# Show ( (-1)^n + (1/n) ) diverges.

1. Sep 28, 2008

### Unassuming

1. The problem statement, all variables and given/known data
Show that the sequence (-1)^n + (1/n) diverges.

2. Relevant equations

3. The attempt at a solution

I have assumed it converges and found the difference between consecutive terms to be "1+ (1/ 2n(n+1) ). I have tried many things but I cannot progress.

2. Sep 28, 2008

### arildno

Well, your expression for the difference is wrong.
Think otherwise:
Your sequence is a sum of two other sequences.
What can you say about the convergens/divergence of those two subsequences?

3. Sep 28, 2008

### Unassuming

They both converge to two different numbers.

4. Sep 28, 2008

### Unassuming

We have not gotten to the subsequences yet so I hesitate to say that this violates the thrm saying that subsequences always converge to the same point as the sequence itself.

This problem is two sections before that theorem and any talk of subsequences. Is there any way to prove it without them?

5. Sep 28, 2008

### HallsofIvy

Staff Emeritus
And what are the two numbers you think they converge to?

6. Sep 29, 2008

### Unassuming

1 and -1? I am beginning to doubt myself.

7. Sep 29, 2008

### danago

Try a few different values of n, and see if that helps visualize what is happening to each term. For the (-1)^n part, can i suggest that you see what happens when n=1,2,3,4...Does it really converge to 1?

EDIT: Sorry misunderstood what you were saying, thought that you meant (-1)^n converges to 1 for all even and odd n.

Last edited: Sep 29, 2008
8. Sep 29, 2008

### Unassuming

I know what it does. I have tried values of n. I have pictured it and I am ready to prove it. Am I missing something here?

For all the even n, it converges to 1 from the right.

For all the odd n, it converges to -1 from the right.

I am trying to prove that it diverges.

9. Sep 29, 2008

### Gib Z

Ok. First - separate them into two separate sequences. It in fact, does NOT converge. You took some partial sums, did the partial sums ever tend to a certain value? The partial sums "oscillates" between 1 and -1 . It doesn't converge. For the purpose of the question, we need only know that is it bounded. I'm sure you can do the rest.

10. Sep 29, 2008

### HallsofIvy

Staff Emeritus
No, neither of them converges to 1 or -1! I have no idea how you could come to that conclusion.

Look at a few terms: {1/n} starts 1, 1/2, 1/3, 1/4, 1/5, ... what does that converge to?

{(-1)n starts -1, 1, -1, 1, -1, 1... what does that converge to? Does it converge at all?

11. Sep 29, 2008

### Gib Z

O wait sorry, I thought we were doing series instead of sequences. Ignore my comment about the boundedness.