Find supremum and infimum of $(1-1/n^2)^n$

Question

I have to find the supremum and infimum of $(1-1/n^2)^n$ where $n$ is a natural number. A hint is given that an inequality helps. I thought that the inequality which could help is Bernoulli's inequality: $(1+x)^n\ge 1+nx$. But this is not helping. Because then I get: $(1-1/n^2)^n\ge 1-1/n$ It helps in finding an infimum but not a supremum.

Answer 1

So since $$\left( 1-\frac{1}{n^2} \right)^n \leq 1$$ and $$\lim_{n\to\infty} \left( 1-\frac{1}{n^2} \right)^n =1$$ you get that $$\sup_n \left( 1-\frac{1}{n^2} \right)^n =1$$