# Derivatives of the Riemann zeta function at $s=0$

It's a curious fact that for $n>0$, $\zeta^{(n)}(0)\approx -n!$. Apostol gave a table for $\frac{\zeta^{(n)}(0)}{n!}$, among other results on $\zeta^{(n)}(0)$ . the sequence :

$$\delta_{n}=\left | \zeta^{(n)}(0)+n! \right|$$

seems to be fast decreasing. What is the upper bound of $\delta_{n}$ ??

Edit: Following the logic in Apostol's paper, $\zeta(s)-\frac{1}{s-1}$ is holomorphic. Thus:

$$\zeta(s)-\frac{1}{s-1}=:A(s)=\sum_{n=0}^{\infty}\frac{A^{(n)}(0)}{n!}s^{n}$$ where : $$\left|A^{(n)}(0)\right|=\delta_{n}=\left|\zeta^{(n)}(0)+n!\right|$$ the expansion converges everywhere. Therefore : $$\limsup_{n\rightarrow\infty}\left(\frac{\delta_{n}}{n!}\right)^{\frac{1}{n}}=0$$ To be exact, I am interested in the limit : $$\limsup_{n\rightarrow\infty}\left(\frac{\delta_{n}}{n}\right)^{\frac{1}{n}}$$ hence the question !!

• This is a related question. – J. M. is a poor mathematician Aug 24 '12 at 13:13
• I updated my answer with two pictures concerning your limit. I hope it will help more, – Raymond Manzoni Aug 24 '12 at 18:15

This should go as another comment to @Raymond Manzoni.
Here is a short routine in Pari/GP how the above coefficients can be computed to high accuracy by a very simple procedure:

   \p 400    \\ \p 200      \\ set precision for dec digits
\ps 256   \\ \ps 128     \\ set number of terms for taylor-series expansion
taylor_eta = sumalt(k=0,taylor((-1)^k*1/(1+k)^x,x))
laurent_zeta = taylor_eta/(1-2*2^-x)
\\-- coeffs = polcoeffs( laurent_zeta + 1/(1-x),256)  \\ extract coeffic
coeffs = Vec ( laurent_zeta + 1/(1-x) )  \\ extract coeffs (update dez 16)
vectorv(12,r,coeffs[r]*(r-1)!)           \\ display the first few coefficients


The first 12 coefficients
$\small \begin{matrix} 0.500000000000 \\ 0.0810614667953 \\ -0.00635645590858 \\ -0.00471116686225 \\ 0.00289681198629 \\ -0.000232907558455 \\ -0.000936825130051 \\ 0.000849823765002 \\ -0.000232431735512 \\ -0.000330589663612 \\ 0.000543234115780 \\ -0.000375493172907 \\ \vdots \end{matrix}$
and that around k=256 see Raymond's answer. Possibly we should increase the internal num-precision even higher to get meaningful digits below the decimal point for that high coefficients.
The computation to 120 good coefficients took only a few seconds with that given precision of 256 dec digits . For 256 good coefficients we need decimal precision \p 400 and much more memory and a couple of seconds more time

obsolete due to update dez 16 having the most simple precudere by "Vec()"
Pari/GP-script for "polcoeffs"

\\ lp: the polynomial or series, local; maxd: option to force length of result-vector
{polcoeffs(lp, maxd=0) = local(llp, lpd, lv, lv1);
llp=Pol(lp);lpd=poldegree(llp);
if(lpd<0,return(vector(maxd)));
lv=vector(lpd+1,k,polcoeff(llp,k-1));
if(maxd>0,lv1=vector(maxd,k,if(k>lpd+1,0,lv[k]));lv=lv1);
return(lv);}
addhelp("polcoeffs","uses a scalar entry containing a polynomial, converts it into a vector of coefficients.")

