Maximum speed of decay

Question

What's the maximum decay rate we can have for $\alpha(n)$ and still have the following expression converge to zero ? $$ \big[ 1- \alpha(n) (1- e^{-\log^{-2}n}) \big]^n$$ Intuitevely, I would say $\log^2n/n$ is the limiting speed since $\alpha(n)(1- e^{-\log^{-2}n})$ shouldn't decay faster than $1/n$ and $1- e^{-\log^{-2}n} \sim \log^{-2}n$. I was wondering how we can formalize this ?

Answer 1

Note that $(1- e^{-\log^{-2}n})$ is monotone decreasing to $0$ for $n$ sufficiently large. Assume that $0\leq\alpha(n)\leq \frac{\log^2n}{n}$ for $n$ sufficiently large, that is, that $\alpha(n)$ goes to $0$ as fast or faster than $\frac{\log^2n}{n}$ and is eventually nonnegative. Then $$\left[1-\alpha(n)(1- e^{-\log^{-2}n})\right]^n \geq \left[1-\frac{\log^2n}{n}(1- e^{-\log^{-2}n})\right]^n\geq \left[1-\frac1n \right]^n$$ for $n$ sufficiently large. Show that the right side converges to $\frac1e$. Thus the left side can't converge to zero by comparison.

Now assume that $0\leq \frac{\log^2n}{n^p}\leq\alpha(n)$ for $n$ sufficiently large and some $p\in(0,1)$. Then $$0\leq\left[1-\alpha(n)(1- e^{-\log^{-2}n})\right]^n \leq \left[1-\frac{\log^2n}{n^p}(1- e^{-\log^{-2}n})\right]^n$$ and the right side goes to $0$ as $n\rightarrow\infty$.

Now one thing I am assuming for the second argument is that we can replace $e^{-\log^{-2}n}$ by $1-\log^{-2}n$ when taking the limit. I wasn't able to figure out the right way to argue that for this particular second comparison. I'll continue to think of a way to clean it up.