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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!

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...and $a_n$ is? –  J. M. Jul 24 '11 at 17:09
    
any sequence with values in the real numbers –  Daniel Jul 24 '11 at 17:11
    
Can be done by contradiction too - suppose the sequence doesn't tend to zero, what does that imply about the squares (there must be elements of the sequence bounded away from zero). –  Mark Bennet Jul 24 '11 at 20:55

5 Answers 5

up vote 10 down vote accepted

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 $a_n \to 0$. $\square$

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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 $|a_n| < |a_n^2|$.

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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 $$a_n^2\to 0 \Leftrightarrow |a_n|\to 0 \Leftrightarrow a_n\to 0.$$

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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.

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Dear John, If $a_n$ does not tend to $0$, it need not be that $|a_n| > m$ for all $n > N$; rather, this needs to hold just for (some $m > 0$ and) infinitely many values of $n$. Regards, –  Matt E Jul 24 '11 at 21:58
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@John R Ramsden: It is good to avoid division if possible, since it can be a treacherous operation. For example, $a_n^2$ could be $0$ when $n$ is even, making $a_n^2/a_n$ meaningless. We could in this problem deal with such $n$ separately, but that is a complication. –  André Nicolas Jul 24 '11 at 23:34

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

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Sorry to ask: What about this hint is "doing it algebraically and only with inequality"? :) Honestly, I don't think using continuity of the square root is very appropriate here. –  t.b. Jul 24 '11 at 22:26

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