Justify an approximation of $\sum_{n=1}^\infty G_n/\binom{\frac{n}{2}+\frac{1}{2}}{\frac{n}{2}}$, where $G_n$ denotes the Gregory coefficients

Question

While I was playing with an integral posted in Math Oveflow, I wondered a different question using a CAS, and one of those answers posted, see [1]. My combination is to create a series involving the so-called Gregory coefficients $G_n$, see this Wikipedia.

Question. Is it possible to justify an (a reasonably good) approximation of $$I=\sum_{n=1}^\infty\frac{G_n}{\binom{\frac{n}{2}+\frac{1}{2}}{\frac{n}{2}}}\,?$$ Thanks in advance.

Thus that I am asking is how one can to justify an approxiation of $I$ using analysis or analytic number theory. There are no more context for this question, only the idea about how to get such good approximation that I evoke (I can get the approximation $I\approx 0.35305$, but I don't know how justify it).

References:

[1] The answer from T. Amdeberhan of A curious sin-integral, MathOverflow (2017).

Answer 1

Here are some results on expressing this series through hypergeometric functions.

First, we rewrite the sum using Gamma functions:

$$I=\frac{\sqrt{\pi}}{2} \sum_{n=1}^\infty \frac{\Gamma \left( \frac{n}{2}+1 \right)}{\Gamma \left( \frac{n}{2}+\frac{3}{2} \right)} G_n=\frac{\sqrt{\pi}}{2} \sum_{n=1}^\infty (-1)^{n+1} \frac{\Gamma \left( \frac{n}{2}+1 \right)}{\Gamma \left( \frac{n}{2}+\frac{3}{2} \right)} |G_n|$$

Here we also use the fact that $G_n$ have alternating signs.

Now the main idea is to separate even and odd terms, which is necessary to be able to use hypergeometric functions. Let's introduce two new series:

$$S_1=\sum_{k=1}^\infty \frac{\Gamma \left(k+ \frac{1}{2} \right)}{\Gamma \left( k+1 \right)} G_{2k-1} >0$$

$$S_2=\sum_{k=1}^\infty \frac{\Gamma \left(k+ 1 \right)}{\Gamma \left( k+\frac{3}{2} \right)} G_{2k} <0$$

$$I= \frac{\sqrt{\pi}}{2} (S_1+S_2)$$

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Let us use one of the integral expressions for $G_n$:

$$G_n=\int_0^1 \binom{x}{n} dx=\int_0^1 \frac{\Gamma(x+1)}{n! \Gamma(x-n+1)}dx$$

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For $S_1$ we have:

$$S_1=\int_0^1 \Gamma(x+1) dx \sum_{k=1}^\infty \frac{\Gamma \left(k+ \frac{1}{2} \right)}{\Gamma \left( k+1 \right) \Gamma(2k) \Gamma(x-2k+2)}$$

Let's call the inner sum $Q_1$ and change the index $k \to k+1$, then:

$$Q_1=\sum_{k=0}^\infty \frac{\Gamma \left(k+ \frac{3}{2} \right)}{\Gamma \left( k+2 \right) \Gamma(2k+2) \Gamma(x-2k)}$$

To find the hypergeometric form of this series, we should consider the ratio of general terms:

$$t_0=\frac{\Gamma \left(\frac{3}{2} \right)}{\Gamma (x)}$$

$$\frac{t_{k+1}}{t_k}=\frac{\left(k+ \frac{3}{2} \right) (x-2k-1)(x-2k-2)}{(k+2)(2k+3)(2k+2)}=\frac{\left(k+ \frac{1}{2}- \frac{x}{2} \right) \left(k+ 1- \frac{x}{2} \right)}{(k+2)} \frac{1}{k+1}$$

By definition this means:

$$Q_1=\frac{\Gamma \left(\frac{3}{2} \right)}{\Gamma (x)} {_2 F_1} \left(\frac{1}{2}- \frac{x}{2},1- \frac{x}{2};2;1 \right)$$

