Nth-term Test of Series

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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
HallsofIvy
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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
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Yes, but isn't (-1)^infinity an indeterminate form?
 
  • #4
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
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
jbunniii
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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 [itex]\lim_{n \rightarrow \infty} a_n \neq 0[/itex], then there's no way the series [itex]\sum_{n = 1}^{\infty} a_n[/itex] can converge, so you're wasting your time with other, more complicated tests.
 
  • #7
jbunniii
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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 [itex]0 / 0[/itex] or [itex]0 \cdot \infty[/itex].

For [itex](-1)^{n+1}[/itex], 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 [itex]n[/itex].

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

A third way to see this is to look at the absolute value: [itex]|(-1)^{n+1}| = 1[/itex] for all [itex]n[/itex], whereas if the sequence converges to zero, so must its absolute value.
 
  • #8
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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 [itex]\lim_{n \rightarrow \infty} a_n \neq 0[/itex], then there's no way the series [itex]\sum_{n = 1}^{\infty} a_n[/itex] can converge, so you're wasting your time with other, more complicated tests.
 
  • #9
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So, could anyone explicitly show me how to apply the nth term test?
 
  • #10
jbunniii
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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.
 
  • #11
SammyS
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[itex]\displaystyle \lim_{n\,\to\,\infty}\,(-1)^{n}[/itex] Does Not Exist, so the limit is certainly not zero.

Therefore, [itex]\displaystyle \sum_{n=1}^\infty\,(-1)^{n\ \ }[/itex] diverges.
 
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  • #12
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Thank you.
 

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