How Do I Apply the Nth-Term Test to the Sequence (-1)^n+1?

[Bashyboy]

Homework Statement

I attached the problem

Homework Equations

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?

 

[HallsofIvy]

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.

 

[Bashyboy]

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

 

[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.

 

[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

 

[jbunniii]

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.

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.

 

[jbunniii]

To answer Bashyboy's original question:

When I try to apply the nth-term test on the sequence, (-1)^n+1, I get an indeterminate form

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.

 

[Mark44]

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.

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

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.

 

[Bashyboy]

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

 

[jbunniii]

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

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

 

[SammyS]

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

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

 

[Bashyboy]

Thank you.