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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.

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

1 Answer 1

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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.

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  • $\begingroup$ How do you know limit can be taken inside exponent? $\endgroup$
    – Harmonic
    Commented Nov 10, 2020 at 8:26
  • $\begingroup$ @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. $\endgroup$
    – balddraz
    Commented Nov 10, 2020 at 8:30

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