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This has been answered before on this site but I don't really understand those answers enough so I am asking again.

Prove that if $\{a_n\}$ converges to $l\in\mathbb{R}$, then $\{a_n^2\}$ converges to $l^2$.

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

This means $\forall\epsilon>0,\exists N_1>0, s.t,\forall n\in\mathbb{N},\text{ if } n>N_1,\text { then } |a_n-l|<\epsilon$

I want to show that $\forall\epsilon>0,\exists N_2>0, s.t,\forall n\in\mathbb{N},\text{ if } n>N_2,\text { then } |a_n^2-l^2|<\epsilon$

Note: $a_n^2$ is the entire sequence, squared, evidently the same as $(a_n)^2$.

Consider $|a_n^2-l^2| = |a_n-l||a_n+l|$.

If $n>N_1$, then $|a_n-l||a_n+l|<\epsilon|a_n+l|$.

How do I continue from here?

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  • $\begingroup$ You might decide to prove the more general $a_n \rightarrow L_a$ and $b_n \rightarrow L_b$ implies $a_n b_n \rightarrow L_a L_b$, from which this immediately follows with $b_n = a_n$. $\endgroup$ Apr 14, 2017 at 22:39
  • $\begingroup$ Or, you could prove that if $f$ is continuous at $L$ and $a_n \to L$, then $f(a_n) \to f(L)$ $\endgroup$
    – user169852
    Apr 14, 2017 at 22:45

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Same approach but slightly shorter.

Suppose $\lim_{n\to\infty}a_n=l$. Let $B=\sup\{\vert a_n+l\vert\}$ for all $n$.

Let $\epsilon>0$ and let $N>0$ such that if $n>N$ then

$$ \vert a_n-l\vert<\frac{\epsilon}{B}$$

Then

$$ \vert a_n-l\vert\cdot\vert a_n+l \vert<\frac{\epsilon}{B}\cdot B$$

$$ \vert a_n^2-l^2\vert<\epsilon$$

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Maybe you forgot that you posted a related question found in here. Let us follow the same technique found in there.

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 $n\in\Bbb N$.

Let $\epsilon>0$. Then we can choose $N\in\Bbb N$ such that for all $n\geq N$, we have $$|a_n-l|<\frac{\epsilon}{M+|l|}.$$ Hence, for all $n\geq N$, we get $$\begin{align} |a_n^2-l^2|&=|a_n-l|\cdot|a_n+l|\qquad\text{then use the Triangle Inequality to get}\\ &\leq |a_n-l|\cdot(|a_n|+|l|)\\ &<\frac{\epsilon}{M+|l|}\cdot \big(M+|l|\big)=\epsilon. \end{align}$$

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  • $\begingroup$ Yeah you know I remembered that after people answered, and I ended up doing the same thing you answered! :) $\endgroup$
    – K Split X
    Apr 15, 2017 at 0:51
  • $\begingroup$ I was just wondering if there was any other way of doing it rather then assuming that every convergent sequence is bounded... for example if this wasn't a sequence question but rather a limit question, how would your proof differ? $\endgroup$
    – K Split X
    Apr 15, 2017 at 0:54
  • $\begingroup$ @K Split X That depends on the question. Let us see then if you have a particular problem. $\endgroup$ Apr 15, 2017 at 0:57
  • $\begingroup$ @K Split X In my opinion, you can not avoid to use the concept of boundedness if you want to prove limit of sequences by using the definition. $\endgroup$ Apr 15, 2017 at 1:02
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Note than $a_n$ converges, so it is bounded, so $\exists M > 0$, $|a_n| \le M$.

Therefore, to be more elegant you could have chosen $\frac{\epsilon}{2M}$, so that, $$ \forall \epsilon \exists N \forall n \ge N , \quad |a_n - l | \le \frac{\epsilon}{2M} $$ which is true since $\frac{\epsilon}{2M}>0$.

By the triangle inequality, $$ |a_n+l| \le |a_n| + |l| \le 2M. $$

Finally, according to what you have written $$ \forall \epsilon > 0 \exists N\forall n \ge N, \quad |a_n^2 - l^2 | < \epsilon $$ which concludes your proof.

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  • $\begingroup$ I'm not sure how you got the $2$ involved? $\endgroup$
    – K Split X
    Apr 14, 2017 at 22:48
  • $\begingroup$ Since $|a_n| \le M$ and $|l| \le M$. Because if the sequence is bounded, the limit of the sequence is bounded by the same constant by the "squeeze/policeman theorem". $\endgroup$ Apr 14, 2017 at 22:50
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At the end of the day, you need to bound something of the form $|a_n^2 - l^2|$, right? Factoring that, it's just $|(a_n - l)(a_n + l)|$. You have a bound on the first term from the fact that $a_n \to l$. For the second part, that's bounded absolutely.

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