Does Convergence of (Sequence)^2 Imply Original Sequence Converges?

[cloud360]

If some (sequence)^2 converges does that mean the original (sequence) always converges?

(using mobile version)

attempted solution:

All i know is if {an} converges to L==> {an}^2 converges to L^2

 

[cloud360]

any help please, i really need. would be very grateful

 

[spamiam]

I think this is false. Try using a sequence that oscillates between values for a counterexample.

 

[cloud360]

but if its true that if {an} converges to L==> {an}^2 converges to L^2, then how can it be false?

going backwards means {an}^2 converges to L^2{an} converges to L


why am i wrong?

 

[Kreizhn]

Consider the sequence $a_n = (-1)^n$ as motivation to what spamiam said. Clearly $a_n^2 = (-1)^{2n} = 1$ is the constant sequence and so converges. However, the original clearly does not converge as it oscillates between -1 and +1 indefinitely.

Edit: Fixed wording

 

[Kreizhn]

As for the statement as how the other could be wrong, this is called the converse and is logically not equivalent to the original statement.

All dogs are mammals is true. The converse is that all mammals are dogs, and this is not true.

 

[spamiam]

The converse of a true statement is not always true. Take the classic Alice In Wonderland example: I always breathe when I sleep, but this doesn't mean I always sleep when I breathe.

In other words, "P implies Q" does not mean that "Q implies P".

In any case, there's a very simple sequence that serves as a counterexample: try using my hint.

 

[nonequilibrium]

Kreizhn: I understand you're trying to help (which is great), but giving the exact answer is not the idea of PF, as that let's the OP simply copy the answer given onto their sheet without possibly understanding it/going through the thought procedure

 

[Kreizhn]

If you are not satisfied with this boring example, try using the Harmonic series instead. Define
[tex] a_n = \frac1n [/itex]
It is very well known that this series does not converge. However,
$$a_n^2 = \frac1{n^2}$$
does converge.

 

[cloud360]

thanks so much :)

 

[Kreizhn]

Kreizhn: I understand you're trying to help (which is great), but giving the exact answer is not the idea of PF, as that let's the OP simply copy the answer given onto their sheet without possibly understanding it/going through the thought procedure

I know what you're trying to say, but at this point the answer has already been made clear. The issue is no longer that the answer has been given, but instead the motivation should be to clearly demonstrate why the solution is the way it is.