1
$\begingroup$

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

  • 0
    but since it is true that $a_n \le a$ we can say that $\sqrt[5]{a_n^4}+\sqrt[5]{a_n^3a}+\sqrt[5]{a_n^2a^2}+\sqrt[5]{a_na^3}+\sqrt[5]{a^4} \le \sqrt[5]{a^4}+\sqrt[5]{a^4}+\sqrt[5]{a^4}+\sqrt[5]{a^4}+\sqrt[5]{a^4}=\sqrt[5]{5\cdot a^4}$, which is a constant independent of n, which also, considering the last line, ends the proof... yes?2012-10-29

1 Answers 1

0

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.

  • 0
    Oh right. Guess it's about time i go sleep. Thanks.2012-10-29