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Let $a=1/2-30.424876126i$ ($i^2=-1$). Then trying to plot $$\operatorname{Re}\sum_{n=1}^{200} \frac{(a\ln x)^n}{n!\, n\zeta (n+1)}$$ in $x$ leads to a very inaccurate jaggy graph possibly caused by some floating point errors:

a:=1/2-30.424876126I
f[x_]:=Sum[(a Log[x])^n/(n! n Zeta[n+1]),{n,1,200}]
Plot[Re[f[x]],{x,1,5}]

graph The beginning of the graph is good, though. How can the errors be mitigated?

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  • $\begingroup$ If your value of a is supposed to be a zero of the zeta function, have you also tried including more decimal values? $\endgroup$ Jul 20 at 23:59
  • $\begingroup$ @zeattledave Combining SetPrecision and WorkingPrecision solved the problem. $\endgroup$
    – Wane
    Jul 21 at 0:54
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Use higher WorkingPrecision:

f[u_] = Sum[u^n/(n n! Zeta[1 + n]), {n, 1, 200}];
a = SetPrecision[1/2 - 30.424876126 I, ∞];  (* make exact value *)
Plot[Re[f[a*Log[x]]], {x, 1, 5}, PlotRange -> All, 
     WorkingPrecision -> 100]

enter image description here

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  • $\begingroup$ The option PlotPoints -> 100 instead of WorkingPrecision -> 100 works too. BTW, the plot is different from yours. $\endgroup$
    – user64494
    Jul 19 at 17:59

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