Prove $\lim_{n\rightarrow \infty} n^{\frac{1}{n}} = 1$

Question

Prove that $$\lim_{n\rightarrow \infty} n^{\frac{1}{n}} = 1$$

Hint: write $$n^{\frac{1}{n}} = 1 + \epsilon_n, 0 < \epsilon_n $$ then $$n = (1+\epsilon_n)^n > 1 + \frac{n(n+1)}{2}\epsilon_n^2$$ by the binomial theorem.

I am not sure how to establish this result. The only manipulation I see is taking $\frac{1}{n}$ on both sides of the inequality in the hint, but I do not see how that can help me establish a result.

Answer 1

If $ n^{\frac{1}{n}} = 1 + \epsilon_n, $ then by taking $n$th powers, we know that $ n = (1 + \epsilon_n)^n $ and so $$ n = (1 + \epsilon_n)^n = 1 + \binom{n}{1} \epsilon_n + \binom{n}{2} \epsilon_n ^2 + \cdots > 1 + \binom{n}{2} \epsilon_n ^2 = 1 + \frac{n(n+1)}{2} \epsilon_n ^2 $$ and therefore.....

$$ 0 < \epsilon_n < \sqrt{ \frac{2(n-1)}{n(n+1)} } $$ But, what happens if we look at the limit as $n \to \infty$? Then, $\epsilon_n \to 0$, and so $n^{\frac{1}{n}} \to 1$.

Answer 2

You can also have $n>\frac{n(n+1)}{2}\epsilon_n^2$, then $\sqrt{\frac{2}{n+1}}>\epsilon_n>0$ which in the limit $n\to\infty$ implies that $\epsilon_n\to0$.

Answer 3

$$ n^{\frac{1}{n}}=e^{ln{n^{\frac{1}{n}}}}=e^{\frac{1}{n}ln(n)}=\\e^{\frac{ln n}{n}}$$ now its suffice to prove $\lim_{n \rightarrow \infty}\frac{ln n}{n}=0$ so $$n \rightarrow \infty \space \space ln n < \sqrt{n}\\ 0\leq\lim_{n \rightarrow \infty}\frac{ln n}{n} \leq \lim_{n \rightarrow \infty}\frac{\sqrt{n}}{n} \rightarrow 0 \\ \lim_{n \rightarrow \infty}e^{\frac{ln n}{n}}=e^0=1$$