# 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.

• I encourage to everyone to read the related Wikipedia dedicated to James Gregory. This numbers seem to me interesting.
– user243301
Feb 21 '18 at 19:55
• Since the asymptotics are in the Wikipedia, I am interested in the strategy to get our approximation, more than the explicit calculations. Many thanks.
– user243301
Feb 21 '18 at 19:56
• You could use the table of Gregory coefficients and the known values of the zeta function to compute the first $5-10$ terms and get some bounds for the sum. Remember that since Gregory coefficients have alternating signs, even and odd terms give you bounds for the limit Feb 21 '18 at 20:06
• Yours are a good remarks, I forgot that the sequence of Gregory coefficients has alternating signs. I am waiting if some user want provide details of your strategy. Any case I take notes of your reasoning in my notebook and thanks @YuriyS
– user243301
Feb 21 '18 at 20:14

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.

• I have already taken a look at all your elegant approaches, and tomorrow I will read them in detail, many thanks for your attention and help.
– user243301
Feb 21 '18 at 22:56
• @user243301, I would advise you to wait a bit before accepting answers. Wait at least a week. I haven't got anything useful so far for computing the series, and if you already have an accepted answer, some other users might feel discouraged from attempting their own answers Feb 22 '18 at 11:04
• I'm sorry, you're right, now I've check the answer it as not accepted, thus my apologizes. I've read your answer and has itself an approximation (your comment about that our series is alternating) and your formulas have a great beauty. I am waiting if some user in next weeks do more feedback, but your answer seems unbeatable.
– user243301
Feb 22 '18 at 11:09
• @user243301, that's ok. I would advise you to attempt some things on your own in the future. At least as far as computing the first few terms of a series (you can use Wolfram Alpha for example or some other tool). While you have great many interesting question here, it's always more satisfying to discover something on your own Feb 22 '18 at 11:11