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For $S_2$ we have:

$$S_2=\int_0^1 \Gamma(x+1) dx \sum_{k=1}^\infty \frac{\Gamma \left(k+ 1 \right)}{\Gamma \left( k+\frac{3}{2} \right) \Gamma(2k+1) \Gamma(x-2k+1)}$$

Let's call the inner sum $Q_2$ and change the index $k \to k+1$, then:

$$Q_2=\sum_{k=0}^\infty \frac{\Gamma \left(k+ 2 \right)}{\Gamma \left( k+\frac{5}{2} \right) \Gamma(2k+3) \Gamma(x-2k-1)}$$

To find the hypergeometric form of this series, we should consider the ratio of general terms:

$$t_0=\frac{1}{2 \Gamma \left(\frac{5}{2} \right) \Gamma (x-1)}$$

$$\frac{t_{k+1}}{t_k}=\frac{\left(k+ 2\right) (x-2k-2)(x-2k-3)}{\left(k+\frac{5}{2} \right)(2k+4)(2k+3)}=\frac{(k+1)\left(k+ \frac{3}{2}- \frac{x}{2} \right) \left(k+ 1- \frac{x}{2} \right)}{\left(k+\frac{3}{2} \right) \left(k+\frac{5}{2} \right)} \frac{1}{k+1}$$

By definition this means:

$$Q_2=\frac{1}{2 \Gamma \left(\frac{5}{2} \right) \Gamma (x-1)} {_3 F_2} \left(1, \frac{3}{2}- \frac{x}{2},1- \frac{x}{2};\frac{3}{2},\frac{5}{2};1 \right)$$

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Now let us collect the results (with some simple transforms based on Gamma function properties):

$$S_1=\frac{\sqrt{\pi}}{2} \int_0^1 x ~{_2 F_1} \left(\frac{1}{2}- \frac{x}{2},1- \frac{x}{2};2;1 \right) dx $$

$$S_2=- \frac{2}{3 \sqrt{\pi}} \int_0^1 x(1-x) ~{_3 F_2} \left(1, \frac{3}{2}- \frac{x}{2},1- \frac{x}{2};\frac{3}{2},\frac{5}{2};1 \right) dx $$

$$I= \frac{\sqrt{\pi}}{2} (S_1+S_2)$$

$$I=\frac{\pi}{4} \int_0^1 x ~{_2 F_1} \left(\frac{1}{2}- \frac{x}{2},1- \frac{x}{2};2;1 \right) dx- \\ - \frac{1}{3 } \int_0^1 x(1-x) ~{_3 F_2} \left(1, \frac{3}{2}- \frac{x}{2},1- \frac{x}{2};\frac{3}{2},\frac{5}{2};1 \right) dx$$

The numerical values of the integrals, computed by Mathematica with working precision set to 100, lead to:

$$I = 0.353056313931150786257841734863968642813 \dots$$

Which is in good agreement with the value proposed by the OP.

The original sum computed for $996-999$ terms leads to:

$$I_{996}=\color{blue}{0.35305}5994793\dots \\ I_{997}=\color{blue}{0.353056}632513\dots \\ I_{998}=\color{blue}{0.35305}5995902\dots \\ I_{999}=\color{blue}{0.353056}631407\dots$$

The arithmetic mean of succesive even and odd terms gives better approximation, but not by much.

So the convergence is really bad. I suspect it could be accelerated by the usual techniques (see https://en.wikipedia.org/wiki/Series_acceleration), meanwhile the integrals provide the accurate value.

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Note: Using another integral formula:

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

leads to integrals of elementary functions for $S_1$ and $S_2$. However, the convergence of the integrals is bad (or at least Mathematica has trouble computing them numerically) and so I haven't been able to confirm the numerical value with sufficient accuracy. So I invite anyone else to check this other way to obtain a better closed form.