Strategy for tackling the $\lim_{n\to+\infty}\frac{(-1)^nn}{(1+n)^n}$

Question

What strategy should I use to calculate this limit? Can I avoid using Hopital?

$$\lim_{n\to+\infty}\frac{(-1)^nn}{(1+n)^n}$$

Thank you in advance.

Answer 1

For $n\ge 2$ we have $(1+n)^n\ge (1+n)^2\ge n^2$ and already $\frac{(-1)^nn}{n^2}=\frac{(-1)^n}n\to 0$.

Answer 2

One possible way is: write $\dfrac{(-1)^nn}{(1+n)^n}=\dfrac{(-1)^n}{n^{n-1}} \dfrac{n^n}{(1+n)^n}=\dfrac{(-1)^n}{n^{n-1}} \dfrac{1}{(1+\dfrac{1}{n})^n}$.

hence we have:

$$ \begin{align} \lim_{n \to \infty} \left[\dfrac{(-1)^nn}{(1+n)^n}\right] &=\lim_{n \to \infty} \dfrac{(-1)^n}{n^{n-1}} \dfrac{1}{\left(1+\dfrac{1}{n}\right)^n}\\\,\\ &=\lim_{n \to \infty} \dfrac{(-1)^n}{n^{n-1}} \cdot \lim_{n \to \infty} \dfrac{1}{\left(1+\dfrac{1}{n}\right)^n}\\ &=0 \cdot \dfrac{1}{e}=0\\ \end{align} $$