sequences and series - If $lim_{n to infty} a_n=L$, then $lim_{n to infty} a_n^{1/k}=L^{1/k}$. {Edited]

I need some help proving the following statement:

[EDIT]: forgot to mention that:

$a_n \ge 0$ , $L \ge 0$.

If the limit of a sequence $\lim_{ \ n \to \infty} a_n=L$,
then, for any $k \in \mathbb{N}$, the limit of the sequence $\lim_{n \to \infty} \sqrt[k]{a_n} = \sqrt[k]{L}$.

I already proved that every sequence $\{a_n\}^k$ converges to $L^k$ by induction step,

but when I tried the same for $\{1/k\}$, it just doesn't work.

Answer

Let $a_{n} \geq 0$ for all $n \geq 1$, so that, if $a_{n} \to L$, then $L \geq 0$. This allows $a_{n}^{1/k}$ to be meaningful for all $n,k \geq 1$.

Let $k \geq 1$. If $L=0$, then the statement is obvious; for, we have $a_{n} < \varepsilon$ iff $a_{n}^{1/k} < \varepsilon^{1/k}$. Suppose $L > 0$. Then there is some $N_{1} \geq 1$ such that $a_{n} > 0$ for all $n \geq N_{1}$. If $n \geq N_{1}$, then

$$|a_{n}^{1/k} - L^{1/k}| = \frac{|a_{n}-L|}{ \sum_{j=0}^{k-1}a_{n}^{\frac{j}{k}}L^{\frac{k-1-j}{k}}}.$$
On the other hand, there is some $N_{2} \geq N_{1}$ such that $n \geq N_{2}$ implies
$|a_{n} - L| < L/2$, implying $L/2 < a_{n}$, implying that

$$\sum_{j=0}^{k-1} a_{n}^{\frac{j}{k}} L^{\frac{k-1-j}{k}} > L^{\frac{k-1}{k}}/ \sum_{j=0}^{k-1}{2^{\frac{j}{k}}} =: M.$$
If $\varepsilon > 0$, then there is some $N_{3} \geq N_{2}$ such that $n \geq N_{3}$ implies $|a_{n}-L| < M\varepsilon$.
We have proved that, for every $\varepsilon > 0$, there is some $N \geq 1$ such that $n \geq N$ implies $|a_{n}^{1/k} - L^{1/k}| < \varepsilon$.