Limit sup/inf $(\frac{1}{n+1}-1)^n$

Question

My suggestion was that $$\varlimsup_{n \to \infty}(\frac{1}{n+1}-1)^n=1$$ and $$\varliminf_{n \to \infty}(\frac{1}{n+1}-1)^n=-1$$ by looking at the even/odd subsequence since obviously $\lim_{n \to \infty}(\frac{1}{n+1}-1)= -1$.

But can I just conclude that $\lim_{n \to \infty}a_n^{2n}=1$ when $\lim_{n \to \infty}a_n=-1$? I feel like that's not very mathematical. I didnt find any rule or anything that would allow that.

Thanks!

Answer 1

$$x_n=(\frac{1}{n+1}-1)^n=(-1)^n(1-\frac{1}{n+1})^n=(-1)^n(1-\frac{1}{n+1})^{n+1} \frac{1}{1-\frac{1}{n+1}}$$

So $\limsup_nx_n=\frac{1}{e}$ and $\liminf_nx_n=-\frac{1}{e}$