Calculating $\limsup\limits_{n \to \infty}a_n$ and $\liminf\limits_{n \to \infty}a_n$ with $a_n=\left( \frac{n+(-1)^n}{n} \right)^n$

Question

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?

Answer 1

Hint. One may recall that $$ \lim_{n \to \infty} \left(1+\frac{x}n\right)^n=e^x, \qquad x \in \mathbb{R}, $$ and one may observe that $$ \left(1-\frac1n\right)^n\le a_n \le \left(1+\frac1n\right)^n, \quad n=1,2,\cdots. $$

Answer 2

$a_n := \left({n+(-1)^n \over n}\right)^n\implies \log a_n=n\log\left(1+\frac{(-1)^n}{n}\right)=n\left(\frac{(-1)^n}{n}+O\left(\frac{1}{n^2}\right)\right)=(-1)^n+O\left(\frac{1}{n}\right)$

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 $\inf a_n=\exp(-1)$.