I'm interested in calculations involving the Euler-Macheroni constant and the so-called Gregory coefficients $G_n$, see the related Wikipedia. Today from an inequality I have got this series


I know how get a lower and an upper bound like these $$0.395908<\sum_{n=1}^\infty|G_n|\log\left(\frac{n+1}{n}\right)<0.439844,\tag{2}$$

or improvements of this inequality.

But I would like to how improve my approximation for such series, I did a draft using partial summation and another using Abel' summation lemma, from which I think that these don't work.

Question. What is a strategy to get and justify a good approximation (four or six right digits) of $$\sum_{n=1}^\infty|G_n|\log\left(\frac{n+1}{n}\right),$$ where $G_n$ denote the Gregory coefficients? Many thanks.

If it is in the literature feel free to answer as a reference request, and I try to find and to read such proposition from the literature.


By the very definition of Gregory coefficients $$ \sum_{n\geq 1}|G_n| z^n = 1+\frac{z}{\log(1-z)} \tag{1}$$ and by their integral representation $|G_n|\sim\frac{1}{n\log^2 n}$ for large values of $n$. Since $\log\left(1+\frac{1}{n}\right)\sim\frac{1}{n}$ the wanted series is absolutely convergent; by Frullani's integral and $(1)$

$$\sum_{n\geq 1}|G_n|\log\left(1+\frac{1}{n}\right) = \int_{0}^{+\infty}\frac{1-e^{-t}}{t}\sum_{n\geq 1}|G_n|e^{-nt}\,dt=\int_{0}^{+\infty}\frac{1-e^{-t}}{t}\left(1+\frac{e^{-t}}{\log(1-e^{-t})}\right)\,dt $$ and by substituting $t=-\log u$ in the last integral we get $$\sum_{n\geq 1}|G_n|\log\left(1+\frac{1}{n}\right) = \int_{0}^{1}\frac{u-1}{u\log u}\left(1+\frac{u}{\log(1-u)}\right)\,du \tag{2}$$ where the RHS of (2) is perfectly manageable through standard numerical routines (Newton-Cotes formulas, Gaussian quadrature or a combination of them). My version of Mathematica returns a $\color{green}{0.4122998}$ as an approximated value of the RHS of $(2)$.

  • $\begingroup$ Many thanks, I am going to study your answer invoking Frullani's theorem and the generating function of Gregory coefficients. Many thanks $\endgroup$ – user243301 Mar 15 '18 at 20:51

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