If $(a_n^2)$ converges to $0$, then $(a_n)$ converges to $0$

Question

I had a problem with this I could prove it by contradiction but, I wonder if you can do only algebraically can be done only with inequality? $$\left( {a_n } \right)^2 \to 0 \qquad\Longrightarrow \qquad \left( {a_n } \right) \to 0 $$ Thanks!

Answer 1

Here's an elementary proof:

By definition, $x_n \to 0$ iff for any $\epsilon > 0$ there exists an $m$ such that $|x_n| < \epsilon$ whenever $n > m$.

Assume $(a_n)^2 \to 0$. Then for any $\epsilon > 0$, we know that $\epsilon^2 > 0$, and thus there exists an $m$ such that $|(a_n)^2| < \epsilon^2$ whenever $n > m$.

But, since $|(a_n)^2| = |a_n|^2$, we can take square roots to see that $|(a_n)^2| < \epsilon^2 \implies |a_n| < \epsilon$.

Thus we have shown that, for any $\epsilon > 0$, there exists an $m$ such that $|a_n| < \epsilon$ whenever $n > m$, and so $a_n \to 0$. $\square$

Answer 2

Hint: First, $a_n\rightarrow 0$ if and only if $|a_n|\rightarrow 0$. Then use the continuity of the square root function on $[0,\infty)$.

Answer 3

Very similar, in fact equivalent, to Ilmari's answer (and probably his is better didactically), but ..

If {$a_n$} doesn't tend to zero then for some sufficiently small constant $m > 0$ we have $m < |a_n|$ for all $n > N$; but then $0 < |a_n| = |a_n^2 / a_n| < a_n^2 / m$, whereas the RHS tends to zero, which is a contradiction.

Answer 4

The following tools are useful for proving the result:

• $a_n\to 0$ if and only if $|a_n|\to 0$
• If $f$ is continuous on the interval $[x,y]$ (where continuity on a closed interval means one sided continuity at the endpoints), and if $a_n\in [x,y]$ for sufficiently large $n$, then $\lim f(a_n)=f(\lim a_n)$.

The first statement is easily proved with an $\epsilon-\delta$ proof (essentially, it is just the definition of convergence). The second statement is a strengthening of the standard result that if $f(x)$ is continuous at $L$ and $a_n\to L$, then $f(a_n)\to f(L)$. Having the stronger statement is necessary if you want to apply the square root function to a sequence (as you need to know that the sequence is positive for things to make sense, and you need to know that having one sided continuity at $0$ doesn't cause problems).

Using the tools, we have that $$a_n^2\to 0 \Leftrightarrow |a_n|\to 0 \Leftrightarrow a_n\to 0.$$

Answer 5

If you mean the squeeze theorem when you ask about doing it by inequalities, that does not work. The problem is that $|a_n^2| < |a_n|$ when $|a_n| < 1$, and if $(a_n^2)$ goes to $0$ then $|a_n|$ will in fact be less than $1$ after some point. So you cannot prove the inequality necessary to use the squeeze theorem, which would be $|a_n| < |a_n^2|$.