• +1 for the effort. I couldn't try your script since polcoeffs doesn't seem to exist in my 2.5.0 release (polcoeff exists). The newly compiled v2.5.3 didn't know polcoeffs either (while polcoeff(...) returns "non existent component in truecoeff"). What does [? polcoeffs] return on your system ? Could it be a 'added' function? – Raymond Manzoni Aug 24 '12 at 17:29
• @RaymondManzoni: upps, yes, I simply made this a user function giving the vector of all "polcoeff()"s because I need that permanently... – Gottfried Helms Aug 24 '12 at 18:00
• Raymond: I've put the code into the answer. – Gottfried Helms Aug 24 '12 at 19:30
• I was trying it and, with your initial settings, found that the value for $n=120$ was correct to $26$ digits (on the $28$ returned) : not bad. I tried with $500$ digits of precision and this time I got $326$ digits for $n=120$ (no idea how many are right :-)). It is a really excellent method ! Many thanks to share it !! – Raymond Manzoni Aug 24 '12 at 19:40

UPDATED:

Since Apostol's table is imprecise for the latest values let's exhibit this partial table of $n!+\zeta^{(n)}(0)$ values obtained with the method proposed by Gottfried Helms in the comments :

$\small \begin{array} {r|l} n&\qquad n!+\zeta^{(n)}(0)\\ \hline 0 & 0.500000000000000000000000000000000000000000000000000000000000000000 \\ 1 & 0.0810614667953272582196702635943823601386025263622165871828484595172 \\ 2 & -0.00635645590858485121010002672996043819899491016091988116986828085776 \\ 3 & -0.00471116686225444776106081336637528546180766829598013289308154130860 \\ 4 & 0.00289681198629204101278047225899433810886006507829657502399066695362 \\ 5 & -0.000232907558454724535985837795819747892057172470502296621517290052364 \\ 6 & -0.000936825130050929504283508545398558763852909268098676811811642454272 \\ 7 & 0.000849823765001669151706027602351218392176760368993802245821950220545 \\ 8 & -0.000232431735511559582855690063716869861547455605351528951730144900587 \\ 9 & -0.000330589663612296445256127250159219129163115391201597238597920006568 \\ 10 & 0.000543234115779708472231988943120310085619430025648031886746513765534 \\ 11 & -0.000375493172907263650467030884105539552908523317127333739022948360384 \\ 12 & -0.0000196035362810139197664840250843355865881821335996260346542408699771 \\ 13 & 0.000407241232563033143432121366810273073439244495052894296377049143472 \\ 14 & -0.000570492013281777715641291383838137142317654464393538891561665994592 \\ 15 & 0.000393927078981204421827660818939487435931013173319003367358811853101 \\ 16 & 0.0000834588058255016817276488047155531844625161484345203508967032195293 \\ 17 & -0.000660943729628596896169402998134057724748414684628214724260392025847 \\ 18 & 0.00102622728654085400217701415546883787759831069743902026886240548348 \\ 19 & -0.000865575776779282991576072414036571104593129616540810229322531122882 \\ 20 & 0.0000192936717837051401063299760357760104805477068753543599966583874264 \\ 21 & 0.00135690605213454946114913783265117619902887065782808784758635491569 \\ 22 & -0.00269215645875329128403425710948994793671854878855377935283522438652 \\ 23 & 0.00305138562124162713884543738615856563404395363868348883899894968459 \\ 24 & -0.00142429184941854585322218679179524558923410706804512920069410425063 \\ 24 & -0.00142429184941854585322218679179524558923410706804512920069410425063 \\ 25 & -0.00270778921288600678819748219175554231288488376985887236498730634210 \\ 30 & -0.0264657041470797526937304048599592953393370731885768642502823064627 \\ 35 & -0.263594454732269692589658594912151283515046273581182559219921957221 \\ 40 & -2.99127389405887676303274513146663241574504274783600393720076526420 \\ 45 & -37.8116918598476995713457928854407359489376750764425231304638226967 \\ 50 & -484.410856973911340196834881321159996957875322777427689682560124377 \\ 55 & -4532.79225770921715189195122554511879361201057777310708972171082184 \\ 60 & 62714.1067695718525498151218611523939474897844785985218047818417901 \\ 65 & 7023172.71452427788836637890070922964875579872202726818830758507697 \\ 70 & 369710251.754342613761487189243065702707445997550023978646801198349 \\ 75 & 16153042555.8916006284817291830191070360270906645699878986789707549 \\ 80 & 615738270543.419763620055014818673603045117612121993882170431591493 \\ 85 & 18734769337973.5357476254698119630570458879847958412519956399551375 \\ 90 & 236370935383452.039427873106518081170156120659521416134138380827174 \\ 95 & -26002457205974856.1210597020683157222183992446452182712157359931706 \\ 100 & -3067048412469082717.13203493456872773456001456014271660974790930507 \\ 105 & -208147105464557539810.933105520613946023324136236314019489461672300 \\ 110 & -9181100257482418076527.78433198963677385024539967354760263208242840 \\ 115 & -51947662171852808135142.6566163041506055684371473226227514782141120 \\ 120 & 40156333121359621232445103.0657969214804033377921435547843142396453 \\ 125 & 4885455264162691954362582051.19629295409841919596706506250394536303 \\ 130 & 326172379219132017786027255436.163662671728811426407157377065370050 \\ 135 & 5681896814647267766788984138309.92777649447549648680365985502173310 \\ 140 & -1823873410669202891713087061952487.29233810951837725134296601143730 \\ 145 & -287161238605183347710570327381611857.621502693613616741540893113635 \\ 150 & -21305861581790622498949173421790799625.1089486390817454023538053647 \\ 155 & 41341935656925531212500416560539095352.2344118482658208289324353056 \\ 160 & 247591097041903905305863994419088881629306.695276383394597551295589 \\ 165 & 35417487509305790307439844806554155410647762.2818942813077054202595 \\ 170 & 1939388852429349721510180790653718054320127522.76657886070312620767 \\ 175 & -219609544533102325798714608918968968215179933676.462881353291615996 \\ 180 & -64398214417872662764963987879167602127249665707913.3748997726013799 \\ 185 & -6471529441461413822723169640664516218513802097544790.17826333568557 \\ 190 & 124737730975894951649278632325321300323483372940042824.738271112913 \\ 195 & 146090125339857661850314283330560855583771401129477483038.196790939 \\ 200 & 21761038288742061134507006188990514804372485347492068735353.4677389 \\ 205 & 448206643590051608263691568113493984443540648811947725902790.626596 \\ 210 & -436802309714509751568738654004051406952276718382033685343775072.767 \\ 215 & -87517428053442479414927505641545087908985720235451301367834785555.4 \\ 220 & -3.84724299091446288828137723409916186345658241907462046206042911305E66 \\ 225 & 1.78354688800770241687161303825386645838232647101391084926254576406E69 \\ 230 & 4.39266696650096770242083480719428532550626963368237730956507167675E71 \\ 235 & 2.44222335896278620212620252751346268294589748965319118864768279107E73 \\ 240 & -1.01131768916824854497126506938489065442700604328045419557651761065E76 \\ 245 & -2.76259374758593015959757159949637125866476599264626984421242976978E78 \\ 250 & -1.51090342799297835940857060215282929045189635533361177391732022698E80 \end{array}$
(updated 3.12.2016, values from index 200 to 250 had to be corrected)

