# Please help me with sequences

1. Sep 9, 2010

### MelanieSwan

1. The problem statement, all variables and given/known data

Give an example of a sequence {an} whose value is 7 for infinitely many values of n, but which does not converge to 7

2. Relevant equations

3. The attempt at a solution

I tried to think about such a sequence but cannot come up with any that satisfies that its value is 7 for infinitely many values of n yet does not converge to 7. The only sequence I can think of is the constant sequence, an =7 but it does converge to 7 so...

I hope I can get some help here. Thanks a lot.

2. Sep 9, 2010

### jgens

From the way you've stated the problem, the sequence {an} need not converge to any real value. That should help.

3. Sep 9, 2010

### Staff: Mentor

The sequence {1, -1, 1, -1, 1, -1, ...} has infinitely many values that are 1, but doesn't converge to 1.

4. Sep 9, 2010

### MelanieSwan

Oh thanks very much. I think I misunderstand the question as "with all values of n". That's why i got stuck. Thanks for your answers :)

5. Sep 9, 2010

### MelanieSwan

so it should be:

an = 7(-1)^n

^^

6. Sep 9, 2010

### jgens

Yes. Good job!

7. Sep 9, 2010

### MelanieSwan

Thank you all very much.

As you my guess I started learning about sequences and haven't got used to them yet. One of the questions I raised during the lecture was that if a sequence {an} converges with n=2 to n=infinity, does it converge with n=1 to n=infinity. Obviously it may not since look at the sequence {an} = 3n/(n-1) for example. This sequence converges with n=2 to n= infinity but with n=1 the sequence is undefined. But how about if a sequence {an} converges with n=1 to n= infinity then does it converge with n=2 to n= infinity. Intuitively I think it's absolutely a YES, but I don't come up with a good proof for myself yet. Could you help?

8. Sep 9, 2010

### jgens

The sequence {an} converges to L means: For every e > 0 we can find an N > 0 such that if n > N, then |an-L| < e. Can you figure out how to make this work?

9. Sep 9, 2010

### MelanieSwan

ok i thought about that but how can I put n=1 to n=infinity in to work using this?

10. Sep 9, 2010

### jgens

You just need to recognize that by choosing a large enough N, it doesn't matter.

Edit: You could define two sequences {an} and {bn} such that an = bn for n > 1 and show that they converge to the same limit if you want to get really formal I suppose.

11. Sep 9, 2010

### Staff: Mentor

A sequence is a function whose domain is the integers (or some subset of them). If {an} is defined by 3n/(n - 1), then pretty obviously, the sequence is not defined for n = 1. This doesn't have anything to do with whether the sequence converges. Convergence has to do with the behavior of the terms in the sequence for large values of n.

12. Sep 9, 2010

### MelanieSwan

this is what I am confused about. Obviously 3n/(n-1) converges at 3, but is it right to say that it converges with n=1 to n=infinity? I think it's must be that 3n/(n-1) converges at 3 with n=2 to n= infinity?

:S

13. Sep 9, 2010

### Staff: Mentor

{3n/(n - 1)} converges to 3. For all the terms to be defined, n must be >= 2. Again, you are confusing the ideas of the domain of a sequence with whether it converges.

Last edited: Sep 9, 2010
14. Sep 12, 2010

### MelanieSwan

sorry I have been away and didn't have the chance to check this thread out. Yes, you're right. I'm confused about that. So that means when n=1, 3n/(n-1) is not defined so we cannot use this sequence as a counter example?

Could you give me some hint on how to work out this problem?

15. Sep 12, 2010

### Staff: Mentor

The sequence {3n/(n - 1)} is NOT a counterexample, if I understand what you're asking, which I think is this:
There are essentially two parts to a sequence: the first few (finitely many) terms, and all the rest (the tail). What happens in the first few terms is completely immaterial to whether the sequence converges (or not). Convergence or divergence are completely controlled by what happens in the tail; i.e., what happens for large n.

In answer to your question, if a sequence $$\{a_n\}_{n = 2}^{\infty}$$ converges, then so does $$\{a_n\}_{n = 1}^{\infty}$$, as long as a1 is defined. If a1 isn't defined, it makes no sense to talk about $$\{a_n\}_{n = 1}^{\infty}$$.

16. Sep 13, 2010

### MelanieSwan

I got it. thanks very much. :)

17. Sep 17, 2010

### MelanieSwan

I am doing my homework and somehow again I couldn't work out this problem. Any help would be much appreciated:

1. The problem statement, all variables and given/known data

Consider the sequences {an} and {bn} where sequence {an} diverges to infinity and the sequence {anbn} converges. Prove that {bn} must converge to zero

2. Relevant equations

3. The attempt at a solution

I tried to use definition of converge and diverse to infinity for this one.
*{an} diverges to infinity so by definition, for any M: M> 0, there exists n* in N such that for any n>n* we'll have an > M
*{anbn} converges so by defnition for each ε >0, there exists n*>= 0 such that for any n > n*, we'll have: |anbn - A| < ε

I'm kinda stuck here. Should I go ahead and solve for bn to come up with something like |bn - 0| < ε or what?

Thanks in advance!

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