Proving that a sequence diverges: $s_n = (\frac 1 n - 1)^n$

Question

$s_n = (\frac 1 n - 1)^n$ $\rightarrow$$\infty$ as $n \rightarrow \infty$

my take

To prove a sequence diverges, need to find $N \in \mathbb{N}$ for all $m>0$ such that $s_n \ge M$ for all $n \ge N$

so given $M \ge 0$, let $N \ge \frac{\ln(M)}{\ln(\frac 1 N - 1)}$ then

ln$(\frac 1 n - 1)^n \ge$ ln $(\frac 1 N - 1)^N \ge$ ln$(e^M) = M$ for all $n \ge N$

Is my proof okay?

Answer 1

Hint. One may observe that, as $n \to \infty$, $$ s_{2n}=\left(\frac1{2n}-1 \right)^{2n}=\left(1-\frac1{2n}\right)^{2n} \to e^{-1} $$ and that $$ s_{2n+1}=\left(\frac1{2n+1}-1 \right)^{2n+1}=\color{red}{-1}\cdot\left(1-\frac1{2n+1}\right)^{2n+1} \to \color{red}{-1}\cdot e^{-1} $$ thus the considered sequence is divergent.