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Suppose $\{a_n\}$ converges to $a \in \mathbb{R}$. Prove $\{a_n^4\}$ converges to $a^4$.

Proof.

Suppose $\{a_n\}$ converges to $a \in \mathbb{R}$.

This means, $\forall \epsilon > 0, \exists N > 0, s.t, \forall n\in \mathbb{N}, if \text{ } n>N, then \text{ } |a_n - a| < \epsilon$ .

I want to show that $\{(a_n)^4\}$ converges to $a^4 \in \mathbb{R}$.

This means, $\forall \epsilon > 0, \exists N > 0, s.t, \forall n\in \mathbb{N}, if \text{ } n>N, then \text{ } |(a_n)^4 - a^4| < \epsilon$ .

Pick $N$ = ___ $ > 0$


I don't know how I can use the given information that the original sequence converges to prove the problem. Any hints?

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    $\begingroup$ If $f$ is a continuous function, then $f(\lim\limits_n x_n) = \lim\limits_n f(x_n)$, so you just need to show that $f(x) = x^4$ is continuous. $\endgroup$ – D_S Mar 15 '17 at 1:34
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    $\begingroup$ To show $f$ is continuous, you can use the fact that $f(x) = x$ is continuous, and the product of two continuous functions is continuous. $\endgroup$ – D_S Mar 15 '17 at 1:34
  • $\begingroup$ Sorry I was busy I didn't have time to go on, please remove the downvote if that was you $\endgroup$ – K Split X Mar 18 '17 at 16:17
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We know that the sequence $\{a_n\}$ has to be bounded and so, we can find $M>0$ such that $|a_n|\leq M$ for all $n\in\Bbb N$. Note that $$x^4-y^4=(x-y)(x^3+x^2y+xy^2+y^3).$$ Let $\epsilon>0$. Then we can choose $N\in\Bbb N$ such that for all $n\geq N$, we have $$|a_n-a|<\frac{\epsilon}{M^3+M^2|a|+Ma^2+|a^3|}.$$ Hence, for all $n\geq N$, we get $$\begin{align} |a_n^4-a^4|&=|a_n-a|\cdot|a_n^3+a_n^2a+a_na^2+a^3|\qquad\text{then use the Triangle Inequality to get}\\ &\leq |a_n-a|\cdot(|a_n|^3+|a_n|^2a+|a_n|a^2+|a^3|)\\ &<\frac{\epsilon}{M^3+M^2|a|+Ma^2+|a^3|}\cdot \big(M^3+M^2|a|+Ma^2+|a^3|\big)=\epsilon. \end{align}$$

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  • $\begingroup$ I don't understand how you got the line: $$|a_n-a|<\frac{\epsilon}{M^3+M^2|a|+Ma^2+|a^3|}.$$ $\endgroup$ – K Split X Mar 18 '17 at 17:27
  • $\begingroup$ Why the $M$ in the deonominator? $\endgroup$ – K Split X Mar 18 '17 at 17:28
  • $\begingroup$ Let me put some ideas now. Suppose we let $$K=M^3+M^2|a|+Ma^2+|a^3|$$ and we have $$|a_n^4-a^4|\leq |a_n-a|\cdot K.$$ What would be your bound for the factor $|a_n-a|$ so that $$|a_n^4-a^4|<\epsilon?$$ $\endgroup$ – Juniven Mar 18 '17 at 17:50
  • $\begingroup$ It would be $$\dfrac{\epsilon}{K}$$ Right? $\endgroup$ – K Split X Mar 18 '17 at 17:51
  • $\begingroup$ Good. Then use $\frac{\epsilon}{K}$ (instead of $\epsilon$) in the definition of $$\lim_{n\to\infty}a_n=a.$$ $\endgroup$ – Juniven Mar 18 '17 at 17:52

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