Does $(1-e^{-y})^n $ converge as $n \to \infty$?

Question

Let $y \in R_{\ge 0}$ . Does $(1-e^{-y})^n $ converge as $n \to \infty$? If yes, what does it converge to?

It is obvious that when $y=0$, it converges to $0$. But how about $y \gt 0$ ?

Answer 1

For $y\ge 0$, we have $0 < e^{-y} \le 1$, so the base $1-e^{-y} \in [0,1)$. Therefore, $\lim\limits_{n\to\infty}(1-e^{-y})^n = 0$.

Answer 2

In fact, the complete discussion (for $y\in \mathbb{R}$) is the following.

1. For $y<-log(2)$ the sequence alternates and grows in absolute value (no limit)
2. For $y=-log(2)$ the sequence is $-1,1,-1\ldots$ (no limit)
3. For $y>-log(2)$ then $|1-e^{-y}|<1$ and the sequence tends to $0$.