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 (https://link.springer.com/article/10.1023/A:1009809424076 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.

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.

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 (https://link.springer.com/article/10.1023/A:1009809424076 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 $t_1\rightarrow\infty$, $t_2\rightarrow z$ (and vice versa). Writing $$a_k=t_1+t_2\Big|_{t_1=-k}=1-\frac{(2k+1)^2}{4(z+k)}, $$ 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.

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 (https://link.springer.com/article/10.1023/A:1009809424076 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.