upper bound for equation Let $0 < p < 1$ be some constant. I am looking for an $M$ such that  
$$f(n) = \left(1-p^{\log{n}}\right)^{n} < M(n)$$
I am looking for a tight bound, something of the form: $2^{-n/\log{ n}}$. I checked in wolfram and it seems to go to $0$ when $n$ increases. I am looking for a bound in $n$.
Thanks a lot for your help!
 A: $1$ is an upper bound for obvious reasons.
A: Assuming that $n$ is a positive integer, the bound would be 1. To see this, think about what happens when you take positive power of a number between $0$ and $1$. It will be again a number between $0$ and $1$.
Edit:
The best bound in $n$ you can possibly get is $M(n)=1$.
To see this, fix $p$ to be $0.1$. The limit $\lim_{n \rightarrow \infty} f(n) = 1$. On the other hand, $\lim_{n \rightarrow \infty} f(n) = 0$ for $p=0.9$.
This is because $1-p^{\log x} = 1-e^{\log x \log p} = 1-e^{\log x^{\log p}} = 1-\frac{1}{x^{-\log p}}$: 


*

*For $-\log(p)>1$, $f(n)$ goes to $0$ (as $n$ aproaches infinity)

*for $0<-\log(p)<1$, $f(n)$ goes to $1$ (as $n$ aproaches infinity)

*for $\log(p)=-1$, $f(n)$ goes to $1/e$ (as $n$ aproaches infinity).
These results can be obtained from the definition of $e$ as a limit of $(1+y)^{1/y}$,  where you take $y=1/x$.
If you allow your bounds to be a function in $p$ as well, you can take for example $M(n,p) = 1-p^{\log n}$, which is exceptionally poor bound for higher $n$. For $p$ such that $\log(p)>-1$, the best upper bound you can get is the maximum of $f$ which is achieved on $(1,\infty)$.
