Calculating $\limsup\limits_{n \to \infty}a_n$ and $\liminf\limits_{n \to \infty}a_n$ with $a_n=\left( \frac{n+(-1)^n}{n} \right)^n$ I need to calculate $\limsup\limits_{n \to \infty} a_n$ and $\liminf\limits_{n \to \infty}a_n$ where $a_{n} = \left( \dfrac{n+(-1)^n}{n}\right)^n, \: n \in \mathbb{N}$
My approach is to calculate the limits for $n$ being even or odd.


*

*$n \pmod 2 \equiv 1$


$$\quad \lim_{n \to \infty} \left(\dfrac{n-1}{n} \right)^n$$


*

*$n \pmod 2 \equiv 0$


$$\quad \lim_{n \to \infty} \left(\dfrac{n+1}{n} \right)^n$$
I don't know how I can solve the two formulas above. How should I continue?

$a_n := \left({n+(-1)^n \over n}\right)^n\implies \log a_n=n\log\left(1+\frac{(-1)^n}{n}\right)=n\left(\frac{(-1)^n}{n}+O\left(\frac{1}{n^2}\right)\right)=(-1)^n+O\left(\frac{1}{n}\right)$
by using that $\log(1+x)=x+O(x^2)$ as $x\to0$ (can prove by looking at Taylor series / integration / various other methods).
So, $a_n=\exp((-1)^n+o(1))$, whence we see that $\sup a_n=\exp(1)$ and that $\inf a_n=\exp(-1)$.
