If a sequence converges to a, prove the squares of its terms converge to a^2 Given:
$$a_n\to a$$ 
 That means :
For all $\epsilon^*:=\frac{\epsilon }{2|a|+1} $  we can find an N such that for all n>N $\left |a_n-a  \right |<\frac{\epsilon }{2|a|+1}$
I have to prove
$$(a_n)^2 \to a^2 $$
At first we guess 
$\left |a_n+a  \right |=\left |a_n-a+a+a  \right |<\frac{\epsilon }{2|a|+1 }+2|a | $
Now we can prove 
$$(a_n)^2 \to a^2 $$
$\left |(a_n)^2-   a^2 \right |=\left |(a_n-a)(a_n+a)\right |=\left |a_n-a  \right |\left |a_n+a  \right |<\frac{\epsilon }{2|a|+1}(\frac{\epsilon }{2|a|+1}+2|a |)<\epsilon^2+\epsilon$
I changed my epsilon and i think it works now??
any suggestion ?  
 A: Given $\varepsilon > 0$, let $\eta := \min\{1, \frac{\varepsilon}{(1 + 2|a|)}\} > 0$.
So since $a_n \to a$, there exists a positive integer $N$ such that $|a_n - a| < \eta$ for all $n \ge N$.
Now if $n \ge N$, then in particular, $|a_n - a| < 1$, which implies
$$
|a_n + a| = |(a_n - a) + 2a| \le |a_n -a| + 2|a| < 1 + 2|a|
$$
Thus
$$
|a_n^2 - a^2| = |a_n + a||a_n - a| < (1 + 2|a|) \frac{\varepsilon}{1 + 2|a|} = \varepsilon$$
Since $\varepsilon$ was arbitrary, $~a_n^2 \to a^2$.
A: I am not sure if this works as a solution, so someone should check this:
Since $a_n \rightarrow a$ we know that for any $\epsilon > 0$ there exists an $N$ such that for all $n \geq N$ we have
$$ \lvert a_n - a \rvert \leq \lvert x_n\rvert + \lvert a\rvert < \epsilon $$
$$ \lvert a_n - a \rvert^2 \leq \left( \lvert x_n\rvert + \lvert a\rvert \right)^2 < \epsilon^2 $$
It follows that
$$ \left( \lvert x_n\rvert + \lvert a\rvert \right)^2 = x_n^2 + 2x_na + a^2 < \epsilon^2 $$
$$ \lvert x_n^2 - a^2 \rvert < x_n^2 + 2x_na + a^2 < \epsilon^2$$
So if $x_n$ converges to a then $ x_n^2 $ converges to $a^2$
