<s>[INST] prove a sequence is bounded above. I want to prove that $x_n=\left(1-\frac{1}{n}\right)^n$ is bounded above.
I know that $b_n=\left(1+\frac{1}{n}\right)^n$ is bounded above, because $b_n\rightarrow \mathcal e$. Indeed, it can be proved that $b_n<3$. Since $b_{n-1}=\frac{n^{n-1}}{(n-1)^{n-1}}$, $x_n=\frac{(n-1)^n}{n^n}=\frac{n-1}{n}.\frac{1}{b_{n-1}}$. So, $\frac{1}{b_{n-1}}>\frac13$, and $(x_n)$ is not bounded above. I commited a mistake, but where?
 [/INST] If $n\ge 1$, you have $0 \le 1-\frac{1}{n} \le 1+\frac{1}{n}$. Hence $(1-\frac{1}{n})^n \le (1+\frac{1}{n})^n$.
Indeed, $(1-\frac{1}{n}) (1+\frac{1}{n}) = 1-\frac{1}{n^2} \le 1$. So you have $(1-\frac{1}{n})^n (1+\frac{1}{n})^n \le 1$, or $(1-\frac{1}{n})^n \le \frac{1}{(1+\frac{1}{n})^n}$.
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