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

Question

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 : $$\lim\sup_{n\rightarrow\infty}\left(\frac{\delta_{n}}{n!}\right)^{\frac{1}{n}}=0$$ To be exact, I am interested in the limit : $$\lim\sup_{n\rightarrow\infty}\left(\frac{\delta_{n}}{n}\right)^{\frac{1}{n}}$$ hence the question !!

Answer 1

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 200    \\ set precision for dec digits
   \ps 256   \\ set number of terms for taylor-series expansion
   powser_eta = sumalt(k=0,taylor((-1)^k*1/(1+k)^x,x)) 
   laurent_zeta = powser_eta/(1-2*2^-x)
   coeffs = polcoeffs( laurent_zeta + 1/(1-x),256)  \\ extract 256 coefficients
   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 \\ ... \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

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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.")

Answer 2

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 :

$ \begin{array} {r|l} n&\qquad n!+\zeta^{(n)}(0)\\ \hline 1 & +0.081061466795327258219670263594382360138602526362217\\ 2 & -0.0063564559085848512101000267299604381989949101609199\\ 3 & -0.0047111668622544477610608133663752854618076682959801\\ 4 & +0.0028968119862920410127804722589943381088600650782966\\ 5 & -0.00023290755845472453598583779581974789205717247050230\\ 6 & -0.00093682513005092950428350854539855876385290926809868\\ 7 & +0.00084982376500166915170602760235121839217676036899380\\ 8 & -0.00023243173551155958285569006371686986154745560535153\\ 9 & -0.00033058966361229644525612725015921912916311539120160\\ 10 & +0.00054323411577970847223198894312031008561943002564803\\ 11 & -0.00037549317290726365046703088410553955290852331712733\\ 12 & -0.00001960353628101391976648402508433558658818213359962\\ 13 & +0.00040724123256303314343212136681027307343924449505289\\ 14 & -0.00057049201328177771564129138383813714231765446439354\\ 15 & +0.00039392707898120442182766081893948743593101317331900\\ 16 & +0.00008345880582550168172764880471555318446251614843452\\ 17 & -0.00066094372962859689616940299813405772474841468462821\\ 18 & +0.0010262272865408540021770141554688378775983106974390\\ 19 & -0.00086557577677928299157607241403657110459312961654081\\ 20 & +0.00001929367178370514010632997603577601048054770687535\\ 21 & +0.0013569060521345494611491378326511761990288706578281\\ 22 & -0.0026921564587532912840342571094899479367185487885538\\ 23 & +0.0030513856212416271388454373861585656340439536386835\\ 24 & -0.0014242918494185458532221867917952455892341070680451\\ 25 & -0.0027077892128860067881974821917555423128848837698589\\ 30&-0.026465704147079752693730404859959295339337073188577\\ 35&-0.26359445473226969258965859491215128351504627358118\\ 40&-2.9912738940588767630327451314666324157450427478360\\ 45&-37.811691859847699571345792885440735948937675076443\\ 50&-484.41085697391134019683488132115999695787532277743\\ 55&-4532.7922577092171518919512255451187936120105777731\\ 60&+62714.106769571852549815121861152393947489784478599\\ 65&+7023172.7145242778883663789007092296487557987220273\\ 70&+369710251.75434261376148718924306570270744599755002\\ 75&+16153042555.891600628481729183019107036027090664570\\ 80&+615738270543.41976362005501481867360304511761212199\\ 85&+18734769337973.535747625469811963057045887984795841\\ 90&+236370935383452.03942787310651808117015612065952142\\ 95&-26002457205974856.121059702068315722218399244645218\\ 100&-3067048412469082717.1320349345687277345600145601427\\ 105&-208147105464557539810.93310552061394602332413623631\\ 110&-9181100257482418076527.7843319896367738502453996735\\ 115&-51947662171852808135142.656616304150605568437147323\\ 120&+40156333121359621232445103.065796921480403337792144\\ 125&+4885455264162691954362582051.1962929540984191959671\\ 130&+326172379219132017786027255436.16366267172881142641\\ 135&+5681896814647267766788984138309.9277764944754964868\\ 140&-1823873410669202891713087061952487.2923381095183773\\ 145&-287161238605183347710570327381611857.62150269361362\\ 150&-21305861581790622498949173421790799625.108948639082\\ 155&+41341935656925531212500416560539095352.234411848266\\ 160&+247591097041903905305863994419088881629306.69527638\\ 165&+35417487509305790307439844806554155410647762.2818942813077054202595\\ 170&+1939388852429349721510180790653718054320127522.76657886070312620767\\ 175&-219609544533102325798714608918968968215179933676.462881353291615995\\ 180&-64398214417872662764963987879167602127249665707913.3748997726013798\\ 185&-6471529441461413822723169640664516218513802097544790.17826333568547\\ 190&+124737730975894951649278632325321300323483372940042824.738170084667\\ 195&+146090125339857661850314283330560855583771401129477555170.288406962\\ 200&+21761038288742061134507006188990514804372485314425207281255.1444267\\ 205&+448206643590051608263691568113493984452862240936661159203824.575986\\ 210&-436802309714509751568738654004052940208588317008859449997875439.739\\ 215&-87517428053442479414927505503691736597016156193237933277989943662.6\\ 220&-3847242990914462888287610077836765513262393392990395867588899745072.\\ 225&+1.78354688800770254378644818320484851438306069887890285327456308 \rm{E}69\\ 230&+4.39266696649098114932308252500592591975090024455299051343992957 \rm{E}71\\ 235&+2.44222360080279317167994898279834010610042190503137084735641980 \rm{E}73\\ 240&-1.01144099944259968270241440744065554134825102647611664424324558 \rm{E}76\\ 245&-2.70449272645088612156914866282417480363449265831314743378453851 \rm{E}78\\ 250&+3.05921285543816770383418771952247557481329946696456084332720656 \rm{E}79\\ \end{array} $

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

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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 $0.4646$.

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//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

Answer 3

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 !?