Rate of convergence of exp(-n)

Question

I am trying to determine the rate of convergence of the sequence $ e^{-n}$ (as $n \to \infty$).

I know that I need to find a sequence $\beta_n$ that converges to 0, that is larger than $\alpha_n - \alpha$ when multiplied by some positive constant $K$. ( $|\alpha_n - \alpha| \leq K|\beta_n|$ ) If this is satisfied, $\alpha_n = O(\beta_n)$.

In other rate of convergence problems, I have seen the Taylor Expansion with remainder of a function used to bound the sequence (specifically with $\sin$ and $\cos$). I'm assuming a similar procedure should be used here, but once I build the $|\alpha_n - \alpha|$ term I get stuck.

$|\alpha_n - \alpha| = | \exp(-n) - 0| = |1 - n + \frac{1}{2}n^2 - \exp(-\epsilon)\frac{n^3}{3!}| \leq \hspace{10px} ???$

I know that the $n^3$ term should grow the fastest, but beyond that I am at a loss as to how I should proceed.

Any help would be greatly appreciated!

Answer 1

The rate of convergence of $a_n:=\exp(-n)$ is itself, unless you have some objection to that. It is an example of exponential decay. The sequence $a_n$ converges to $0$ faster than $n^{-k}$ for any integer $k$ using a ratio test.

Answer 2

$$|e^{-n} -0|\leq \frac{1}{2^n} \leq \frac{1}{n}$$