uplakanimkk

2022-07-15

Evaluating:
$\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n+1}\left(\omega +\nu {\right)}^{\left(n+1\right)}{z}^{\nu }{}_{2}{F}_{1}\left(1,\omega +\nu +1;n+2;1-z\right)$

Expert

So, we have
$\begin{array}{rl}{S}_{n}& =\frac{1}{n+1}\left(\omega +\nu {\right)}^{\left(n+1\right)}{z}^{\nu }\frac{\mathrm{\Gamma }\left(n+2\right)}{\mathrm{\Gamma }\left(n-\omega -\nu +1\right)}\frac{1}{{n}^{\omega +\nu +1}}\left(1+\mathcal{O}\phantom{\rule{negativethinmathspace}{0ex}}\left(\frac{1}{n}\right)\right)\\ & =\left(\omega +\nu {\right)}^{\left(n+1\right)}{z}^{\nu }\frac{\mathrm{\Gamma }\left(n+2\right)}{\mathrm{\Gamma }\left(n-\omega -\nu +1\right)}\frac{1}{{n}^{\omega +\nu +2}}\left(1+\mathcal{O}\phantom{\rule{negativethinmathspace}{0ex}}\left(\frac{1}{n}\right)\right)\\ & ={z}^{\nu }\left(\omega +\nu {\right)}^{\left(n+1\right)}\frac{1}{n}\left(1+\mathcal{O}\phantom{\rule{negativethinmathspace}{0ex}}\left(\frac{1}{n}\right)\right)\\ & =\left(-1{\right)}^{n+1}\frac{{z}^{\nu }}{\mathrm{\Gamma }\left(-\omega -\nu \right)}\frac{\mathrm{\Gamma }\left(n-\omega -\nu \right)}{n}\left(1+\mathcal{O}\phantom{\rule{negativethinmathspace}{0ex}}\left(\frac{1}{n}\right)\right).\end{array}$
If $\omega +\nu$ is not a non-negative integer, ${S}_{n}$ will tend to infinity in absolute value and will oscillate in sign.

Do you have a similar question?