How to prove that if $a_n$ converges to $L$ then $(a_n)^{1/m}$ converges to $L^{1/m}$ I'm trying to prove the following:

If $a_n$ converges to $L$, then $(a_n)^{1/m}$ converges to $L^{1/m}$.

I could use some help, thanks.
 A: First, I'm assuming that $m$ is a natural number as this is usually the first thing you try to prove about expressions like this (which is what you seem to be going for).
Assume by way of contradiction that $a_n^{1/m}$ does not converge to $L^{1/m}$. Then there exists an infinite subsequence of $a_n^{1/m}$ (let us call it $a_{n_i}^{1/m}$) and $\epsilon_0>0$ such that
$$|a_{n_i}^{1/m}-L^{1/m}|>\epsilon_0$$
for all $i\in \mathbb{N}$. Additionally, since $a_{n_i}^{1/m}\neq L^{1/m}$, an infinite  subsequence of this subsequence must always be greater than or less than $L^{1/m}$. Assume the former. That is, there is a subsequence $a_{n_{i_j}}^{1/m}$ such that
$$a_{n_{i_j}}^{1/m}-L^{1/m}>\epsilon_0$$
for all $j\in \mathbb{N}$. For the sake of notation, call this subsequence $a_{n_{i_j}}^{1/m}=b_j^{1/m}$. Note that since $b_j$ is a subsequence of $a_n$, $b_j$ converges to $L$ from above ($b_j>L$ for all $j$). Now, for all $j\in\mathbb{N}$
$$b_j^{1/m}-L^{1/m}>\epsilon_0$$
$$b_j^{1/m}>\epsilon_0+L^{1/m}$$
$$b_j>(\epsilon_0+L^{1/m})^m.$$
Using the binomial theorem, this can be expanded to
$$b_j>\sum_{s=0}^m\binom{m}{s}\epsilon_0^{m-s}L^{s/m}=L+\sum_{s=0}^{m-1}\binom{m}{s}\epsilon_0^{m-s}L^{s/m}$$
$$b_j-L>\sum_{s=0}^{m-1}\binom{m}{s}\epsilon_0^{m-s}L^{s/m}$$
$$|b_j-L|>\sum_{s=0}^{m-1}\binom{m}{s}\epsilon_0^{m-s}L^{s/m}$$
However, the expression on the right is a positive number that doesn't depend on $j$. This contradicts the fact that $b_j$ approaches $L$. We conclude that $a_{n}^{1/m}$ converges to $L^{1/m}$.
Now, if instead there is an infinite subsequence $a_{n_{i_j}}^{1/m}$ such that for some $\epsilon_0>0$
$$L^{1/m}-a_{n_{i_j}}^{1/m}>\epsilon_0$$
for all $j\in\mathbb{N}$. Again, for the sake of notation, call this subsequence $a_{n_{i_j}}^{1/m}=b_j^{1/m}$. Note that since $b_j$ is a subsequence of $a_n$, $b_j$ converges to $L$ from below ($b_j<L$ for all $j$). We proceed in much the same way as above
$$L^{1/m}-b_j^{1/m}>\epsilon_0$$
$$L>(\epsilon_0+b_j^{1/m})^m=\sum_{s=0}^m\binom{m}{s}\epsilon_0^{m-s}b_j^{s/m}=b_j+\sum_{s=0}^{m-1}\binom{m}{s}\epsilon_0^{m-1-s}b_j^{s/m}$$
$$L-b_j>\sum_{s=0}^{m-1}\binom{m}{s}\epsilon_0^{m-1-s}b_j^{s/m}=\sum_{s=1}^{m-1}\binom{m}{s}\epsilon_0^{m-1-s}b_j^{s/m}+\epsilon_0^{m-1}>\epsilon_0^{m-1}$$
$$|L-b_j|>\epsilon_0^{m-1}$$
Again, the expression on the right is a positive number that doesn't depend on $j$. This contradicts the fact that $b_j$ approaches $L$. We conclude that $a_{n}^{1/m}$ converges to $L^{1/m}$.
Having exhausted both cases, we conclude if $a_n\to L$, then $a_n^{1/m}\to L^{1/m}$.
A: For simplicity, assume that $a_n\rightarrow L>0$ and $m\in \mathbb N.$ One needs to prove that $$\forall \epsilon>0,\exists N~{\rm such ~that ~}|a_n^{\frac 1 m}-L^{\frac 1 m}|<\epsilon,\forall n>N.$$ For this, let $\epsilon>0$ be given. Since $a_n\rightarrow L,$ $\exists N_1$ such that $a_n>\frac 1 2L,\forall n>N_1.$ Hence $$|a_n-L|=|(a_n^{1/m})^m-(L^{1/m})^m|$$
$$=|a_n^{1/m}-L^{1/m}|\cdot |(a_n^{1/m})^{m-1}+(a_n^{1/m})^{m-2}L^{1/m}+\cdots+(L^{1/m})^{m-1}|$$
$$\geq |a_n^{1/m}-L^{1/m}|\cdot m\left(\frac L 2\right)^{\frac {m-1}m}~{\rm if~}n>N_1.$$  Hence $$|a_n^{1/m}-L^{1/m}|\leq \frac 1m\left(\frac 2 L\right)^{\frac{m-1}m}|a_n-L|,\forall n>N_1.$$ Since $a_n\rightarrow L$, there exists $N_2$ such that $$|a_n-L|<m\left(\frac L 2\right)^{\frac {m-1}m}\epsilon,\forall n>N_2.$$ Letting $N=\max(N_1,N_2),$ it follows that $$|a_n^{1/m}-L^{1/m}|< \frac 1m\left(\frac 2 L\right)^{\frac{m-1}m}\cdot m\left(\frac L 2\right)^{\frac {m-1}m}\epsilon=\epsilon,\forall n>N,$$ as required. QED
A: If $a_n\rightarrow 0$ this is obvious, hence take $a_n\rightarrow L>0$. 
Since $a_n\rightarrow L,$ $\exists n_0$ such that $a_n>\frac 1 2L,\forall n>n_0.$
It follows $\frac{a_n-L}L > -1, \forall n>n_0 $ and using Bernoulli Inequality we get
$$a^{1/m}(1+\frac{a_n-L}{na_n}) \leq a^{1/m}(1+a^{1/m}\frac{a_n-L}L)^{1/m} \leq a^{1/m}(1+\frac{a_n-L}{nL})$$ 
