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I wanted to numerically verify the validity of the formula for the first Stieltjes constant

$$\gamma_1=-\frac12\sum_{n=0}^\infty\frac1{n+1}\sum_{k=0}^n\binom{n}{k}(-1)^k\log^2(k+1)$$

 -1/2*Sum[1/(n + 1)*Sum[(-1)^k*Binomial[n, k]*Log[k + 1]^2, {k, 0, n}], {n, 0, Infinity}]

The correct value is

 StieltjesGamma[1]//N
 (* -0.0728158 *)

If I enter 40 terms of the series, the result is still OK

 N[-1/2*Sum[1/(n + 1)*Sum[(-1)^k*Binomial[n, k]*Log[k + 1]^2, {k, 0, n}], {n, 0, 40}]]
 (* -0.0623308 *)

With 200 terms, however, the result runs away to wrong values

 N[-1/2*Sum[1/(n + 1)*Sum[(-1)^k*Binomial[n, k]*Log[k + 1]^2, {k, 0, n}], {n, 0, 200}]]
 (* 3.28992*10^42 *)

However, if I use ListPlot, the graph for 200 terms (and even 400 terms!) is fine without having to enter any additional parameters.

 ListPlot[Table[-1/2*Sum[1/(n + 1)*Sum[(-1)^k*Binomial[n, k]*Log[k + 1]^2, {k, 0, n}], {n, 0, m}], {m, 1, 200}]]

ListPlot

What is the problem? After all, ListPlot must also somehow calculate the values for the graph numerically. Does anyone know how to do this?

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  • $\begingroup$ Looks strange. Might it be a bug? $\endgroup$ – Αλέξανδρος Ζεγγ Jun 7 at 11:49
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    $\begingroup$ I experimented with this representation of $\gamma_1$ some time back; I've found that this one (and the more general $\gamma_k$ expression this was derived from) has poor numerical properties. (I finally settled on using Abel-Plana instead.) Nevertheless: -NSum[(-1)^k*Binomial[n, k]*Log[k + 1]^2/(n + 1), {n, 0, ∞}, {k, 0, n}, Method -> {"WynnEpsilon", Degree -> 1, "ExtraTerms" -> 25}, NSumTerms -> 55, WorkingPrecision -> 45]/2 $\endgroup$ – J. M.'s technical difficulties Jun 7 at 12:01

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