$a_n\ge0, l\ge0$If $\lim_{n\rightarrow\infty}a_n=l$ Then $\lim_{n\rightarrow\infty}a^2_n=l^2$
Proving $\lim_{n\rightarrow\infty}a_n=l\Rightarrow \lim_{n\rightarrow\infty}a^2_n=l^2$
-
0Oh, I didn't seem to understand what the problem was till now. Right on it. – 2012-11-14
3 Answers
Here is a trick, which will help you solve this problem.
$a_n^2 - l^2 = (a_n - l)^2 + 2l(a_n - l)$ using this you should be able to show that $a_n^2 - l^2 \rightarrow 0$.
EDIT: To elaborate further, by the triangle inequality we now know that $\lvert a_n^2 - l^2\rvert \leq \lvert (a_n - l)^2 \rvert + \lvert 2l(a_n-l)\rvert \hspace{10mm} (1)$ So we just need to show that each of the two terms on the right get small as $n$ gets large. You should be able to make both of them small by using the fact that $a_n \rightarrow l$.
FINAL EDIT: To see a complete proof. Fix $\epsilon > 0$. Choose $N_0$ so that for $n\geq N_0$, $\lvert a_n - l\rvert \leq \sqrt{\epsilon};$ and $N_1$ large so that for $n\geq N_1$, $\lvert a_n - l \rvert \leq \frac{1}{\lvert 2 l \rvert} \epsilon.$Then if $n\geq\max(N_0,N_1)$, the inequality (1) above gives us $\lvert a_n^2 - l^2 \rvert \leq \epsilon + \epsilon = 2 \epsilon$ and since $\epsilon$ was arbitrary we conclude that $\lvert a_n^2 - l^2\rvert \rightarrow 0$.
-
0@DevenWare Can I just ask of you one last thing? Can you please $c$omplete your proof, so I'll see how it should look? Ma$n$y thanks. – 2012-11-14
Since $a_n \to l$, there is a $n_1$ such that $|a_n-l| < 1$ for $n > n_1$.
So, for $n > n_1$, $|a_n^2-l^2| = |(a_n-l)(a_n+l)| < |a_n-l|(2|l|+1) $.
Let $n_2$ be such that $n_2 > n_1$ and $|a_n-l| < \epsilon/(2|l|+1)$ for $n > n_2$. Then $|a_n^2-l^2| < \epsilon$ for $n > n_2$, so $\lim_{n \to \infty} a_n^2 = l^2$.
Suppose: $\lim_{n\rightarrow\infty}a_n^2\not\rightarrow l^2$ Exists an $\epsilon>0$ such that: $|n^2-l^2|\ge\epsilon \Leftrightarrow l^2-\epsilon\ge n^2 \ge l^2 + \epsilon$
Left side of the equation is false, because if it were true, $\epsilon<0$. Which is false.
Right side is false because:$n^2-l^2\ge\epsilon \Leftrightarrow -l^2 \ge \epsilon-n^2 \ge 0$
Which is again false, because $n,l\ge0$
EDIT: If this is wrong, please let me know why.