# Show that $\lim_{n\to \infty} n^{1/n} = 1$. Is this convergence proof correct?

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?

• No... you want $n$ large and yet you are supposing $n\lt\epsilon$ for arbitrarily small $\epsilon$. This does not work Jan 24, 2022 at 17:49
• 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.
– Mark
Jan 24, 2022 at 17:49
• 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? Jan 24, 2022 at 17:49
• 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 $n<\epsilon.$ Jan 24, 2022 at 17:50
• "$n < \epsilon$" 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. Jan 24, 2022 at 18:16