What is the name of this sequence that approximates $\pi$?

Question

I came across this sequence $$\sum_{n=1}^{\infty}\frac{2^n}{2n\choose n}n^m=a_{m}+b_{m}\pi$$ where $a_m$ and $b_m$ are rational with $\lim_{m\to\infty}\frac{a_m}{b_m}=\pi$. I cannot figure out where it comes from or if it has a name. Any ideas?

Answer 1

Write it as $$S_m=\sum_{n=1}^{\infty}\frac{2^n}{2n\choose n}n^m=a_{m}+\frac \pi 2 \, b_{m}$$

The first $a_m$ are $$\{1,3,11,55,355,2807,26259,283623,3473315,47552791,719718067\}$$ and this is sequence $A180875$ in $OEIS$.

Quoting $OEIS$, these are "Lehmer's coefficients stemming from an inverse binomial coefficient series"

The first $b_m$ are $$\{1,2,7,35,226,1787,16717,180560,2211181,30273047,458186752\}$$ and this is sequence $A014307$ in $OEIS$.

If you look in the formula sections, you will find approximations given in year $2014$ by Vaclav Kotesovec, namely $$a_m \sim \frac {\pi m^{m+1} }{\sqrt{2} e^m \log ^{m+\frac{3}{2}}(2) }\qquad \qquad b_m\sim\frac{\sqrt{2} (m+1)^{m+1}}{e^{m+1} \log ^{m+\frac{3}{2}}(2) }$$

So, using these approximations $$2\frac { a_m} {b_m}\sim \pi \,e \,\left(\frac m {m+1} \right)^{m+1}$$ which is not good for an estimation of the error.

For $0 \leq m \leq 15$, a quick and dirty linear regression gives $(R^2=0.979)$ gives $$\log \left(\left| 2\frac{ a_m}{b_m}-\pi \right| \right)=-2.14718 m-1.37498$$ Increasing the range of $m$ and searching for nice looking numbers in inverse symbolic calculators $$\left| 2\frac{ a_m}{b_m}-\pi \right| \sim e^{-e/2} \left(\frac{\log (2)}{2 \pi }\right)^m$$

Edit

Formally, in terms of the generalized hypergeometric function $$S_m=\, _{(m+1)}F_m\left(2,\cdots,2;1,\cdots,1,\frac{3}{2};\frac{1}{2}\right)$$

Answer 2

A little internet search led me to this page. Here someone plays a little around with some series represantion of trigfunctions using the central binomial coefficient:

$$ C = \binom{2n}{n} = \frac{(2n)!}{(n!)^2}$$

Note: These are called central binomial coefficients because they form the central column in Pascal's triangle.

After some playful integration and other mathematical magic one can show by differentiating $u \frac{d}{du} \arcsin^2(\sqrt u)$ that:

$$\pi = -2 + \sum_{n=1}^\infty\frac{2^{n+1}}{C}$$

As far as I can tell the formula in question, i.e.

$$ \forall k\geq0, \sum_{n=1}^{\infty} n^k \frac{2^n}{C} = a_k \pi + b_k, \text{where } \frac{b_k}{a_k} \rightarrow \pi$$

is this an open conjecture (Conjectur 16 here). I personally would assume the arrow stans for "as $k \rightarrow \infty$", yet i am not really sure. And apparently this conjecture stems from the mind of Géry Huvent, a french mathematician it seems to me. (Huvent on Research Gate)

Since his publications are written in french I personally know only very, very basic french, I wasn't able to find a publication which mentions this conjecture or how he had come up with it. But maybe someone more skilled in the french language can find such a paper...

Or his (I think) personal webpage, which is sadly also in french, could be of assistance.

Answer 3

Too long for a comment.

In comments, @JeanMarie gave a link to this answer from @Felix Marin

$$S_1=\sum_{n = 1}^{\infty}{n\,x^{n} \over {2n \choose n}}=x\,\frac{6 \sqrt{(4-x) x}+4(x+2)\tan ^{-1}\left(\sqrt{\frac{x}{4-x}}\right) } { (x-4)^2 \sqrt{(4-x) x}}=f(x)$$ which can also write $$f(x)=\frac{6 x}{(x-4)^2}+\frac{4 (x+2) \sqrt{x} }{(4-x)^{5/2}}\,\csc ^{-1}\left(\frac{2}{\sqrt{x}}\right)$$

If we differentiate both side with respect to $x$ $$S_2=\sum_{n = 1}^{\infty}{n^2\,x^{n} \over {2n \choose n}}=x\,f'(x)$$ $$S_3=\sum_{n = 1}^{\infty}{n^3\,x^{n} \over {2n \choose n}}=x \big[x\,f'(x)]'$$ $$S_4=\sum_{n = 1}^{\infty}{n^4\,x^{n} \over {2n \choose n}}=x\Big[x \big[x\,f'(x)]'\Big]'$$

Making $x=2$ gives the results in the question. But, we can play with the value of $x$ and, as shown here, obtain the general result

$$S_m=\sum_{n = 1}^{\infty}{n^m\,x^{n} \over {2n \choose n}}=R_{3,m}+R_{4,m}\,\,\sqrt{\frac{x}{4-x}}\,\, \sin ^{-1}\left(\frac{\sqrt{x}}{2}\right)$$ where $R_{3,m}$ and $R_{4,m}$ are rational numbers.

Have a look here for the final version of the paper by Dyson, Frankel and Glasser.

Edit

We could even start with $$S_0=\sum_{n = 1}^{\infty}{x^{n} \over {2n \choose n}}=\frac x{4-x}+\frac{4 \sqrt{x}}{(4-x)^{3/2}}\sin ^{-1}\left(\frac{\sqrt{x}}{2}\right)$$ and since $$S_m= x S'_{m-1}$$ we can generate similar sequence for several values of $x$ using the particular values of the sine function.