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

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.

• If a sequence is convergent the its $\lim \sup$ is same as its limit. Commented Nov 10, 2020 at 8:02
• @KaviRamaMurthy How do I know if the sequence given here is convergent? Commented Nov 10, 2020 at 8:11
• @Harmonic My answer below shows how sequence converges. Commented Nov 10, 2020 at 8:15

## 1 Answer

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.

• How do you know limit can be taken inside exponent? Commented Nov 10, 2020 at 8:26
• @Harmonic The function $f(x) = x^{-p}$ is continuous for all $p \in \mathbb Z$ and all $x > 0$ . This allows the exchange of exponents. Commented Nov 10, 2020 at 8:30