Justify an approximation of $\sum_{n=1}^\infty G_n\zeta(n+1)$, where $G_m$ denotes the so-called Gregory coefficients

Question

I would like to know how get a good approximation (four or six right digits) of the series $$\sum_{n=1}^\infty G_n\zeta(n+1),\tag{1}$$ where $\zeta(s)$ denotes the Riemann zeta function and $G_n$ the $n$th Gregory coefficient, see this Wikipedia.

This is just a curiosity. This kind of series $(1)$ arises when one take the derivative of the generating function (that is the Maclaurin series of the logarithm in the first paragraph of previous Wikipedia), perform the specialization $z=\frac{1}{k}$ and after that we multiply our resulting equation by $\frac{(-1)^k}{k^2}$ we take the sum over integers $k\geq 2$.

Question. How do you justify from a good strategy the first few digits of $$\sum_{n=1}^\infty G_n\zeta(n+1)\,?$$ Many thanks.

I think that maybe can be some interesting strategy to get and justify such approximation.

Answer 1

First, let's compute the first few partisal sums:

$$S_N=\sum_{n=1}^N G_n\zeta(n+1)$$

$$\begin{array}( N & S_N \\ 4 & 0.7673924262 \\ 5 & 0.7400290549 \\ 6 & 0.7591042373 \\ 7 & 0.7447159200 \end{array}$$

So we can assume the rough approximation for the series to be:

$$\sum_{n=1}^\infty G_n\zeta(n+1) \approx \frac{3}{4}$$

---

Now we can try to get some analytical results.

Let us use an integral formula for the Gregory coefficients:

$$G_n=(-1)^{n+1} \int_0^\infty \frac{dx}{(1+x)^n (\ln^2 x+\pi^2)}$$

And the series definition for the Zeta function:

$$\zeta(n+1)=\sum_{k=1}^\infty \frac{1}{k^{n+1}}$$

Then the series becomes:

$$\sum_{n=1}^\infty G_n\zeta(n+1)=\int_0^\infty \frac{dx}{(\ln^2 x+\pi^2)} \sum_{n=1}^\infty \sum_{k=1}^\infty \frac{(-1)^{n+1}}{(1+x)^n}\frac{1}{k^{n+1}}$$

The double sum can be rearranged:

$$- \sum_{k=1}^\infty \frac{1}{k} \sum_{n=1}^\infty \frac{(-1)^n}{(1+x)^n k^n}=\sum_{k=1}^\infty \frac{1}{k} \frac{\frac{1}{(1+x) k}}{1+\frac{1}{(1+x) k}}=\sum_{k=1}^\infty \frac{1}{k} \frac{1}{1+(1+x)k}$$

So we have:

$$\sum_{n=1}^\infty G_n\zeta(n+1)=\sum_{k=1}^\infty \frac{1}{k(k+1)} \int_0^\infty \frac{dx}{(\ln^2 x+\pi^2)(1+ \frac{k}{k+1}x)}$$

For large $k$ the integral is:

$$\int_0^\infty \frac{dx}{(\ln^2 x+\pi^2)(1+ x)}=G_1=\frac{1}{2}$$

For $k=1,2,\dots$ the integral is slightly greater than $1/2$ (for $k=1$ it's about $0.88539$). Thus, a good bound from below for the series will be:

$$\sum_{n=1}^\infty G_n\zeta(n+1)> \frac{1}{2} \sum_{k=1}^\infty \frac{1}{k(k+1)}=\frac{1}{2}$$

But that's nothing new, since we already have a better approximation from the first few terms.

---

Let's get back to the sum in terms of $k$ and find its 'closed form':

$$\sum_{k=1}^\infty \frac{1}{k} \frac{1}{1+(1+x)k}=\psi \left( \frac{2+x}{1+x} \right)+\gamma$$

Where we have the digamma function and the Euler's constant.

Now the series becomes:

$$\sum_{n=1}^\infty G_n\zeta(n+1)=\int_0^\infty \left(\psi \left( \frac{2+x}{1+x} \right)+\gamma \right) \frac{dx}{\ln^2 x+\pi^2}$$

Note that we can't separate the integrals inside because they do not converge on their own.

---

Let's try another way. This time we will use the integral representation for the Zeta function as well:

$$\zeta(n+1)=\frac{1}{n!} \int_0^\infty \frac{y^n}{e^y-1} dy$$

Thus:

$$G_n \zeta(n+1)=\frac{1}{n!} \int_0^\infty \int_0^\infty \frac{(-1)^{n+1}y^n}{(1+x)^n}\frac{dx dy}{(e^y-1)(\ln^2 x+\pi^2)} $$

So:

$$\sum_{n=1}^\infty G_n\zeta(n+1)=\int_0^\infty \int_0^\infty \frac{dx dy}{(e^y-1)(\ln^2 x+\pi^2)} \sum_{n=1}^\infty \frac{1}{n!} \frac{(-1)^{n+1}y^n}{(1+x)^n}$$

The series inside is:

$$\sum_{n=1}^\infty \frac{1}{n!} \frac{(-1)^{n+1}y^n}{(1+x)^n}=1-\exp \left(-\frac{y}{1+x} \right)$$

So we have a double integral representation for our series:

$$\sum_{n=1}^\infty G_n\zeta(n+1)=\int_0^\infty \int_0^\infty \frac{1-\exp \left(-\frac{y}{1+x} \right)}{(e^y-1)(\ln^2 x+\pi^2)}dx dy$$

Or we can rewrite it by scaling $y$:

$$\sum_{n=1}^\infty G_n\zeta(n+1)=\int_0^\infty \int_0^\infty \frac{(1+x)(1-e^{-y})}{(e^{(1+x)y}-1)(\ln^2 x+\pi^2)}dx dy$$

So far no simple ways to integrate over $x$ or $y$ to get a single integral didn't come up. I'll see what I can do later.