I fear that this will grow without bounds even if much slower than $n!$ but an asymptotic formula could be conjectured from these values !

Concerning the limit : $$\lim\sup_{n\rightarrow\infty}\left(\frac{\delta_{n}}{n}\right)^{\frac{1}{n}}$$

I can only show you the 'brainy' picture obtained for values of $n$ from $1$ to $250$ :

The largest value obtained is near $2.047$ but this doesn't seem to stop.

Note that this is nearly the same picture than for $\ \lim\sup_{n\rightarrow\infty}\left(\delta_{n}\right)^{\frac{1}{n}}\$ (division by $n$ doesn't matter much).

If we observe that the real takeoff of $\delta_n$ waits until $n=25$ then a not too bad approximation of the previous curve is : $$f(n)=\frac{\sqrt[3]{n-17}}3$$ represented here (for $n$ from $17$ to $250$) :

I tried to divide $\delta_n$ by different expressions in your limit and found : $$\ \lim\sup_{n\rightarrow\infty}\left(\dfrac{\delta_{n}}{\sqrt[3]{n!}}\right)^{\frac{1}{n}}\$$

with the interesting 'saturation' near $0.4646$.

//Scripts used (pari/gp) :

//Method proposed by Gottfried Helms (precomputed Stieltjes table)
zs(n)=(-1)^n*sum(k=0,#Stieltjes-n-1,Stieltjes[k+n+1]/k!)

//Direct evaluation of the nth derivative at z (ep= 1E-50 or less)
zp(z,n,ep)=sum(k=0,n,(-1)^k*binomial(n,k)*zeta(z+(n-2*k)*ep))/(2*ep)^n

• Besides the link to the older question provided by @J.M here is the formula which uses the Stieltjes-constants to compute that values much accurately. Here is a formula for Pari/GP: sum(k=0,max_k,Stieltjes[k+n]/k!) where n is the index in your above table. Sthe Stieljes-constants can be found online to several hundreds digits precision (but that is not even needed here). In case it is of interest: I've discussed that zeta(0)-derivatives in playing with the gamma: go.helms-net.de/math/musings/UncompletingGamma.pdf from page 18 – Gottfried Helms Aug 24 '12 at 14:32
• I have a table of the 78 first Stieltjes constants (to 256 digits) (from S. Plouffe I think) and I'll try your methods. Many thanks @Gottfried ! – Raymond Manzoni Aug 24 '12 at 14:50
• @Gottfried: I think that the real interest of the OP is the convergence of $$\sum_{k=1}^\infty K_{2k} \frac{B_{2k}}{(2k)!} s^{2k}$$ with $$K_n:=\frac{n!+\zeta^{(n)}(0)}n$$ – Raymond Manzoni Aug 24 '12 at 18:29
• @MohammadAlJamal: From the Asymptotic approximation of $B_n$ and after simplification we get an asymptotic behavior : $2\sum_{k=1}^\infty K_{2n} \left(\frac s{2\pi}\right)^{2n}$. Since $K_n$ seems to be of order $\sqrt[3](n!)a^n$ I fear (as you did) that the radius of convergence will be zero... – Raymond Manzoni Aug 24 '12 at 19:07
• The computation of the long table seems to have indufficient decimal digits precision. I've recomputed that table using internal precision \p 600 where visible differences begin at index 200, increase via index 225, 240 and at index 250 even the first digit is wrong. Would you mind to insert better values? (I could just provide the current results if you do not want to reopen your Pari/GP-session and "allocatemem" up to 128 MB ) – Gottfried Helms Dec 3 '16 at 16:12

consider the original problem : $$f(s)=-\psi\left(1-\frac{1}{s}\right)-K_{0}-\frac{K_{1}}{2}s-\sum_{n=1}^{\infty}\frac{K_{2n}B_{2n}}{(2n)!}s^{2n}$$ Where :

$K_{0}=1.825593297777$

$K_{1}=1+\zeta^{(1)}(0)$

$K_{n}=\frac{n!+\zeta^{(n)}(0)}{n}$

We use the asymptotic expansion of $\psi(x)$: $$\psi(1+x)=\frac{1}{x}+\psi(x)=\ln(x)+\frac{1}{2x}-\sum_{n=1}^{\infty}\frac{B_{2n}}{2nx^{2n}}$$ therefore-naively speaking-: $$f(s)=-K_{0}-\frac{\zeta^{(1)}(0)}{2}s+\ln(-s)-\sum_{n=1}^{\infty}\frac{B_{2n}}{(2n)!}\left(\frac{\zeta^{(2n)}(0)}{2n} \right )s^{2n}$$ and we got rid of $(2n)!$ in $K_{2n}$. a couple of questions remain: is the asymptotic expansion of the digamma function exact!? what's the domain of convergence of the expansion!? and what's the radius of convergence of our last expansion !?

• Since few people search questions in answers don't hope too many reactions (it breaks the Q->A principle favored here !). Anyway the two series are only asymptotic (from my previous link) so that $=$ should be replaced by $\sim$ and the radius of convergence will be $0$ (removing the $2n!$ will make the convergence worse as in $\sum \frac {\zeta(kx)}k$). Asymptotic expansions are very powerful when we use only the first terms! Your problem seems to be asymptotic versus analytic. You could ask a question clarifying this. – Raymond Manzoni Aug 25 '12 at 20:21