Example
Let \(a > 0\). Show that \(\lim\limits_{n \rightarrow \infty} a^{1/n} = 1\)

Proof

If \(a = 1\), then \(\lim\limits_{n \rightarrow \infty} 1^{1/n} = 1\) and we are done. Now, consider two cases:
Case 1: \(a > 1\). Then we know that \(\frac{1}{n} > \frac{1}{n+1}\). Then, since \(a > 1\), \(a^x\) is increasing so we must have

$$ \begin{align*} a^{1/(n+1)} \leq a^{1/n} \end{align*} $$

But this shows that \(a_n\) is decreasing. Moreover, since \(a > 1\), then

$$ \begin{align*} a &> 1 \\ a^{1/n} &\geq 1^{1/n} = 1. \\ \end{align*} $$

So \(a_n\) is bounded below by \(1\). Thus, \(a_n\) is both decreasing and bounded below by \(1\).

Case 2: \(0 < a < 1\). We also know that \(\frac{1}{n} > \frac{1}{n+1}\). But \(0 < a < 1\) and \(a^x\) is decreasing. Therefore

$$ \begin{align*} a^{1/(n+1)} \geq a^{1/n} \end{align*} $$

Thus, \(a_n\) is increasing. It is bounded above by \(1\) since

$$ \begin{align*} 0 &< a < 1 \\ 0 &< a^{1/n} < 1^{1/n} = 1. \\ \end{align*} $$

Finally, by the Monotone Convergence Theorem, the limit exists. To find the limit observe that

$$ \begin{align*} a^{1/2n} &= (a^{1/n})^{1/2} \\ \lim\limits_{n \rightarrow \infty} a^{1/2n} &= \lim\limits_{n \rightarrow \infty} (a^{1/n})^{1/2} \\ L &= \lim\limits_{n \rightarrow \infty} (a^{1/n})^{1/2} \quad \text{(since $2n > n$)} \\ L &= \lim\limits_{n \rightarrow \infty} \sqrt{a^{1/n}} \\ \end{align*} $$

Recall from Homework 3, that if \(\lim\limits_{n\to\infty} a_n = a\) and \(a\ge 0\), then \(\lim\limits_{n\to\infty} \sqrt{a_n} = \sqrt{a}\). In this case, we know that \(a^{1/n}\) converges to \(L\) so this means that \(\lim\limits_{n \rightarrow \infty} \sqrt{a^{1/n}} = \sqrt{L}\). Thus

$$ \begin{align*} L &= \sqrt{L} \\ L^2 - L &= 0 \\ L(L - 1) &= 0. \end{align*} $$

But \(L\) can’t be zero since if \(a > 1\), then \(a^{1/n} \geq 1\). Similarly, if \(0 < a < 1\), then \(a^{1/2} \geq a\). Therefore, \(L = 1\) as desired. \(\ \blacksquare\)

References

  • Lecture Notes by Professor Chun Kit Lai