Let $0<p=p(n)<1$ with $p=o(1)$. For which $p$ is it true that $1-(1-p)^n \sim pn$?
With $\sim$ I mean that they are asymptotically the same, so $\frac{1-(1-p)^n}{pn}\rightarrow 1$, or at least that $\frac{1-(1-p)^n}{pn}\rightarrow c$ for some constant $c$.
Which kind of series expansion do I need for that?
Thanks for any hints!
Before his edit
Pretty sure it only works for functions $p(n)=o(1)$ (using little-$o$ notation). Otherwise, $p(n)=\Omega(1)$, so $1-(1-p(n))^n=\Omega(p(n)^n)$. Assuming $p \rightarrow \infty$, we know $\frac{p(n)^n}{np(n)} \rightarrow \infty$.
After his edit
I found that for my ratio, $p(n) \sim n^{\frac{1}{n-1}}$ yields $p^n \sim np$. However, it's possible $1-(1-p)^n \neq O(p^n)$. It's fairly easy to see that $1-(1-p)^n \sim (1+p)^n= O(\binom{n}{\lfloor n/2 \rfloor}p^n)$, and $(1+p)^n=\Omega(p^n)$.