@J.M.:thanks again! Because a powerseries allows a simple representation of the derivative I use the stieltjes-powerseries expression for the zeta-function. Then we have a very simple, very quickly converging, numerical procedure: $ \zeta^{(p)}(0) = (-1)^p \cdot \sum_{k=0}^{\infty} ({\gamma_{p+k} \over k! }) -p! $ where $ [\gamma_0, \gamma_1, \gamma_2,...] \approx [ 0.5777,-0.072,-0.0096,...] $. Then the log-sums $[L_1,L_2,L_3,...] \approx [0.9189,-2.00635,6.00471,...] $$ (Hope I have all this correct so far) $\endgroup$ May 17 '11 at 11:19
  • $\begingroup$ I'll accept this answer because it gave the helpful link to the zeta-connection. Q2 concerning the correct application of the integral in the Ramanujan-summation is still open, but maybe better posed in an extra question which shall be focused on that. Many thanks to @J.M. again ! $\endgroup$ May 19 '11 at 4:27
5
$\begingroup$

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} $

$\endgroup$
1
$\begingroup$

Often looked at this question wondering if my reply might be useful as i think why you got the difference of 1.

What i think is that the calculations you did were correct, you just calculated the wrong regularization.

$$\sum_{n=1}^{\infty} ln(1+n)=\sum_{j=1}^{\infty}\sum_{n=1}^{\infty}(-1)^{j+1}\frac{n^{j}}{j} \ne\sum_{j=1}^{\infty}\frac{(-1)^{j+1}\zeta(-j)}{j}=\sum_{j=1}^{\infty}\frac{\zeta(-j)}{j}=\ln\bigg(\frac{\sqrt{2\pi}}{e}\bigg)$$

I had this is some [old notes] which i decided to type out, I used the product function. Few notation flaws, I've the habit to use the factorial.

$$\prod_{n=1}^{p}1+n=exp\bigg(\sum_{m=1}^{\infty}\frac{(-1)^{m+1}\sum_{n=1}^pn^m}{m}\bigg) $$

Just so that at j=-1 it should be seen as the limit of j goes to -1, so that it cancels out the factorial at -1! because zeta(1)/(-1)!=-1. $$\sum_{n}n^m=\sum_{j=-1}^{m;\infty}-\zeta(-j)n^{m-i} \frac{m!}{(m-j)!(j)!}$$

$$\prod_{n=1}^{p}1+n=exp\bigg(\bigg|_{n=0}^{n=p}\sum_{m=1}^{\infty}\frac{(-1)^{m+1}\sum_{j=-1}^{m-1;\infty}-\zeta(-j)n^{m-j} \frac{m!}{(m-j)!(j)!}}{m}\bigg|\bigg) $$ $$=exp\bigg(\bigg|_{n=0}^{n=p}\sum_{m=1}^{\infty}(-1)^{m+1}\sum_{j=-1}^{m;\infty}-\zeta(-j)n^{m-j} \frac{(m-1)!}{(m-i)!(j)!}\bigg|\bigg) $$

Sum it diagonally so you sum sit over the same power of n. $$=exp\bigg(\bigg|_{n=0}^{n=p}\sum_{j=-1}^{\infty}\sum_{c=1-j}^{\infty}(-1)^{c+j}\zeta(-j)n^{c} \frac{(c+j-1)!}{(c)!(j)!}\bigg|\bigg)$$

If $c<0$ all coefficients will be 0 (if $m>0$, and $m>=1$ so you even got a little space sort to speak). Now let's split it up, because the first 2 lines at i=-1 and i=0 are a bit tricky.

$$=exp\bigg(\bigg|_{n=0}^{n=p}\bigg(\sum_{j=1}^{\infty}\sum_{c=0}^{\infty}(-1)^{c+i}\zeta(-i)n^{c} \frac{(c+j-1)!}{(c)!(j)!}\bigg)+\bigg(\sum_{c=2}^{\infty}(-1)^{c-1}\zeta(1)n^{c} \frac{(c-2)!}{(c)!(-1)!}\bigg)$$

$$+\bigg(\sum_{c=1}^{\infty}(-1)^{c}\zeta(0)n^{c} \frac{(c-1)!}{(c)!(0)!}\bigg)\bigg|\bigg) $$

