# Prove that if $\left\{ x_{n}\right\}$ converges then $\left\{ \left(x_{n}\right)^{2}\right\}$ converges.

Good night. I have a problem with this problem. I tried the following:

Proof:

Let $\left\{ x_{n}\right\}$ be a convergent sequence. By definition:

$\mid x_{n}-x\mid<\epsilon$

Then:

$\mid\mid x_{n}\mid-\mid x\mid\mid<\mid x_{n}-x\mid$

$(\mid\mid x_{n}\mid-\mid x\mid\mid)^{2}<(\mid x_{n}-x\mid)^{2}$

$\mid(\mid x_{n}\mid-\mid x\mid)^{2}\mid<(\mid x_{n}-x\mid)^{2}$

$\mid\mid x_{n}\mid^{2}-\mid\left(x_{n}\right)\left(x\right)\mid+\mid x\mid^{2}\mid<(\mid x_{n}-x\mid)^{2}$

Suppose that $x_n\to x$, for some $x\in\mathbb{R}$. Also suppose you have $\epsilon>0$.

Now, because $\{x_n\}$ is a convergent sequence we have that ${x_n}$ is bounded and so $\exists M\in\Bbb {R^+}:\forall n\in\Bbb N \ \ |x_n|\le M$ (and so $x\le M$) and because $x_n\to x$ $\exists n_0\in\Bbb N:\forall n\geq n_0 \ \ |x_n-x|<\frac{\epsilon}{2M}.$

Then, for all $n\geq n_0$ we have that $|(x_n)^2-x^2|=|x_n-x||x_n+x|\le (|x_n|+|x|)|x_n-x|\le 2M|x_n-x|\le 2M\cfrac{\epsilon}{2M}=\epsilon$

and so $(x_n)^2\to x^2$.

• +1 This was the exact proof I had already lined up myself. You beat me by seconds! – NoseKnowsAll Jul 2 '16 at 22:08
• Thanks man, but i don't understand the $2M\mid x_{n}-x\mid\leq2M\frac{\epsilon}{2M}=\epsilon$ why e/2m? – Bvss12 Jul 3 '16 at 1:56
• Yeah. my wrong, check again the edit. – richarddedekind Jul 3 '16 at 13:04

Hint: if $a=\lim_{n\to \infty}a_n$ exists and $b=\lim_{n\to \infty}b_n$ exists, then $\lim_{n\to \infty}a_nb_n$ exists and equals $ab$.

• I think a delta-epsilon hint would be more useful. – ReverseFlow Jul 2 '16 at 21:31
• @ReverseFlow I think using the algebra of limits when applicable saves a lot of work and also sheds more light into the underlying structure. While a delta-epsilon approach is mostly always tailored to a particular example, this kind of technique can be used in much broader settings. – Fimpellizieri Jul 2 '16 at 21:35
• I agree. Based on his work it seems that the delta-epsilon approach is what is being asked. If he is allowed to use lemmas, like the ones you mention, then there really isn't much to do. – ReverseFlow Jul 2 '16 at 21:38
• Yeah, but your hint is as fundamental as the problem above – richarddedekind Jul 2 '16 at 21:39

We have $$\forall { \varepsilon }_{ 1 }>0\quad ,\exists { n }_{ { \varepsilon }_{ 1 } },n\ge { n }_{ { \varepsilon }_{ 1 } }\quad \mid x_{ n }-x\mid <{ \varepsilon }_{ 1 }$$ and we have to show $$\lim _{ n\rightarrow \infty }{ { x }_{ n }^{ 2 } } =a$$ where $a$ is equal to $x=\sqrt { a } ,$ $$\quad \mid { x }_{ n }^{ 2 }-a\mid =\left| \left( { x }_{ n }+\sqrt { a } \right) \left( { x }_{ n }-\sqrt { a } \right) \right| =\left| { x }_{ n }+\sqrt { a } \right| \left| { x }_{ n }-\sqrt { a } \right| <{ \varepsilon }_{ 1 }\left| { x }_{ n }+\sqrt { a } \right| =\\ ={ \varepsilon }_{ 1 }\left| { { x }_{ n }-\sqrt { a } +2\sqrt { a } } \right| <{ \varepsilon }_{ 1 }\left| { x }_{ n }-\sqrt { a } \right| +2={ \varepsilon }_{ 1 }^{ 2 }+2\left| { \varepsilon }_{ 1 }\sqrt { a } \right|$$ and ${ \varepsilon }_{ 2 }={ \varepsilon }_{ 1 }^{ 2 }+2\left| { \varepsilon }_{ 1 }\sqrt { a } \right|$ so we have $\forall { \varepsilon }_{ 2 }>0\quad ,\exists { n }_{ { \varepsilon }_{ 2 } },n\ge { n }_{ { \varepsilon }_{ 2 } }$

$$|{ x }_{ n }^{ 2 }-a\mid <{ \varepsilon }_{ 2 }$$

Suppose that the sequence $\{x_n\}$ converges to $a$. We claim that the squared sequence converges to $a^2$. Note that $$|x_n^2-a^2|=|x_n-a||x_n+a|.$$ We bound $|x_n+a|$ by requiring that $|x_n-a|<1$ so that $|x_n+a|=|x_n-a+2a|<1+2|a|$ . Given $\epsilon>0$ choose $N$ s.t. $n\geq N$ implies that $|x_n-a|<\min(1, \epsilon/(1+2|a|))$ and the result follows.