# Nth-term Test of Series

1. Oct 9, 2012

### Bashyboy

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

2. Relevant equations

3. The attempt at a solution
When I try to apply the nth-term test on the sequence, (-1)^n+1, I get an indeterminate form; so, I try to raise e to the ln[(-1^n+1)], but I run into the same problem. How do I take the limit, in this instance, in order for the nth-term test to decide anything?

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2. Oct 9, 2012

### HallsofIvy

Staff Emeritus
What do you mean by the "nth term test"? One theorem says that if the terms do not go to 0, then the series cannot converge.

3. Oct 9, 2012

### Bashyboy

Yes, but isn't (-1)^infinity an indeterminate form?

4. Oct 9, 2012

### Taiki_Kazuma

You are correct that the alternating series of the form (-1)^n diverges.

So, this series for sine is just a different form of an alternating series.

Now, the nth-term test is a waste of time. This would be a last resort test, because there are so many other (more simplicities) ways to identify convergence/divergence.

The nth-term test is when you right out the first 4 or 5 terms and the last few nth-terms.

Example:
sin(pi) + sin(3/2 pi) + sin(2 pi) + sin(5/2 pi) + ... + sin([2n-7]/2 pi) + sin([2n-5]/2 pi) + sin([2n-3]/2 pi) + sin([2n-1]/2 pi)

Simplifying:
0 - 1 + 0 + 1 + ... + sin([2n-7]/2 pi) + sin([2n-5]/2 pi) + sin([2n-3]/2 pi) + sin([2n-1]/2 pi)

Therefore, per the nth-term test, the series oscillates between 1 and -1. Therefore, the given series diverges.

5. Oct 9, 2012

### Taiki_Kazuma

Sorry. I guess I didn't show my last step.

the series further simplifies to:
-1 + 1 + ...
Which is a basic alternating series of (-1)^(n)
This series is a known to diverge. No further justifications needed

6. Oct 9, 2012

### jbunniii

That's not what I would call the nth term test. The usual meaning is described here:

http://en.wikipedia.org/wiki/Term_test

and is in fact the FIRST test you should apply when checking whether a series converges or not. Namely, if $\lim_{n \rightarrow \infty} a_n \neq 0$, then there's no way the series $\sum_{n = 1}^{\infty} a_n$ can converge, so you're wasting your time with other, more complicated tests.

7. Oct 9, 2012

### jbunniii

No, an indeterminate form is something like $0 / 0$ or $0 \cdot \infty$.

For $(-1)^{n+1}$, it's simple to see that this sequence diverges: it oscillates between +1 and -1, so there can't be any value that all of the sequence terms will be near for arbitrarily large $n$.

Another way to see this is that the sequence contains a subsequence (the odd values of $n$) that converges to 1, and another subsequence (the even values of $n$) that converges to -1. This cannot happen with a convergent sequence.

A third way to see this is to look at the absolute value: $|(-1)^{n+1}| = 1$ for all $n$, whereas if the sequence converges to zero, so must its absolute value.

8. Oct 9, 2012

### Staff: Mentor

I heartily agree. It's very easy to apply this test, so if the test is inconclusive, you haven't wasted much time.

9. Oct 9, 2012

### Bashyboy

So, could anyone explicitly show me how to apply the nth term test?

10. Oct 9, 2012

### jbunniii

I showed you three different, but equivalent, ways in my previous message.

11. Oct 9, 2012

### SammyS

Staff Emeritus
$\displaystyle \lim_{n\,\to\,\infty}\,(-1)^{n}$ Does Not Exist, so the limit is certainly not zero.

Therefore, $\displaystyle \sum_{n=1}^\infty\,(-1)^{n\ \ }$ diverges.

Last edited: Oct 9, 2012
12. Oct 9, 2012

Thank you.