Sequence (a_n)^2 ->0 implies (a_n) ->0 ?

[kingwinner]

Homework Statement

Let a_n be an infinite sequence.

1) Is it true that a_n² ->0 as n->∞
=> a_n ->0 as n->∞ ?

2) Is it also true that a_n ->0 as n->∞
=> a_n² ->0 as n->∞ ?

Homework Equations

N/A

The Attempt at a Solution

Both SEEM to be true to me, but I am not sure why. What is the simplest way to explain these? I want to understand intuitively.

Any help is appreciated!

 

[HallsofIvy]

Saying that $a_n\to 0$ means that, given any $\epsilon> 0$ there exist N such that if n> N, $|a_n- 0|= |a_n|< \epsilon$.

Saying that $a_n^2\to 0$ means that, given any $\epsilon> 0$ there exist N such that if n> N, $|a_n^2- 0|= |a_n^2|< \epsilon$.

If you know that $a_n\to 0$ then, given any $\epsilon> 0$ there exist N such that if n> N, $|a_n|< \sqrt{\epsilon}$. From that it follows that, for n> N, $|a_n^2|< \epsilon$.

Conversely, if you know that [itxex]a_n^2\to 0[/itex] then, given any $\epsilon> 0$, there exist N such that if n> N, $|a_n^2|< \epsilon^2$. From that it follows that, for n> N, $|a_n|< \epsilon$.

More generally, if $a_n\to a$ and f(x) is continuous in some neighborhood of a, then $f(a_n)\to f(a)$.

 

[kingwinner]

More generally, if $a_n\to a$ and f(x) is continuous in some neighborhood of a, then $f(a_n)\to f(a)$.

Thanks! But I have a question, then.
a_n² ->0 as n->∞
=> a_n ->0 as n->∞

But the square root function is NOT continuous at 0, and it's definitely NOT continuous in a neighbourhood of 0, so the theorem does not apply??
It's weird...

 

[Dick]

Halls said 'more generally'. sqrt is one-sided continuous at 0, from the positive side. If you don't have special theorems about one sided continuity, then just pay attention to his suggestions about an epsilon delta proof. Isn't that what you meant about "SEEM to be true"?

 

[kingwinner]

I have a question about the proof.

Saying that $a_n\to 0$ means that, given ANY $\epsilon> 0$ there exist N such that if n> N, $|a_n- 0|= |a_n|< \epsilon$.

Because of the word "ANY", we can take epsilon here to be $\sqrt{\epsilon}$, and there would still exist some N, right?

 

[Dick]

You can if you know as epsilon approaches zero, sqrt(epsilon) approaches zero. If you are ok with that, go for it.