1

So, I was trying to prove that $n^{\frac{1}{n}} \rightarrow 1$ where $n \in \mathbb{N}$.

Here's what I did:

Since $n^{1/n}>1$, let us suppose $n=(1+k)^n$ for $n>1$ and some $k>0$.

Now, I used the binomial expansion and wrote: $$\begin{align} n&=1+nk+\frac{n(n-1)}{2}k^2+....+k^n \geq1+\frac{n(n-1)}{2}k \\ &\Rightarrow n-1 \geq \frac{n(n-1)}{2}k^2 \\ &\Rightarrow k^2 \leq \frac{2}{n} \end{align}$$

So now, we can always find an $n>0$ such that $\frac{2}{\epsilon^2}<n\Rightarrow \frac{2}{n}<\epsilon^2$.

Now, as $|n^{\frac{1}{n}}-1|\geq0$, we have,

$n^{\frac{1}{n}}-1=k\leq(\frac{2}{n})^{\frac{1}{2}}<\epsilon$

And in this way, I proved that $n^{\frac{1}{n}}\rightarrow1$.

I wanted to know how can I prove this without using Binomial expansion. I tried to do this using the Bernoulli's Inequality, but couldn't get too far.

Any help/hint would be highly appreciable.

DeBARtha
  • 609
  • You could consider it as a subsequence of the function $f:[1,\infty)\to\mathbb{R}$, $x\mapsto x^{1/x}$. Using logarithmic differentiation, you can show this function is decreasing for $x>e$, and it is clearly bounded below by zero, so it has a horizontal asymptote. Then you can use LHR to show that this asymptote is $1$. – Integrand Oct 14 '20 at 18:22
  • Had a nice solution but the problem closed all-of-a-sudden! – mjw Oct 14 '20 at 18:56
  • The power series $\sum_{n=1}^\infty z^n$ is a geometric series and is known to converge for $|z|<1.$ Differentiating, the $n^\textrm{th}$ term is $nz^{n-1}$. Since this series also has radius of convergence $R=1$, $\lim \sup |a_n|^{1/n} = n^{1/n} =1.$ – mjw Oct 14 '20 at 18:57
  • @mjw thanks for the hint...it's really nice – DeBARtha Oct 15 '20 at 03:11

1 Answers1

2

I would look at $\log\left(n^{\frac{1}{n}}\right) = \frac{1}{n}\log n$. This quantity converges to zero as $n\rightarrow\infty$, so the original expression must converge to $1$.

sven svenson
  • 1,415
  • 5
  • 14