<s>[INST] If $a_n=n\left(1-\frac{1}{n}\right)^{n[\log n]}$, prove $1\leqslant \liminf a_n$ and $\limsup a_n\leqslant e.$ 
Prove that:
  $$\limsup n\left(1-\frac{1}{n}\right)^{n[\log n]} \leqslant e$$
  and 
  $$\liminf n\left(1-\frac{1}{n}\right)^{n[\log n]} \geqslant 1.$$

Attempt. Since for $n>2$: 
$$1<[\log n]<\log n+1\leqslant n,$$ we would get:
$$n\left(1-\frac{1}{n}\right)^{n[\log n]}\leqslant n\left(1-\frac{1}{n}\right)^n \to +\infty,$$
$$n\left(1-\frac{1}{n}\right)^{n[\log n]}\geqslant n\left(1-\frac{1}{n}\right)^{n^2} \to 0,$$
so the above estimates are not convenient.
Thanks for the help.
 [/INST] First, note that $\log n-1 < [\log n] \leq \log n$. Now, $0<1- \frac{1}{n}<1$, so we have
$$ n \left( 1- \frac{1}{n} \right)^{n \log n} \leq n \left( 1- \frac{1}{n} \right)^{n [\log n]} < n \left( 1- \frac{1}{n} \right)^{n (\log n - 1)}. $$
We now want to use the fact that $\left(1- \frac{1}{n}\right)^{n} \to e^{-1}$ as $n \to \infty$ to simplify the above inequalities. To make the computation easier, write this as $n\log(1- \frac{1}{n}) \to -1$, which can be sharpened to $n\log(1- \frac{1}{n}) = -1 - \frac{1}{2n} + O\left( \frac{1}{n^2} \right)$ using the Taylor series for $\log(1-x).$
Now, 
$$\log \left(n \left( 1- \frac{1}{n} \right)^{n \log n} \right) = \log n \left( 1 + n \log \left(1 - \frac{1}{n} \right) \right) = \log n\left(- \frac{1}{2n} + O\left( \frac{1}{n^2}\right) \right) \to 0$$
as $n \to \infty$, and similarly
$$ \log \left( n \left( 1- \frac{1}{n} \right)^{n (\log n - 1)} \right) = \log n \left( 1 + n \log \left(1 - \frac{1}{n} \right)  \right) - n \log\left( 1- \frac{1}{n} \right) \to 1$$
as $n\to\infty.$ Taking the exponentials of both of these inequalities gives the desired result.
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