product converging to one

Question

I have a question concerning an infinite product. Suppose $x_n$ is a sequence of positive real numbers. My intuition says that $$\lim_n(1-\exp(-x_n))^n=1$$ for any sequence $x_n=n^\alpha$ with $\alpha \in (0,1)$ a constant. How can I show this? Is there a condition on $x_n$ such that the limit holds for general $x_n$?

Answer 1

We have $$\lim_{n\rightarrow\infty}\left(1-\frac{1}{e^{n^{\alpha}}}\right)^{n}=\exp\left(\lim_{n\rightarrow\infty}n\log\left(1-\frac{1}{e^{n^{\alpha}}}\right)\right) $$ and $$ \lim_{n\rightarrow\infty}n\log\left(1-\frac{1}{e^{n^{\alpha}}}\right)=\lim_{n\rightarrow\infty}\frac{\log\left(1-\frac{1}{e^{n^{\alpha}}}\right)}{\frac{1}{n}} $$ then by Hospital's rule $$=\lim_{n\rightarrow\infty}-\frac{n^{\alpha+1}\alpha}{e^{n^{\alpha}}-1}=0 $$ then your claim. Note that this proof holds for all $\alpha>0 $.

Answer 2

Since $\lim_{n\to\infty}(1-\frac{1}{n})^n=e^{-1}$, we just need $exp(x_n)$ converge to $\infty$ faster than $n$, in other words, $x_n$ converges faster than $\ln{n}$. Your example certainly fits this criteria.

Should point out though, when $x_n$ converges at the same speed as $\ln{n}$, coefficient/constant matters, e.g., $x_n=\ln(n)-1$ works, while $x_n=\ln{x}$ does not.