Limit of $\sqrt[n]a$ for $0

Question

I am trying to prove that the sequence $(a_n)_{n\geq1}$ where $a_n = \sqrt[n]a$ for $0<a<1$ converges to $1$. So that $$\lim_{n\to\infty}\sqrt[n]a=1$$

I have managed to prove it for $1 \lt a$, using the fact that : $$\sqrt[n]a = \theta_n + 1 \implies a = (\theta_n + 1)^n \implies a = \sum_{k = 0}^n \binom{n}{k} \theta_n^k \gt \binom{n}{1}\theta_n=n\theta_n \implies \theta_n \lt \frac{a}{n}$$

Thus setting $n_\epsilon = \frac{a}{\epsilon}$, we have $\forall n \gt n_\epsilon, \theta_n \lt \epsilon \implies \lvert \sqrt[n]a - 1 \rvert < \epsilon$. Thus $\lim_{x\to\infty}\sqrt[n]a=1$.

However, this method does not work for $0<a<1$ as $\theta_n < 0$. I have tried finding a solution using $\sqrt[n]a = 1- \theta_n$, however this introduces a $(-1)^k$ in the sum. I have also tried setting $a = \frac{1}{b}$ with $1<b$, however this method is not satisfactory as it used reasoning in theorems outside of the ones allowed. Finally, I also know that I could probably use the monotone convergence theorem, but I ran into some trouble showing that $\sqrt[n]a < \sqrt[n+1]a$.

I would like a hint for the direction to take to show the above limit for $0<a<1$, preferably using the method above. Thanks for reading !

Answer 1

Let $0<a<1$;

$a^{1/n} =\dfrac{1}{1+e_n}$, where $e_n >0;$

$a=\dfrac{1}{(1+e_n)^n};$

$\dfrac{1}{a}=(1+e_n)^n>1+ne_n;$

$ne_n < \dfrac{1}{a}-1$ $(>0);$

$0<e_n<(1/n)(\dfrac{1}{a}-1);$

Sandwich.

Answer 2

Here is an alternative approach:

\begin{align*} \lim_{n\to\infty}\sqrt[n]{a} & = \lim_{n\to\infty}\exp\left(\ln(\sqrt[n]{a})\right)\\\\ & = \lim_{n\to\infty}\exp\left(\frac{\ln(a)}{n}\right)\\\\ & = \exp\left(\lim_{n\to\infty}\frac{\ln(a)}{n}\right)\\\\ & = \exp(0) = 1 \end{align*}

Hopefully this helps !

Answer 3

Yet another alternative approach:

1. Show $\lim_{n \to \infty} \sqrt[n]{a} \le 1$: \begin{align} 0 < a < 1 \implies \sqrt[n]{a} < \sqrt[n]{1}=1 \quad\forall n \end{align} since $\sqrt[n]{-}$ is strictly increasing on $\mathbb{R}^+$.

2. Show $\lim_{n \to \infty}\sqrt[n]{a} \ge 1$: Suppose $\sqrt[n]{a} = 1 - \epsilon$ for some small positive $\epsilon$. Since \begin{align} \sqrt[2n]{a} = \sqrt{\sqrt[n]{a}}=\sqrt{1-\epsilon} > 1 - \epsilon = \sqrt[n]{a} ,\end{align} we can always choose a higher $n$ to obtain a term larger than $1 - \epsilon$. Therefore, the limit is $\ge 1$.

Combining these makes $\lim_{n \to \infty}\sqrt[n]{a} = 1$ for $0 < a < 1$. \begin{align} \end{align}

Answer 4

Using the comment of @A.Γ., you find : $$a<1 \implies 1 < \frac{1}{a} \implies \lim_{n\to\infty} \sqrt[n]{\frac{1}{a}} = 1 \implies \frac{1}{\lim_{n\to\infty} \sqrt[n]a} = 1 \implies \lim_{n\to\infty} \sqrt[n]a = 1$$