Fill in the zeta's $$=exp\bigg(\bigg|_{n=0}^{n=p}\bigg(\sum_{j=1}^{\infty}\sum_{c=0}^{\infty}(-1)^{c+j}\zeta(-j)n^{c} \frac{(c+j-1)!}{(c)!(j)!}\bigg)+\bigg(\sum_{c=2}^{\infty}(-1)^{c}n^{c} \frac{(c-2)!}{(c)!}\bigg)$$ $$+\bigg(\sum_{c=1}^{\infty}(-1)^{c}\zeta(0)n^{c} \frac{(c-1)!}{(c)!(0)!}\bigg)\bigg|\bigg)= \exp{\bigg((1)+(2)+(3)\bigg)}$$

$$\bigg(\sum_{c=2}^{\infty}(-1)^{c-1}n^{c} \frac{(c-2)!}{(c)!}\bigg)=(n+1)\ln(n+1)-n \tag 1$$ $$\bigg(\sum_{c=1}^{\infty}(-1)^{c}\zeta(0)n^{c} \frac{(c-1)!}{(c)!(0)!}\bigg)=-\zeta(0)ln(n+1)=\frac{\ln(n+1)}{2}\tag 2$$ $$\sum_{n=0}^{\infty} c^{-n}\frac{(n+s)!}{(s)!(n)!}=\bigg(\frac{c}{c-1}\bigg)^{s+1} $$ $$\bigg(\sum_{j=1}^{\infty}\sum_{c=0}^{\infty}(-1)^{c+j}\zeta(-j)n^{c} \frac{(c+j-1)!}{(c)!(j)!}\bigg)=\sum_{j=1}^{\infty}\frac{(-1)^j\zeta(i)}{j(n+1)^{j}}\tag 3$$ Combine (1)+(2)+(3)

$$=exp\bigg(\bigg|_{n=0}^{n=p}\bigg(\sum_{j=1}^{\infty}\frac{(-1)^j\zeta(-j)}{j(n+1)^{j}}\bigg)+\big((n+1)\ln(n+1)-n\big)+\big(\frac{\ln(n+1)}{2}\big)\bigg|\bigg) $$

$$=exp\bigg(\bigg(\sum_{j=1}^{\infty}\frac{(-1)^j\zeta(-j)}{j(p+1)^{j}}\bigg)-\bigg(\sum_{j=1}^{\infty}\frac{(-1)^j\zeta(-j)}{j}\bigg)+\big((p+1)\ln(p+1)-p\big)+\big(\frac{\ln(p+1)}{2}\big)\bigg) $$

$$=(p+1)^{p+1}e^{-p}(p+1)^{1/2}\exp\bigg(\bigg(\sum_{j=1}^{\infty}\frac{(-1)^j\zeta(-j)}{j(p+1)^{j}}\bigg)-\bigg(\sum_{j=1}^{\infty}\frac{(-1)^j\zeta(-j)}{j}\bigg)\bigg) =$$ $$=\prod_{n=1}^{p}1+n=(p+1)!$$

Now let's swap p+1=n because that reads nice. $$n!=\bigg(\frac{n}{e}\bigg)^{n}\big(e\sqrt{n}\big)* exp\bigg(\bigg(\sum_{j=1}^{\infty}\frac{(-1)^j\zeta(-j)}{j(n)^{j}}\bigg)-\bigg(\sum_{j=1}^{\infty}\frac{(-1)^j\zeta(-j)}{j}\bigg)\bigg) $$

Now you already see what you missed was this factor 'e'. Luckily this contains a simple delta function. First let's remove the alternating part.

$$n!=\bigg(\frac{n}{e}\bigg)^{n}\big(e\sqrt{n}\big)* exp\bigg(\bigg(\sum_{j=1}^{\infty}\frac{\zeta(-j)}{j}\bigg)-\bigg(\sum_{j=1}^{\infty}\frac{\zeta(-j)}{j(n)^{j}}\bigg)\bigg) $$

Now we use (n+1)!/n!=n+1

$$n+1=(1+1/n)^{n+1/2}*\frac{n+1}{e}*exp\bigg(\bigg(\sum_{j=1}^{\infty}\frac{\zeta(-j)}{jn^{j}}\bigg)-\bigg(\sum_{j=1}^{\infty}\frac{\zeta(-j)}{j(n+1)^{j}}\bigg)\bigg)$$

$$exp\bigg(\bigg(\sum_{j=1}^{\infty}\frac{\zeta(-j)}{j(n+1)^{j}}\bigg)-\bigg(\sum_{j=1}^{\infty}\frac{\zeta(-j)}{j(n)^{i}}\bigg)\bigg)=\frac{(1+1/n)^{n+1/2}}{e}$$

