_{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: $$\pi = 4 + \sum_{n\ge1}\frac1{n!}\left(\frac1{n+\lambda}-\frac4{2n+1}\right)\left(\frac{(2n+1)^2}{4(n+\lambda)}-n\right)_{n-1}\tag1$$ 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 $(1)$ can be rewritten in terms of fractional binomial coefficients: $$\pi=4-4\sum_{n\ge1}(-1)^n\left(\frac{2n}{4n^2-4\lambda+1}-\frac1{2n+1}\right)\binom{-\frac{4(1-\lambda)n+1}{4n+4\lambda}}n.\tag2$$

_{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: $$\pi=4-4\sum_{n\ge1}(-1)^n\left(\frac{2n}{4n^2-4\lambda+1}-\frac1{2n+1}\right)\binom{-\frac{4(1-\lambda)n+1}{4n+4\lambda}}n.\tag2\label{474141_2}$$