Solve $\sum_\limits{n=-\infty}^0\mathrm P_n^n(z)$ with Associated Legendre P functions of type 1

Question

Here is a simple looking sum which should have an alternate form since it is just a double hypergeometric series with the associated Legendre P function of type 1 $\mathrm P_a^b(z)$

The definitions for the Legendre P function of the first kind uses a regularized Gauss hypergeometric function, Pochhammer symbol, and incomplete beta function:

$$
\begin{split}
\mathrm P_v^u(z)=\left(\frac{1+z}{1-z}\right)^\frac u2\, & _2\tilde{ \mathrm F}_1\left(-v,v+1;1-u;\frac{1-z}2\right)\\
&\Downarrow\\
\sum_{n=0}^\infty \mathrm P_{-n}^{-n}(z)=&\sum_{n=0}^\infty \frac{(1+z)^\frac n2\,_2\mathrm F_1\left(1-n,n;n+1;\frac{1-z}2\right)}{(1-z)^\frac n2 n!}\\
=&\sum_{n=0}^\infty \frac{(1+z)^\frac n2}{(1-z)^\frac n2 } \sum_{k=0}^\infty \frac{(1-n)_k(n)_k(1-z)^k}{2^k\Gamma(k+n+1)k!} \\
= &\frac1\pi\sum_{n=0}^\infty \sum_{k=0}^\infty \frac{\sin(\pi n)(k-n)!(1-z)^{k-\frac n2} (1+z)^\frac n2 }{2^k(k+n)k!} \\
=&\sum_{n=0}^\infty \frac{2^n(1-z^2)^\frac n2 \mathrm B_\frac{1-z}2(n,n)}{\Gamma(n)}
\end{split}$$

Here is an integral representation:

$$
\begin{split}
\mathrm B_z(a,b) = &\int_0^z t^{a-1}(1-t)^{b-1} dt\\
& =\int_{z^{-1}}^\infty t^{-a-b}(t-1)^{b-1} dt\\
&\overset{|z\in\Bbb R|<\infty}{=}\int_0^\infty (t+z^{-1})^{-a-b}(t+z^{-1}-1)^{1-b} dt \\
\implies &\mathrm B_z(a,a) = \int_0^\infty (t+z^{-1})^{-2a}(t+z^{-1}-1)^{1-a} dt
\end{split}
$$
Therefore we may or may not have:
$$
\begin{split}
\sum_{n=0}^\infty \mathrm P^{-n}_{-n}(z) &=\sum_{n=0}^\infty \frac{2^n(1-z^2)^\frac n2 \mathrm B_\frac{1-z}2(n,n)}{\Gamma(n)}\\
& = \sum_{n=0}^\infty \frac{2^n(1-z^2)^\frac n2 \int_0^\infty \left(t+ \frac 2{1-z}\right)^{-2n}\left(t+\frac 2{1-z}-1\right)^{1-n} dt }{\Gamma(n)}\\
& =\int_0^\infty \frac{2e^\frac{2\sqrt{1-z^2}}{\left(t+\frac2{1-z}\right)^2\left(t+\frac2{1-z}-1\right)}\sqrt{1-z^2}}{\left(t+\frac2{1-z}\right)^2}dt
\end{split}
$$
We cannot use an Appell Series since there is a negative coefficient of the sum index in the gamma function argument. Here is an approximated plot:

$\sum\limits_{n=-\infty}^0 \mathrm P_n^n(z)$:

There is a discontinuity at $z=1$. The sum terms themselves are just rational functions of $z$. What is an evaluation of the sum? Please correct me and give me feedback.

Answer 1

$$\sum_{n=0}^\infty \text P_{-n}^{-n}(z)=\sum_{n=0}^\infty \frac{(1+z)^\frac n2\,_2\text F_1\left(1-n,n;n+1;\frac{1-z}2\right)}{(1-z)^\frac n2 n!}$$

I think this should be $$\sum_{n=0}^\infty \text P_{-n}^{-n}(z)=\sum_{n=0}^\infty \frac{(1\color{red}-z)^\frac n2\,_2\text F_1\left(1-n,n;n+1;\frac{1-z}2\right)}{(1\color{red}+z)^\frac n2 n!}$$

From here, we have

$$\begin{align}\sum_{n=0}^\infty \text P_{-n}^{-n}(z)&=1+\sum_{n=1}^\infty \frac{2^n \text B_\frac{1-z}2(n,n)}{(n-1)!(1-z^2)^{n/2}} \\\\&=1+\sum_{n=1}^\infty \frac{2^n}{(n-1)!(1-z^2)^{n/2}}\int_{0}^{(1-z)/2}t^{n-1}(1-t)^{n-1}dt \\\\&=1+\int_{0}^{(1-z)/2}\frac{1}{t(1-t)}\sum_{n=1}^\infty \frac{1}{(n-1)!}\bigg(\frac{2t(1-t)}{\sqrt{1-z^2}}\bigg)^n dt \\\\&\color{red}{=}1+\int_{0}^{(1-z)/2}\frac{1}{t(1-t)}\cdot\frac{2t(1-t)}{\sqrt{1-z^2}}e^{2t(1-t)/\sqrt{1-z^2}} dt \\\\&=1+\frac{2}{\sqrt{1-z^2}}\int_{0}^{(1-z)/2}\bigg(e^{2/\sqrt{1-z^2}}\bigg)^{t(1-t)} dt\end{align}$$ where $\displaystyle\sum_{n=1}^{\infty}\dfrac{x^n}{(n-1)!}=xe^x$ was used in the equality in red.

WolframAlpha says $$\int a^{t(1-t)}dt=\frac{\sqrt{\pi}\ a^{1/4}\operatorname{erf}\bigg(\frac 12 (2 t - 1) \sqrt{\ln a}\bigg)}{2\sqrt{\ln a}}+C$$

So, we finally get $$\sum_{n=0}^\infty \text P_{-n}^{-n}(z)=1+\frac{\sqrt{\pi}}{\sqrt 2 (1-z^2)^{1/4}}\ e^{\frac{1}{2\sqrt{1-z^2}}}\bigg(\operatorname{erf}\bigg(\frac{-z}{\sqrt 2 (1-z^2)^{1/4}}\bigg)-\operatorname{erf}\bigg(\frac{-1}{\sqrt 2 (1-z^2)^{1/4}}\bigg)\bigg)$$