Now we have the delta function:

$$\sum_{n=1}^{p}\sum_{j=1}^{\infty}\frac{\zeta(-j)}{j(n+1)^{j}}-\frac{\zeta(-j)}{j(n)^{j}}=\sum_{j=1}^{\infty}\frac{\zeta(-j)}{j(p+1)^{j}}-\frac{\zeta(-j)}{j(1)^{j}}=\sum_{n=1}^{p}(n+1/2)\ln(1+1/n)-1$$

And as p goes to infinity (could be done cleaner) we get the convergent series: $$\sum_{j=1}^{\infty}\frac{-\zeta(-j)}{j}=\sum_{n=1}^{\infty}\bigg((n+1/2)\ln(1+1/n)-1\bigg)=1-\ln\bigg(\sqrt{2\pi}\bigg)$$ $$\sum_{j=1}^{\infty}\frac{\zeta(-j)}{j}=\ln\bigg(\frac{\sqrt{2\pi}}{e}\bigg)$$

This is what you did calculate here I think, and why regularization is tricky^^. And if we fill it in so that we get "stirlings approximation" for the factorial we get:

$$n!=\bigg(\frac{n}{e}\bigg)^{n}\big(\sqrt{2\pi n}\big)* exp\bigg(\sum_{j=1}^{\infty}\frac{-\zeta(-j)}{j(n)^{j}}\bigg) $$

$\endgroup$
2
  • $\begingroup$ Gerben - thank you very much. This is a lot of stuff, to which I'll come later; at the moment I've my head with another question (fixed points of the exp) and shall look at your ideas in a couple of days. $\endgroup$ May 18 '20 at 1:07
  • $\begingroup$ It's much of the same old I shared before. Just this time with justification. Calculations are rather straightforward (which is what I like about this derivation). And that I think why you were 1 off. $\endgroup$
    – Gerben
    May 18 '20 at 2:14
1
$\begingroup$

When I saw this (nine years after its posting), I thought of the derivative of the zeta function. In this answer, it is shown that $\zeta(s)$ is the constant in the Euler-Maclaurin Sum Formula below $$ \sum_{k=1}^nk^{-s}=\frac1{1-s}n^{1-s}+\overset{\substack{\text{constant}\\\downarrow\\[3pt]{}}}{\zeta(s)}+\frac12n^{-s}+O\!\left(n^{-1-s}\right)\tag1 $$ Taking the negative of the derivative with respect to $s$ gives $$ \sum_{k=1}^n\log(k)\,k^{-s}=\frac{\log(n)\,n^{1-s}}{1-s}-\frac{n^{1-s}}{(1-s)^2}-\zeta'(s)+\frac12\log(n)\,n^{-s}+O\!\left(\log(n)\,n^{-1-s}\right)\tag2 $$ Setting $s=0$ gives $$ \sum_{k=1}^n\log(k)=\log(n)\,n-n-\zeta'(0)+\frac12\log(n)+O\!\left(\log(n)\,n^{-1}\right)\tag3 $$ $(3)$ is the logarithm of Stirling's Formula when we set $\zeta'(0)=-\frac12\log(2\pi)$.

Thus, I would expect that regularized, we would get $$ \sum_{k=1}^\infty\log(k)=\frac12\log(2\pi)\tag4 $$

$\endgroup$
4
  • $\begingroup$ Thanks, robjohn, for your msg. After your first version with the negative number as result, I thought: what a nice bunch of flowers gives this question's solutions - positive, negative, reduced by one,.. :-) Happy to see after the correction the positive value. I'll try to reproduce your initial equation (1) - but not today, I'm absorbed with the other problem in my mind (about periodic points of the exp-function) $\endgroup$ May 18 '20 at 14:05
  • $\begingroup$ How do you handle formally your equation at (3), for negative n? $\endgroup$
    – Gerben
    May 21 '20 at 2:44
  • $\begingroup$ @Gerben: These are asymptotic series for $n\to\infty$. We are not using them for negative $n$. $\log(k)$ and $k^{-s}$ are not real for $k\lt0$. $\endgroup$
    – robjohn
    May 21 '20 at 3:22
  • $\begingroup$ How would you describe it minus infinity. Also can we (aka is it possible) to figure an exact series representation so that it's not asymptotic but precise. $\endgroup$
    – Gerben
    May 21 '20 at 9:29

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.