For integers $n\geq 1$, we denote with $G_n$ the $nth$ Gregory coefficient, see the definition from this Wikipedia. I was searching about the asymptotic behaviour of summatory functions related to these numbers, because I am interested about the asymptotic behaviour of $$\int_1^x \frac{\sum_{n\leq t}G_n}{t^2}dt,\tag{1}$$ as $x\to\infty$.

My main purpose is to know the difficulty to calculate such asymptotic behaviour, because I want to do a comparison with the so-called Weak Mertens conjecture ( I do not even know if the asymptotics in calculations related to Gregory numbers are so deep as the Prime Number Theorem, but these numbers, Gregory numbers, seem have importance in number theory).

I know from Abel's identity $$\int_1^x\frac{\sum_{n\leq t}G_n}{t^2}dt=\sum_{n\leq x}\frac{G_n}{n}-\frac{1}{x}\sum_{n\leq x}G_n\tag{2}$$ when $x> 1$. And also can write statements using summation by parts (using the formula 2.3(i) NIST DLMF), for instance $$\sum_{j=1}^{n-1}\frac{G_j}{j}=\frac{1}{n}\sum_{k=1}^{n-1}G_k+\sum_{j=1}^{n-1}\left(\frac{1}{j}-\frac{1}{j+1}\right)\left(\sum_{k=1}^j G_k\right).\tag{3}$$ Additionally one can do specializations of Lemma 1 from [1], I say specializations for arithmetic functions $a(n)=G_n$, or well for $a(n)=\frac{G_n}{n}$ to set more related formulas.

Question. Is it possible set the asymptotic behaviour of $$\int_1^x \frac{\sum_{n\leq t}G_n}{t^2}dt,$$ as $x\to\infty$? You can provide hints and if is required some specific asymptotic for Gregory coefficients you can refer those literature. (Thus I am not specially interested in the details, only a draft to know how much difficult is solve this problem) Many thanks.

Also feel free if you want to do a comparison between our problem, problem in previous Question, and the Weak Mertens conjecture (I say if you know dilucidate that our prolem isn't an unsoved problem so difficult than the cited conjecture).


[1] Apostol, Introduction to Analytic Number Theory, UTM Springer (1976), Lemma 1 of section 13.2.


Since $$\sum_{n\geq 1}G_n z^n = \frac{z}{\log(1+z)}-1\tag{1}$$ we also have $$ \sum_{n=1}z^{n-1}\sum_{m=1}^{n}G_m = \frac{1}{(1-z)\log(1+z)}-\frac{1}{z(1-z)}\tag{2} $$ and by multiplying both sides by $-\log(z)$ and performing $\int_{0}^{1}(\ldots)\,dz $ we get:

$$\begin{eqnarray*} \sum_{n=1}\frac{1}{n^2}\sum_{m=1}^{n}G_m &=& \int_{0}^{1}\frac{\log(z)}{1-z}\left(\frac{1}{z}-\frac{1}{\log(1+z)}\right)\,dz\\(\text{Schröder's})&=&\int_{0}^{+\infty}\frac{\zeta(2)-\text{Li}_2\left(-\frac{1}{1+x}\right)}{(x+2)\left(\pi^2+\log^2 x\right)}\,dx\\&\approx& 0.78027750661.\tag{3}\end{eqnarray*} $$

Many series involving Gregory coefficients can be directly computed from the generating function $(1)$. For instance: $$ \sum_{n\geq 1}\frac{G_n}{n} = \int_{0}^{1}\left(\frac{1}{\log(1+z)}-\frac{1}{z}\right)\,dz = -\gamma+\text{li}(2)\tag{4} $$ where $\gamma$ is the Euler-Mascheroni constant and $\text{li}(a)=\int_{0}^{a}\frac{dt}{\log t}$ is the logarithmic integral.

  • $\begingroup$ Now I am stuck, I am going to try understantd your answer. Many thanks. $\endgroup$ – user243301 Jun 27 '17 at 17:13

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