An equivalence? An equivalence??

[evagelos]

An equivalence??

The two statements :

1)A sequence diverges to infinity .


2)The sequence is unbounded from above.

Are equivalent??

If yes how do we prove it?

If no give a counter example

 

[LCKurtz]

What have you tried? I would try proving it both directions and if I have trouble with one or both directions start looking for a counterexample.

 

[evagelos]

I can prove that :

Divergence to infinity implies unboundness ,but not the opposite .

Any ideas for a good counter example??

 

[jgens]

What about the sequence \{a_n\} defined by a_n = n\sin{(n)}? It certainly isn't bounded above and I wouldn't say that it diverges to infinity, but rather that its limit as n becomes arbitrarily large doesn't exist.

 

[evagelos]

Very difficult to say

Can you prove:

1) nsin(n) is not bounded from above ,and

2) does not diverge to infinity?

 

[jgens]

Whether or not I can prove them is not the issue here since you're the one trying to find a counter-example for a sequence which is unbounded above but does not diverge to infinity. Why don't you show us what you've tried?

 

[LCKurtz]

You might find n\sin{\frac {n\pi}2} easier to work with.

 

[evagelos]

Whether or not I can prove them is not the issue here since you're the one trying to find a counter-example for a sequence which is unbounded above but does not diverge to infinity. Why don't you show us what you've tried?

When i say "can you prove it" ,it means that :i cannot prove it and i am asking you if can you prove it for me .

You suggested a counter example that i cannot justify as the right one and i am asking for your help

Is that clear now

 

[l'Hôpital]

When i say "can you prove it" ,it means that :i cannot prove it and i am asking you if can you prove it for me .

You suggested a counter example that i cannot justify as the right one and i am asking for your help

Is that clear now

This is your problem, not his. We are here to give hints and help, and not do your problem.

His suggestion works, but it's slightly odd. There are simpler ones.

 

[HallsofIvy]

You might find n\sin{\frac {n\pi}2} easier to work with.

This suggestion is excellent! Try it, evagelos. Just write down a few dozen values in that sequence and see what happens.

 

[evagelos]

This suggestion is excellent! Try it, evagelos. Just write down a few dozen values in that sequence and see what happens.

If i am not mistaken the sequence is the the following:

1,0,-3,0,5,0,-7,0,9,0,-11,0,13,0,-15,0,17........

But since we want the sequence to be bounded from above we must consider the sequence

|nsin\frac{n\pi}{2}|

Now to prove that this the right counter example we must prove that:

1) For all ε>0 there exists a natural No n , and |nsin\frac{n\pi}{2}|>\epsilon

Which means that the sequence is not bounded from above .(intuitively this looks correct)


2) There exists an ε>0 ,such that for all natural Nos k there exists a natural No n\geq k ,such that ;

|nsin\frac{n\pi}{2}|\leq\epsilon

Which means that the sequence does not diverge to infinity.

You agree with the above??

 

[HallsofIvy]

Yes, that is correct.

 

[evagelos]

Yes, that is correct.

What n would you suggest for either cases ??

 

[rasmhop]

What n would you suggest for either cases ??

If you want it to be large try to find a large n such that sin(n\pi/2) is 1 or -1. Here large means >epsilon.

If you want it small try to find an n such that sin(n\pi/2)=0.

From your sequence you should be able to guess what kind of n makes it large and what kind of n makes it small.