Show that $\lim_{n\to \infty} n^{1/n} = 1$.

My attempt

Let $a_{n} = n^{1/n}$.

$|a_{n}-1| = |n^{1/n}-1| < n^{1/n} \leq n$.

Consider $|a_{n}-1| < \varepsilon$.

or $n<\varepsilon$.

or $n > 1/\varepsilon$.

Let $m$ be any integer greater than $1/\varepsilon$. Then for $\varepsilon>0$, there exists a positive integer $m$ such that $|a_{n}-1| < \varepsilon$, for all $n\geq m$.

Therefore, $\lim_{n\to \infty} n^{1/n} = 1$.

Is this proof correct?

  • 2
    $\begingroup$ No... you want $n$ large and yet you are supposing $n\lt\epsilon$ for arbitrarily small $\epsilon$. This does not work $\endgroup$
    – FShrike
    Jan 24, 2022 at 17:49
  • 2
    $\begingroup$ Not even close. $n<\epsilon$ is not equivalent to $n>\frac{1}{\epsilon}$, but to $\frac{1}{n}>\frac{1}{\epsilon}$. Anyway, if $\epsilon<1$ then $n<\epsilon$ clearly never happens. $\endgroup$
    – Mark
    Jan 24, 2022 at 17:49
  • $\begingroup$ It doesn't look right...at all. How can you prove that $\;m>\frac1\epsilon\implies |a_n-1|<\epsilon\;$ ?How deos this follow from what you did? $\endgroup$
    – DonAntonio
    Jan 24, 2022 at 17:49
  • 1
    $\begingroup$ No. First, you want to show $|a_n-1|<\epsilon,$ not assume it. But also, $a<b$ and $a<c$ tells you nothing about whether $b=c,$ $b<c$ or $c<b,$ so it is unclear how you get $n<\epsilon.$ $\endgroup$ Jan 24, 2022 at 17:50
  • 2
    $\begingroup$ "$n < \epsilon$" should be your tip off that it couldn't possibly work. $n$ gets arbitrarily large but we need $\epsilon$ to be arbitrarily small.... But so far as I can tell the only reason you would think that $n < \epsilon$ is because $|a_n-1|<\begin{cases}\epsilon\\ n\end{cases}$ and you conclude $n < \epsilon$. But $M<\begin{cases}U\\ S\end{cases}$ doesn't tell us anything about how $U$ and $S$ relate. $\endgroup$
    – fleablood
    Jan 24, 2022 at 18:16


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