How to calculate this limit related to hypergeometric functions

Question

How to calculate this limit

$$ \lim_{n\rightarrow+\infty} np^n \sum_{k=n}^{+\infty} \frac1k \binom{k-1}{n-1} (1-p)^{k-n} ,\quad\,where\;0<p<1 $$

Answer 1

For $|z|<1$, using the binomial series, we have $$\sum_{k=n}^{\infty}\binom{k-1}{n-1}\frac{z^k}{k}=\int_0^z\sum_{k=n}^{\infty}\binom{k-1}{n-1}x^{k-1}\,dx=\int_0^z\frac{x^{n-1}\,dx}{(1-x)^n}=\int_0^{z/(1-z)}\frac{y^{n-1}\,dy}{1+y},$$ the last equality is after substituting $x=y/(1+y)$. Thus, if $a:=(1-p)/p$, the given limit is $$\lim_{n\to\infty}\frac{n}{a^n}\int_0^a\frac{y^{n-1}\,dy}{1+y}=\underbrace{\frac{1}{1+a}}_{=p}+\underbrace{\lim_{n\to\infty}\int_0^a\frac{(y/a)^n}{(1+y)^2}\,dy}_{=0}$$ (after integration by parts). Thus, the answer is simply $\color{blue}{p}$.

Answer 2

Assuming that the formula is $$S_n= n\,p^n \sum_{k=n}^{+\infty} \frac1k\, \binom{k-1}{n-1} \,(1-p)^{k-n}$$ $$S_n=n\,p^n\,\frac{\, _2F_1(n,n;n+1;1-p)}{n}=p^n\, _2F_1(n,n;n+1;1-p)$$ which rewrite $$S_n=p \, _2F_1(1,1;n+1;1-p)$$ which seems to tend to $p$.

Computed for $n=10^4$ with limited precision $$\left( \begin{array}{cc} p & S_{10000} \\ 0.1 & 0.100009 \\ 0.2 & 0.200016 \\ 0.3 & 0.300021 \\ 0.4 & 0.400024 \\ 0.5 & 0.500025 \\ 0.6 & 0.600024 \\ 0.7 & 0.700021 \\ 0.8 & 0.800016 \\ 0.9 & 0.900009 \end{array} \right)$$