Divergence of the harmonic series and the Cauchy criterion imply $\lim_{n\to\infty} \sqrt[n]n=1$ Why the divergence of the harmonic series and the Cauchy criterion immediately imply $\displaystyle\lim_{n\to{+}\infty}{\sqrt[n]{n}}=1$?
Thanks for your help
 A: The Cauchy root test says that $\limsup_{n\to\infty} \sqrt[n] {|a_n|}<1$ implies (absolute) convergence of $\sum_{n=1}^\infty a_n$.
Since we know that $\sum_{n=1}^\infty \frac1n$ is divergent, we conclude  $\limsup_{n\to\infty} \sqrt[n] {\frac1n}\ge1$, i.e. $\liminf_{n\to\infty} \sqrt[n] n\le1$. In fact, $\sqrt[n]n\ge1$ for all $n$ implies $\liminf_{n\to\infty} \sqrt[n] n=1$.
Therefore, if you knew that $\sqrt[n]n$ converges to some limit, then this limit must equal the $\limsup$, i.e. 1.
A: As Hagen already said it suffices to show, that $\sqrt[n] n$ converges. (Then by $\liminf_{n\to\infty} \sqrt[n] n=1$ the assertion follows.)
I think for the Cauchy criterion it was meant to show, that $\sqrt[n] n$ is a Cauchy sequence and hence converges.
But here you can easily check that the sequence is monotonously decreasing (for $n\geq 2$). Since (the sequence is bounded from below by $1$ or) $\liminf_{n\to\infty} \sqrt[n] n=1$ this would imply convergence.
For all $n\geq 2$ we have
$$ (n+1)^n = \sum\limits_{k=0}^{n}{\binom{n}{k}n^k}<n^n\sum\limits_{k=0}^{n}{\binom{n}{k}}=2^n\cdot n^n\leq n^n\cdot n^n=n^{n+1} $$
and hence
$$ \sqrt[n+1]{n+1}< \sqrt[n] n.$$
A: I'm not sure what you can do with the divergence of the harmonic series, but a much simpler approach is to notice the following: $n^{\frac{1}{n}} = e^{\ln(n^{\frac{1}{n}})} = e^\frac{\ln(n)}{n}$. Now, $$\lim_{n\to\infty}{e^{ \frac{\ln(n)}{n}}} = e^{\lim_{n\to\infty}\frac{\ln(n)}{n}} = e^0 = 1.$$ The last step follows by L'Hopital's rule on $\frac{\ln(n)}{n}$. I'm sorry if this solution isn't  what you were looking for and there is some other way you had in mind to solve this problem.
