# Homework Help: How do you test the convergence of i^n (imaginary number) or (-1)^n?

1. Sep 14, 2005

### Agnostic

I was thinking alternating series test, but cant that only be used for testing for convergence, not for testing divergence?

2. Sep 14, 2005

### Hurkyl

Staff Emeritus
How about the divergence test? :tongue2:

3. Sep 14, 2005

### LeonhardEuler

You'll have to be more specific. The convergence of what? The sequence of terms, $a_n=i^n$, the series, $\sum_{n=0}^{\infty}i^n$, or some sequence or series that just has $i^n$ as one term? I can tell you right now, the first two diverge.

4. Sep 14, 2005

### Agnostic

The divergence test doesnt work for indeterminant forms.

5. Sep 14, 2005

### Agnostic

I must prove that $\sum_{n=0}^{\infty}i^n$ diverges.

Or rather, I must prove wheter or not its convergent of divergent. I know its divergent, but I still have to prove it.

6. Sep 14, 2005

### LeonhardEuler

$i^n$ is not an indeterminant form.

7. Sep 14, 2005

### Hurkyl

Staff Emeritus
There isn't an indeterminant form involved. :tongue2:

8. Sep 14, 2005

### LeonhardEuler

Oh, that's simple. Use the divergence test as Hurkyl suggested. Do the terms in the series approach zero as n approaches infinty?

9. Sep 14, 2005

### Agnostic

The limit of $i^n$ as n goes to infinity doesnt have a solution does it?

10. Sep 14, 2005

### LeonhardEuler

Right, exactly!

11. Sep 14, 2005

### Agnostic

No, they alternate ... i,-1,-i,1,i,-1,...

12. Sep 14, 2005

### Agnostic

But that is not the same as saying indeterminant form?

13. Sep 14, 2005

### LeonhardEuler

So the limit doesn't exist. Then can it equal zero?

14. Sep 14, 2005

### LeonhardEuler

No, an indeterminant form is something like $\frac{0}{0}$, or $\frac{\infty}{\infty}$, or $1^{\infty}$. A non-existant limit is just a non-existant limit.

15. Sep 14, 2005

### Hurkyl

Staff Emeritus
An indeterminate form is a "limit form" for which we don't have enough information to say what the limit really is, or if it exists. For example, $\lim_{x \rightarrow 0} x/x$ has the indeterminate form 0/0.

16. Sep 14, 2005

### Agnostic

Its usually always silly mistakes :D
Thanks alot guys..