Possible new series for $\pi$

Question

In a recent (unfortunately over-hyped) preprint by Saha and Sinha, Field theory expansions of string theory amplitudes (arXiv:2401.05733), they present the following series for $\pi$: $$\pi = 4 + \sum_{n=1}^\infty {1\over n!} \biggl({1\over n+\lambda} - {4\over 2n+1}\biggr)\biggl({(2n+1)^2 \over 4(n+\lambda)} - n \biggr)_{n-1},$$ where $\lambda$ is an arbitrary complex number and the Pochhammer symbol $(x)_n := x(x+1)\cdots(x+n-1)$.

Setting aside the unfortunate press coverage, as well as the question of the significance of this formula for $\pi$, what I'm wondering is whether it is new. The authors are physicists, and searching the literature for this type of thing is not always easy, so I figured that MathOverflow would be a natural place to ask.

A related, and perhaps easier, question is whether there are other known series for $\pi$ that involve a complex parameter $\lambda$ in the summand, but where the sum of the series is independent of the value of the parameter.

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EDIT: Aninda Sinha asked essentially the same question on math.SE on March 8, 2024. See also the Numberphile interview of Sinha and Saha.

Answer 1

As it has a free parameter $λ$ Ι have tried to prove it by the WZ method. However it seems that the function $G(n, λ)$ inside the summation is not hypergeometric in its two symbols $n$ and $\lambda$; that is, $G(n+1,\lambda)/G(n,\lambda)$ and $G(n,\lambda+1)/G(n,\lambda)$ are not rational functions.

I have not seen this formula for $\pi$ before in the literature despite that I know many other formulas. Although the series is not good for computing $\pi$ (contrary to what is stated in the press), it could have some interest. In the paper there is also a formula of the same style for $\zeta(2)$.Taking the limit of it as $λ \to \infty$ we obtained a well-known alternating series with a better convergence.

Answer 2

Jolley, Summation of Series, contains many "series that involve a parameter $\lambda$ in the summand, but where the sum of the series is independent of the value of the parameter." Here are just a few (the sum isn't generally $\pi$): $$ \sum_{n=1}^{\infty}(-1)^{n-1}{(2n-2)!\over n!(n-1)!}{x^n(x+\lambda)^n\over\lambda^{2n-1}}=x\tag{412} $$

$$ \sum_{k=1}^{\lambda}\cos{(2k-1)\pi\over2\lambda+1}={1\over2}\tag{424} $$

$$ \sum_{k=1}^{(n-1)/2}{1\over\sin^2k\lambda}={n^2-1\over6}{\rm\ where\ }n{\rm\ is\ odd}\tag{439} $$

$$ \sum_{n=1}^{\infty}{(-1)^{n-1}\over2n-1}\cos(2n-1)\lambda={\pi\over4} {\rm\ where\ }-{\pi\over2}<\lambda<{\pi\over2}\tag{506} $$

$$ \sum_{n=0}^{\infty}{\sin(n+{1\over2})\lambda\over n+{1\over2}}={\pi\over2}{\rm\ where\ }0<\lambda<\pi\tag{550} $$

Answer 3

This is just a note that the case $\lambda=1/2$ is nothing more than the arcsine representation of $\pi$.

In this case, the identity becomes $$\pi=4+\sum_{n\ge1}\frac1{n!}\left(-\frac2{2n+1}\right)\left(\frac12\right)_{n-1}$$ or equivalently, \begin{align}\pi&=4-2\sum_{n\ge0}\frac1{4^n(n+1)(2n+3)}\binom{2n}n\\&=4-2\left(\sum_{n\ge0}\frac1{4^n(n+1)}\binom{2n}n-2\sum_{n\ge0}\frac1{4^n(2n+3)}\binom{2n}n\right).\end{align} Using $(1-x)^{-1/2}=\sum_{n\ge0}(x/4)^n\binom{2n}n$, we have \begin{align}\sum_{n\ge0}\frac1{4^n(n+1)}\binom{2n}n&=\int_0^1\frac1{\sqrt{1-x}}\,dx=2\\\sum_{n\ge0}\frac1{4^n(2n+3)}\binom{2n}n&=\int_0^1\frac{x^2}{\sqrt{1-x^2}}\,dx=\frac\pi4\end{align} so the identity is true.

Answer 4

This is NOT an answer. This type of series involving Pochammer symbols $(a)_n$ with $a$ a rational function of $n$ is totally new to me. Reading the paper, I summarize below what I extracted (and simplified) from it:

(1) For $p\in{\mathbb Z}_{\ge-1}$ we have $$\dfrac{\Gamma(s_1)\Gamma(s_2)}{\Gamma(s_1+s_2-p)}=(-1)^p\sum_{n\ge0}\dfrac{1}{n!}\left(\dfrac{1}{n+s_1}+\dfrac{1}{n+s_2}-\dfrac{1}{n+\lambda}\right)\left(1-\lambda+\dfrac{(\lambda-\ s_1)(\lambda-s_2)}{n+\lambda}\right)_{n+p}\;.$$ The formula for $\pi$ follows by choosing $s_1=s_2=1/2$ and $p=-1$, but in my opinion the formula with $p=0$ is slightly more elegant.

