Property of the n-th Root of a Positive Convergent Sequence

Be $k \in \mathbb{R}, k > 1$, and $(a_n) \subset \mathbb{R}^+$ a convergent sequence with $\lim_{n \to \infty} a_n = a \in \mathbb{R}^+$. We want to prove that the sequence $(\sqrt[k]{a_n})$ is convergent and that $\lim_{n \to \infty} \sqrt[n]{a_n} = (\sqrt[n]{\lim_{n \to \infty} a_n})$.

First of all, let’s prove that $(\sqrt[n]{a_n})$ is convergent, as $(a_n)$ is convergent, we know that for all $\epsilon > 0$ there exists an $n_0 \in \mathbb{N}$ such that for all $n \in \mathbb{N}$ with $n \geq n_0$ we have that $|a_n - a| < \epsilon$. We can find a value $M$ such that $0 < M < a$, and also we can look for an $n_1 \in \mathbb{N}$ such that for all $n \in \mathbb{N}$ with $n \geq n_1$ we have that $a_n > M$.

Now, let’s consider the sequence $(\sqrt[k]{a_n})$, we know that $a_n - a = (\sqrt[k]{a_n})^k - (\sqrt[k]{a})^k = (\sqrt[k]{a_n} - \sqrt[k]{a})((\sqrt[k]{a_n})^{k-1} + (\sqrt[k]{a_n})^{k-2}\sqrt[k]{a} + \ldots + (\sqrt[k]{a})^{k-1})$. Then we can operate as follows:

$\sqrt[k]{a_n} - \sqrt[k]{a} = \frac{a_n - a}{(\sqrt[k]{a_n})^{k-1} + (\sqrt[k]{a_n})^{k-2}\sqrt[k]{a} + \ldots + (\sqrt[k]{a})^{k-1}}$

We know that $\sqrt[k]{M} < \sqrt[k]{a_n} < \sqrt[k]{a}$, then we can say that $\sqrt[k]{M}^{2} < \sqrt[k]{a_n}^{2} < \sqrt[k]{a}^{2}$, and so on, until we get that $\sqrt[k]{M}^{k-1} < \sqrt[k]{a_n}^{k-1} < \sqrt[k]{a}^{k-1}$. So we can operate as follows:

$\frac{1}{(\sqrt[k]{a_n})^{k-1} + (\sqrt[k]{a_n})^{k-2}\sqrt[k]{a} + \ldots + (\sqrt[k]{a})^{k-1}} < \frac{1}{(\sqrt[k]{M})^{k-1} + \sqrt[k]{M})^{k-1} + \ldots + (\sqrt[k]{M})^{k-1}} = \frac{1}{k\sqrt[k]{M}^{k-1}}$

Then we can say that $\sqrt[k]{a_n} - \sqrt[k]{a} = \frac{a_n - a}{(\sqrt[k]{a_n})^{k-1} + (\sqrt[k]{a_n})^{k-2}\sqrt[k]{a} + \ldots + (\sqrt[k]{a})^{k-1}} \leq \frac{\epsilon}{k\sqrt[k]{M}^{k-1}}$

As $(a_n - a)$ is a null sequence, we can say that $(\sqrt[k]{a_n})$ is convergent.