Possible new series for $\pi$

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.

EDIT: Aninda Sinha asked essentially the same question on math.SE on March 8, 2024. See also the Numberphile interview of Sinha and Saha.
• You might also want to keep an eye on math.stackexchange.com/questions/4937730/… (which at least as of yet has no answers) ; I didn't ask it there but I'm also curious as to whether this formula is fundamentally new; I'll be keeping an eye on this too. Commented Jun 26 at 2:16
• Not quite an arbitrary complex number: it needs to avoid the poles at negative integers. Also worth noting (as they do in the paper) that the $\lambda \to \infty$ limit gives the Leibniz formula, which may be useful in literature searches. Commented Jun 26 at 10:40
• For a series that contains Pochhammer symbol of rational function of $n$ see Theorem 7.1 in this article link.springer.com/article/10.1023/A:1009809424076
– Nemo
Commented Jun 28 at 17:49
• @PeterTaylor I think one needs $\Re(\lambda)>-1$ for the series to converge. Commented Jun 29 at 9:43
• @Timothy_Chow, with $-s_1=\nu$ and $-s_2=1-\nu$ and $\nu\in\mathbb{Q}$ an infinite bi-parametric series for $\pi$ with the algebraic factor $\sin(\nu\pi)$ is got. $\nu=1/6$ gives another rational series. Since this is a sui generis Beta function expansion, a lemniscate constant series with $s_1=s_2=-1/4$ is got, as well as expressions for other trascendental constants like $\Gamma(1/3)$ with $s_1=s_2=-1/3$. The paper also brings a 2nd Beta function expansion with a different rational argument for Pochhamers. One can apply the same approach as above to get other infinite families of series Commented Jul 5 at 17:26

As it has a free parameter $$ฮป$$ ฮ have tried to prove it by the WZ method. However it seems that the function $$G(n, ฮป)$$ 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 $$ฮป \to \infty$$ we obtained a well-known alternating series with a better convergence.

• I am provisionally accepting this answer because it is probably the best answer we can hope for. Of course if someone eventually finds a very similar identity in the literature then I may have to change my mind. Commented Jun 28 at 15:42
• My understanding was that as $\lambda\to\infty$ you get the Madhava series $\sum(-1)^n\frac4{2n+1}$, which converges quite slowly (more slowly than the SahaโSinha series does for small values of $\lambda$). Is that wrong? Commented Jul 22 at 12:56

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 $$\pi$$): $$\sum_{n=1}^{\infty}(-1)^{n-1}{(2n-2)!\over n!(n-1)!}{x^n(x+\lambda)^n\over\lambda^{2n-1}}=x\tag{412}$$

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

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

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

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

• Yes Gerry, those are other nice examples of series involving a free parameter $ฮป$ that cannot be proved by the WZ method because the function $G(n, ฮป$ inside the summand is not hypergeometric in it's two symbols $n$ and $ฮป$. Commented Jun 26 at 10:16
• Such examples are ample as coming from Fourier series expansion. Commented Jun 26 at 13:42

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.

• Did you find any other (even as some bizarre hypergeometric functions) ? Commented Jun 26 at 13:48
• $$\frac{2x \sqrt{1-x^2} +8 \sin ^{-1}\left(\frac{\sqrt{1-\sqrt{1-x^2 }}}{\sqrt{2}}\right)-2 \sin ^{-1}(x)}{x}$$ Commented Jun 26 at 13:56
• @ClaudeLeibovici, the value $\lambda=\frac14$ also looks interesting because it allows some simplification: we get $$\frac{\pi}{4} = \sum_{n=0}^\infty \frac{- 1}{2n(2n+1)} \binom{\frac{4n^2}{4n+1}}{n}$$ Commented Jun 26 at 14:06
• @PeterTaylor. Thanks. What is interesting is thé diagonal part makes "most" of thé job. Cheers Commented Jun 26 at 14:44
• Thanks @TimothyChow and JesusGuillera, I have posted my question separately: mathoverflow.net/q/474141/113397 Commented Jun 29 at 0:39

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 $$j\ge3$$.

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 $$$$\label{pf}\frac{(t_1+t_2+1)_n}{(t_1)_{n+1}(t_2)_{n+1}}=\sum_{k=0}^n A_k\left(\frac{1}{t_1+k}+\frac 1{t_2+k}-\frac 1{\lambda+k}\right), \qquad (1)$$$$ where $$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.

• @JorgeZuniga: I just solve the first equation for $t_2$. It gives $$t_2=\frac{4\lambda-4t_1\lambda-1}{4(\lambda-t_1)}.$$ I then plug in $t_1=-k$ and $t_2=(4\lambda+4k\lambda-1)/4(\lambda+k)$ in $t_1+t_2$ and simplify. Commented Jul 2 at 16:22
• @Hjalmar_Rosengren. Ok, Thanks a lot. Commented Jul 2 at 16:41

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.

A related, and perhaps easier, question is whether there are other known series for ๐ that involve a complex parameter ๐ 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.

• this might be a stupid question (it is far from what I do). But if you take $g$ to be a step function and approximate it with smooth functions and then take the limit shouldn't you get a sum? Then couldn't you increase the number steps and get in the limit a series? Though, I am not sure this is interesting, Commented Jul 20 at 11:56
• @YiftachBarnea a limit of sums is not always a sum. In this case itโll be an integral. By the way, the integration identity is proved precisely by starting with a discrete summation identity and taking a limit more or less as you suggest. Commented Jul 20 at 15:47
• Thanks, I guess the intuition was right, but its direction was wrong. :-) Commented Jul 20 at 16:01
• In second thought, here is a slightly less stupid idea. Fix g, then integrate from 0 to 1/2, then from 1/2 to 3/4, then from 3/4 to 7/8, and so on. You get a series that depends on g. Again, no idea if it is interesting. Commented Jul 20 at 16:22

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 $$0 and $$\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.