I'm looking for pointers as to how to evaluate the asymptotic behavior of $\sum_{k=1}^{n}\left(1-p^{k}\right)^{n-k}$ for large $n$ where $0
Any help is much appreciated.
I'm looking for pointers as to how to evaluate the asymptotic behavior of $\sum_{k=1}^{n}\left(1-p^{k}\right)^{n-k}$ for large $n$ where $0
Any help is much appreciated.
Note that $f(k,n) = (1-p^k)^{n-k}$ is an increasing function of $k$ for fixed $n$. For $k = \frac{\log n}{\log(1/p)} + t$ Maple says we have, as $n \to \infty$, $f(k,n) \approx e^{-p^t} + O(1/n)$. This goes exponentially to $1$ as $t \to +\infty$ and very rapidly to $0$ as $t \to -\infty$, so I suspect the sum will be $n - \frac{\log n}{\log(1/p)} + O(1)$ as $n \to \infty$ (i.e. the error made by approximating $f(k,n)$ by $0$ for $k < \frac{\log n}{\log(1/p)}$ and $1$ for $k > \frac{\log n}{\log(1/p)}$ will be bounded as $n \to \infty$).
$f(n,k) \approx 1$
for $k > \frac{\log n}{\log(1/p)}$ – 2012-03-08– 2012-03-09
$p^k$
decrease pretty fast to $0$, hence $1-p^k$ increase pretty fast to $1$, so the terms is "some what just below" $(1-p^n)$ - which would lead to the geometric sum $\sum_1^n(1-p^n)^{n-k}=\sum_0^{n-1}(1-p^n)^k=(1-(1-p^n)^n/p^n$. – 2012-03-08