(2) $$\psi(1+x)+\gamma=x\sum_{n\ge1}\dfrac{1}{n!}\left(\dfrac{2n+x}{n^2+nx}-\dfrac{\ 1}{n+\lambda}\right)\left(1-\dfrac{\lambda(x+n)}{n+\lambda}\right)_{n-1}\;.$$

Choosing $x=-1/2$ gives a formula for $\log(2)$ similar (but simpler) to that for $\pi$.

(3) $$\zeta(2)=-\sum_{n\ge1}\dfrac{(n+2\lambda)}{\lambda n^2n!}\left(-\dfrac{n\lambda}{n+\lambda}\right)_n\;.$$

as well as similar but more complicated formulas for $\zeta(j)$ for $j\ge3$.

Answer 5

This is an answer to the related post Proof of "Possible new series for $\pi$" without use of physics but it seems relevant also for this question. My guess is that this formula is new as it stands, but it follows easily from classical results.

I found an elementary proof using partial fractions. This is motivated by Nemo's observation that very similar-looking series appear in the work of Schlosser (Section 7). The matrix inversion used by Schlosser is closely related to partial fractions, see e.g. https://www.arxiv.org/abs/math/0309358.

Fix $\lambda$ and let $t_1$ and $t_2$ be variables satisfying $$t_1+t_2-\frac{t_1t_2}{\lambda}=1-\frac{1}{4\lambda}. $$ By elementary facts about partial fraction expansions, we can write \begin{equation}\label{pf}\frac{(t_1+t_2+1)_n}{(t_1)_{n+1}(t_2)_{n+1}}=\sum_{k=0}^n A_k\left(\frac{1}{t_1+k}+\frac 1{t_2+k}-\frac 1{\lambda+k}\right), \qquad (1)\end{equation} where $$A_k= \frac{(t_1+k)(t_1+t_2+1)_n}{(t_1)_{n+1}(t_2)_{n+1}}\Bigg|_{t_1=-k}.$$ The last term in (1) makes each term vanish in the limit $t_1\rightarrow\infty$, $t_2\rightarrow \lambda$ (and vice versa). Writing $$a_k=t_1+t_2\Big|_{t_1=-k}=1-\frac{(2k+1)^2}{4(\lambda+k)}, $$ this can be simplified to $$A_k=\frac{(a_k+1)_{k-1}(-1)^k}{k!(n-k)!(a_k+n+1)_{k}}. $$

Specializing $t_1=t_2=1/2$ in (1) gives $$\frac{(2)_n}{(1/2)_{n+1}^2}=\sum_{k=0}^n \frac{(a_k+1)_{k-1}(-1)^k}{k!(n-k)!(a_k+n+1)_{k}}\left(\frac{4}{2k+1}-\frac 1{\lambda+k}\right). $$ We now multiply both sides by $n!$ and let $n\rightarrow\infty$. The left-hand side can be written $$\frac{\Gamma(n+2)\Gamma(n+1)}{\Gamma(n+3/2)^2}\cdot\frac{\Gamma(1/2)^2}{\Gamma(2)}\rightarrow \frac{\Gamma(1/2)^2}{\Gamma(2)}=\pi.$$ On the right, we have the factor $$\frac{n!}{(n-k)!(a_k+n+1)_{k}}=(-1)^{k}\frac{(-n)_{k}}{(a_k+n+1)_{k}}\rightarrow 1. $$ This gives (it should not be hard to justify taking the limit) \begin{align*}\pi&=\sum_{k=0}^\infty\frac{(-1)^k}{k!}\left(\frac{4}{2k+1}-\frac 1{\lambda+k}\right)\left(2-\frac{(2k+1)^2}{4(\lambda+k)}\right)_{k-1}\\ &=\sum_{k=0}^\infty\frac{1}{k!}\left(\frac 1{\lambda+k}-\frac{4}{2k+1}\right)\left(\frac{(2k+1)^2}{4(\lambda+k)}-k\right)_{k-1} , \end{align*} which is the desired identity.

More generally, one could start with variables satisfying $$t_1+t_2-\frac{t_1t_2}{\lambda}=x+y-\frac{xy}{\lambda}, $$ take the partial fraction expansion of $$\frac{(t_1+t_2-p)_n}{(t_1)_{n+1}(t_2)_{n+1}}, $$ specialize $t_1=x$, $t_2=y$ and finally let $n\rightarrow\infty$. I checked that this gives equation (4) in the paper by Sana and Sinha (where $x=\alpha-s_1$, $y=\alpha-s_2$). Presumably other identities in their paper follow by further variations of the same argument.

