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