# Show that: $\lim\limits_{n\to\infty} \sqrt[n]n = 1$

First of all, I already know the common proof for this limit. My question concerns a specific proof that I could not deal with.

It starts with defining a sequence $\{x_n\}_{n=1}^{\infty}$ with the general term $x_n=\sqrt[n]{n}-1$ and then shows that this sequence converges to $0$.

We have $n=(1+x_n)^n\geq \frac{n(n-1)}{2}x_n^2$

Up to this point everything is clear. However, the rest of the proof is left to the reader, which I failed to do so.

Could you proceed or give some tips?

By the way, here is the link to the proof.

If you know that

$\frac{n(n-1)}{2}x_n^2 \le n$,

you can rearrange things a bit:

$x_n^2 \le \frac{2}{n-1}$.

This proof is actually way quicker and easier than the one I had to learn when I was taking analysis, but they had a quota of suffering that needed filling.

In response to comment: since the square root function is monotone and continuous, you find that

$0 \le |x_n| \le \sqrt{\frac{2}{n-1}}$,

now use the squeeze theorem.

• I'd like some clarification. Why does convergence of $x_n^2$ imply convergence of $x_n$? – user123641 Feb 5 '14 at 1:07
• the classical proof $\sqrt[n]n= exp(ln(n)/n)$ with $ln(n)/n\to 0$ is actually more straightforward in my opinion. But I agree this is a clever trick :) – Denis Feb 5 '14 at 1:41
• @dkuper when I took that class we introduced the exponential function later, so we couldn't use that argument. It is indeed even simpler though. – korrok Feb 5 '14 at 3:05

From your inequality you get $x_n^2\leq \frac2{n-1}$ so $x_n\to 0$.