Converging vs Diverging Series: Understanding the Root Test

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Homework Help Overview

The discussion revolves around the convergence and divergence of two series, specifically focusing on the application of the root test and the alternating series test. The subject area is series convergence in calculus.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the divergence of the series involving alternating terms and question the application of the root test. There is discussion about the behavior of the nth term and its limit as n approaches infinity.

Discussion Status

Participants are actively engaging with the problem, questioning the assumptions behind the tests applied. Some guidance is offered regarding the conditions necessary for the alternating series test to be valid, and there is recognition of the failure of certain series to converge based on their terms.

Contextual Notes

There is mention of the series not converging to zero, which is a critical aspect in determining convergence. The original poster expresses urgency and confusion regarding the answers provided in the answer key.

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URGENT - Please help - easy series

How can I show that the series from 1 to infinity of

(-1)^(n+1) * n / sqrt(n^(2)+2) diverges instead of converging abs/conditionally?



Also, for the series from 1 to infinity of:

(-1)^(n+1)/(2)^(1/n)


I applied the root test and came out with:

lim n --> infinity of 1^n/2 = 1/2


yet, the answer key says the series diverges...can anyone explain this?
 
Last edited:
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The nth term of either of those series doesn't even converge to zero.
 
try the alternating series test.
 
Dick said:
The nth term of either of those series doesn't even converge to zero.

if you just try to simplify and do the limits the denomintor goes to 1 and the top alternates from -1 to 1 and back and forth.
 
Antineutron said:
if you just try to simplify and do the limits the denomintor goes to 1 and the top alternates from -1 to 1 and back and forth.

Well, yeah. Isn't that what I said?
 
I see, the limit of 1/(2)^1/n goes to 1, so it fails to go to 0 and hence diverges.

and the first can be simplified to n/n = 1, which doesn't go to 0, so it diverges...

Is it that simple?
 
It is that simple. When are applying a test like the alternating series test, make sure that ALL of the premises apply. Of course, just because a test doesn't apply, doesn't make the series diverge. But any series that for large n looks like +1,-1,+1,-1... does not converge.
 
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