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Mathematica 12 evaluates the following sequence of limits thus (tested up to $s=100$) $$\underset{x\to \,1^+}{\text{lim}}\;\left(\frac{\partial ^n }{\partial x^n}\left(\frac{1}{1-x}+\sum _{k=1}^{\infty } \frac{1}{k^x}\right)\right)=(-1)^n\,\gamma _n\tag{1}$$

where $\gamma _n$ are the Stieltjes constants starting with $\gamma _0$ the Euler-Mascheroni Constant.

Wikipedia states that the Stieltjes constants can be found in the limit

$$\underset{m\to \infty }{\text{lim}}\left(\sum _{k=1}^m \left(\frac{\log ^n(k)}{k}-\frac{\log ^{n+1}(m)}{n+1}\right)\right)=\gamma _n\tag{2}$$

which Mathematica has trouble calculating.

I have tried and failed to use (1) to calculate the well known limit for the Euler-Mascheroni constant $\underset{m\to \infty }{\text{lim}}\left(\sum _{k=1}^m \frac{1}{k}-\log (m)\right)=\gamma_0$ . I can see that $\frac{1}{1-x}$ gives a geometric progression, but I do not know how to proceed from there, other than recasting as limit to infinity thus:

$$\underset{m\to \infty }{\text{lim}}\left(\sum _{k=1}^{\infty } \frac{1}{k^{\frac{1}{m}+1}}+\frac{1}{1-\left(\frac{1}{m}+1\right)}\right)=\underset{m\to \infty }{\text{lim}}\left(\zeta \left(\frac{m+1}{m}\right)-m\right)=\gamma_0$$

I can't even see a way of relating one formula to the other, that is (1) to (2) or visa versa.

James Arathoon
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  • With $n = 0$, the first limit doesn't exist, since $\lim_{x \to 1^-}f(x) = \infty$, and $\lim_{x \to 1^+}f(x) = -\infty$. But $\lim_{\epsilon \to 0} \frac{f(1-\epsilon)+f(1+\epsilon)}{2}=\gamma_0$. – Varun Vejalla Apr 25 '21 at 18:56
  • @VarunVejalla: for (1) $n=0$ function is not differentiated of course. The two terms head to $-\infty$ and $+\infty$ respectively as $x \to 1^+$. I am confused by what you say. I will clarify limit to $1^+$ to avoid issues of analytic continuation of Zeta Function which Mathematica assumes by default. – James Arathoon Apr 25 '21 at 19:32
  • Found: Sondow, Jonathan (1998). "An antisymmetric formula for Euler's constant". Mathematics Magazine. 71. pp. 219–220. Which will help in the case of the Euler-Mascheroni Constant. – James Arathoon Apr 25 '21 at 19:46
  • I hope this can help you. – TravorLZH Jul 02 '21 at 05:05

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