Proof of "Possible new series for $\pi$" without use of physics

Question

_{Related post: The post Possible new series for $\pi$ is about whether the identity is new, so to avoid confusion I was advised to ask this question separately.}

I am looking for a proof of the following identity that does not require knowledge of physics: $$\pi = 4 + \sum_{n\ge1}\frac1{n!}\left(\frac1{n+\lambda}-\frac4{2n+1}\right)\left(\frac{(2n+1)^2}{4(n+\lambda)}-n\right)_{n-1}\tag1\label{474141_1}$$ for any $\operatorname{Re}\lambda>-1$ where $(x)_n= x(x+1)\cdots(x+n-1)$ is the Pochhammer symbol.

The current proof relies on various quantum field theory concepts. So far, only the case $\lambda=1/2$ is resolved, as the term inside the Pochhammer being independent of $n$ makes the calculations more straightforward.

It may help that \eqref{474141_1} can be rewritten in terms of fractional binomial coefficients: $$\pi=4-4\sum_{n\ge1}(-1)^n\left(\frac{2n}{4n^2-4\lambda+1}-\frac1{2n+1}\right)\binom{-\frac{4(1-\lambda)n+1}{4n+4\lambda}}n.\tag2\label{474141_2}$$

Answer 1

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. this paper.

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