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Prove that if $a_n$ is a nonnegative sequance and: $$\lim_{n\to \infty} a_n=a$$ then $$\lim_{n\to \infty} \sqrt[5]{a_n}=\sqrt[5]{a}$$

I tried to do this using the definition of the limit of a sequence, but I am not entirely sure whether my reasoning is logically correct (btw. I don't know how to make this left/right arrow):

$$ \lim_{n\to \infty} a_n=a \iff \forall\epsilon>0\ \exists n_0\in N \ \forall n>n_0 \ \ |a_n-a|< \epsilon$$

$$ \lim_{n\to \infty} a_n=a \iff \forall\epsilon>0\ \exists n_0\in N \ \forall n>n_0 \ \ |(\sqrt[5]{a_n}-\sqrt[5]{a})(\sqrt[5]{a_n^4}+\sqrt[5]{a_n^3a}+\sqrt[5]{a_n^2a^2}+\sqrt[5]{a_na^3}+\sqrt[5]{a^4})|< \epsilon$$

$$ \lim_{n\to \infty} a_n=a \iff \forall\epsilon>0\ \exists n_0\in N \ \forall n>n_0 \ \ |(\sqrt[5]{a_n}-\sqrt[5]{a})|< \frac{\epsilon}{\sqrt[5]{a_n^4}+\sqrt[5]{a_n^3a}+\sqrt[5]{a_n^2a^2}+\sqrt[5]{a_na^3}+\sqrt[5]{a^4}}$$

$$\lim_{n\to \infty} \sqrt[5]{a_n}=\sqrt[5]{a} \iff\forall\epsilon_1>0\ \exists n_1\in N \ \forall n>n_1 \ \ |\sqrt[5]{a_n}-\sqrt[5]{a}|< \epsilon_1 $$

Where $\epsilon_1$ is what we received from the previous line

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\Leftrightarrow – Emily Oct 29 '12 at 21:36
@EdGorcenski \iff is more economical than \Leftarrow\Rightarrow or \Leftrightarrow – Pedro Tamaroff Oct 29 '12 at 21:38
Did you try using the continuity of $x^{\frac 1 5}$? – Pedro Tamaroff Oct 29 '12 at 21:39
I am not entirely sure what you mean by "the continuity of $x^\frac{1}{5}$" – Max Oct 29 '12 at 21:43
For $a \ne{0}$ is necessary to prove that denominator ${\sqrt[5]{a_n^4}+\sqrt[5]{{a_n^3}{a}}+\sqrt[5]{a_n^2 a^2}+\sqrt[5]{a_n a^3}+\sqrt[5]{a^4}}$ is bounded by a constant independent of $n$. – M. Strochyk Oct 29 '12 at 21:51
up vote 0 down vote accepted

You are almost there but let me write in a slightly better fashion.

First note that $a \geq 0$. Now split the case when $a=0$ and $a>0$. The case $a=0$ is trivial. Choose $\epsilon = \epsilon_1^5$.

For $a>0$, lower bound $\sqrt[5]{a_n^4}+\sqrt[5]{a_n^3a}+\sqrt[5]{a_n^2a^2}+\sqrt[5]{a_na^3}+\sqrt[5]{a^4}$ by $$\sqrt[5]{(a- \epsilon)^4}+\sqrt[5]{(a- \epsilon)^4}+\sqrt[5]{(a- \epsilon)^4}+\sqrt[5]{(a- \epsilon)^4}+\sqrt[5]{(a- \epsilon)^4} = 5 (a - \epsilon)^{4/5}$$ and choose $\epsilon$ such that $$\dfrac{\epsilon}{5 (a - \epsilon)^{4/5}} < \epsilon_1$$ and now you are done.

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but how do we know that $a_n$ is greater than $a-\epsilon$? isn't it true only when $a_n<a$ because then we get $|a_n-a|<\epsilon \iff a_n-a>-\epsilon \iff a_n>a-\epsilon$ otherwise $a_n<a+\epsilon$. What am i missing? – Max Oct 29 '12 at 23:00
If $\vert a_n - a \vert < \epsilon$, then $a- \epsilon < a_n < a + \epsilon$. – user17762 Oct 29 '12 at 23:02
Oh right. Guess it's about time i go sleep. Thanks. – Max Oct 29 '12 at 23:06

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