Answer 6

This is a non-answer-followup to the non-answer of @HenriCohen:

Noting that $$\tag{*}\label{eq:*} 1-\lambda+\frac{(\lambda-s_1)(\lambda-s_2)}{n+\lambda}= \frac{(n+s_1) (n+s_2)}{n+\lambda} -n-s_1-s_2+1$$ and (note the big confusion regarding the notation of factorial powers) using the (falling) factorial power $a^{(n)}=a(a-1)\cdots(a-n+1)$ instead of the Pochhammer symbol aka. rising factorial power $(a)_{n}=a(a+1)\cdots(a+n-1)$, Henri's first equation simplifies to \begin{align} \frac{\Gamma(s_1)\Gamma(s_2)}{\Gamma(s_1+s_2-p)} =(-1)^p \sum_{n\ge 0}\frac{1}{n!}&\left(\frac{1}{n+s_1}+\frac{1}{n+s_2}-\frac{1}{n+\lambda}\right) \times{}\\ \tag{1}\label{eq:1} &{}\times\left(\frac{(n+s_1) (n+s_2)}{n+\lambda} -(1+s_1+s_2)\right)^{(n+p)}. \end{align} Note that \begin{align}\tag{**}\label{eq:**} \frac{\partial}{\partial n}\ln\frac{(n+s_1) (n+s_2)}{n+\lambda} = \frac{1}{n+s_1}+\frac{1}{n+s_2}-\frac{1}{n+\lambda}. \end{align} Similarly, Henri's second equation can be written as (I use $s_2$ instead of $x$ and add some zeroes) \begin{align} \frac{\psi(1+s_2)+\gamma}{s_2}= \frac{H_{s_2}}{s_2}= \sum_{n\ge 1}\frac{1}{n!}&\left(\frac{1}{n+0}+\frac{1}{n+s_2}-\frac{1}{n+\lambda}\right)\times{}\\ \tag{2}\label{eq:2} &{}\times\left(\frac{(n+0)(n+s_2)}{n+\lambda} -(1+0+s_2)\right)^{(n-1)}, \end{align} with the harmonic numbers $H_n$ (analytically continued).

The two equations are related through a Laurent series expansion of \eqref{eq:1} at $p=-1$ around $s_1=0$ according to \begin{align}\tag{3}\label{eq:3} -\frac{\Gamma(s_1)\Gamma(s_2)}{\Gamma(s_1+s_2+1)} = -\frac{1}{s_1 s_2}+\frac{H_{s_2}}{s_2} + \mathcal{O}(s_1), \end{align} where the divergent term $\frac{-1}{s_1 s_2}$ is equal to the $n=0$ term in \eqref{eq:1}. The constant term at $s_1=0$ in this expansion is the rhs of \eqref{eq:2}.

Finally, Henri's third equation is given by the limit $s_2\to0$ of \eqref{eq:2}.

Maybe, equation \eqref{eq:**} sheds some light onto the discussion.

Answer 7

A related, and perhaps easier, question is whether there are other known series for 𝜋 that involve a complex parameter 𝜆 in the summand, but where the sum of the series is independent of the value of the parameter.

I have something that more or less qualifies if we allow a slightly loose interpretation of the question. In my 2004 paper "Explicit formulas for hook walks on continual Young diagrams" I proved the following identity: $$ \pi = \int_0^1 \left[ \cos\left(\frac{\pi g(x)}{2}\right) x^{-(1+g(x))/2} (1-x)^{-(1-g(x))/2} \exp\left( \frac12 \int_0^1 \frac{g(u)-g(x)}{u-x} du \right) \right] dx. $$ (See the second page of the Introduction.) Here, the "parameter" is an arbitrary smooth function $g:[0,1]\to(-1,1)$, not a single complex number. And the expression is not a sum, it's an integral. But it is a formula for $\pi$, and it's valid independently of the value of the parameter.

Answer 8

This answers the final question about other parametric similar series for $\pi$. Working on the Saha & Sinha paper, it is possible to get a bi-parametric generalization of a related formula. If $s\in\mathbb{Q}$ with $0<s<1$ and $\lambda\notin\mathbb{Z}_{\le0}$, then

$$ \pi=\sin\pi s\cdot\sum_{n=0}^\infty\frac{1}{n!}\cdot\left(\frac{1}{n+s}+\frac{1}{n+1-s}-\frac{1}{n+\lambda}\right)\cdot\left(1-\lambda+\frac{(\lambda-s)(\lambda-1+s)}{\lambda+n}\right)_n $$ which is a logarithmically behaved series that can be summed up using some efficient acceleration method. For instance, H. Cohen's $\Delta$-Euler-MacLaurin. Note that this is a rational expression just for $s=\frac{1}{2},\frac{1}{6},\frac{5}{6}$. For other rational $s$ values $\sin\pi s$ gives an algebraic factor.