Calculate limsup of $\frac{1}{n^{p/n}}$

Question

I am trying to find-

$\displaystyle\limsup_{n\to\infty}\Big|\frac{1}{n^{p/n}}\Big|$ where $p\in\mathbb{Z}$.

I am not sure how to calculate that. All I have proved is that $\displaystyle\lim_{n\to\infty}n^{1/n}=1$. I am not sure how to proceed.

Answer 1

Well, if you know $\lim_{n \to \infty} n^{1/n} = 1$ then for $p \in \mathbb Z$, $$ \lim_{n \to \infty} \frac{1}{n^{p/n}} = \lim_{n \to \infty} (n^{1/n})^{-p} = \big(\lim_{n \to \infty} n^{1/n}\big)^{-p} = 1^{-p} = 1 $$ So, the sequence $(1/n^{p/n})_{n = 1}^\infty$ converges to $1$ and so the $\limsup$ must also be this same value.