Calc II - Alternating Series Test/Limits

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Homework Help Overview

The discussion revolves around the convergence or divergence of an alternating series, specifically the series ∑ (-1)^(n-1)e^(2/n) from n=1 to infinity. Participants are exploring the application of the Alternating Series Test (A.S.T.) to determine the behavior of the series.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to confirm that the series is decreasing and questions how to prove the limit of e^(2/n) as n approaches infinity. Some participants discuss the continuity of the function e^w and its implications for the limit as w approaches 0.

Discussion Status

Participants are actively engaging with the problem, with some providing insights into the continuity of the function and its limits. There is a recognition of the limit approaching 1, but uncertainty remains regarding why it does not approach zero.

Contextual Notes

There is mention of the original poster's difficulties with limit calculations due to time since their last calculus course, which may affect their understanding of the concepts involved.

jimbit
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Hello PF,

I've got a homework question I'm having some trouble with regarding series, particularily alternating series.

The question asks you to test the series for convergence or divergence for an alternate series by using the A.S.T. :


∑ (-1)n-1e2/n
n=1


Homework Equations



The Alternating Series Test states that a series, an, is converging if
(a) an+1 ≤ an (i.e. the series is decreasing) for all n, and
(b) the limit of the an = 0 as n → ∞

The Attempt at a Solution



In my attempt I have confirmed (a) by using the first derivative test to show that the series is decreasing:

Let f(x)= e2/x
f'(x) = -2e2/n / x2

which is going to be negative for all x > 1, thus satisfying the first condition.

The part where I am having trouble is confirming that the limit of e2/x equals/does not equal 0 as x approaches infinity (the second requirement for the A.S.T.).
Looking at the graph of e2/x, the limit seems to be approaching 1, but I do not know to prove this/which method to use.

It's been some years since I have taken Calc I and as a result, I'm having some diffculty with computing limits, though I am understanding the concepts of convergance/divergence well enough.

Thanks for reading and for any assistance that you might be able to give!
 
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jimbit said:
Hello PF,

I've got a homework question I'm having some trouble with regarding series, particularily alternating series.

The question asks you to test the series for convergence or divergence for an alternate series by using the A.S.T. :


∑ (-1)n-1e2/n
n=1


Homework Equations



The Alternating Series Test states that a series, an, is converging if
(a) an+1 ≤ an (i.e. the series is decreasing) for all n, and
(b) the limit of the an = 0 as n → ∞

The Attempt at a Solution



In my attempt I have confirmed (a) by using the first derivative test to show that the series is decreasing:

Let f(x)= e2/x
f'(x) = -2e2/n / x2

which is going to be negative for all x > 1, thus satisfying the first condition.

The part where I am having trouble is confirming that the limit of e2/x equals/does not equal 0 as x approaches infinity (the second requirement for the A.S.T.).
Looking at the graph of e2/x, the limit seems to be approaching 1, but I do not know to prove this/which method to use.

It's been some years since I have taken Calc I and as a result, I'm having some diffculty with computing limits, though I am understanding the concepts of convergance/divergence well enough.

Thanks for reading and for any assistance that you might be able to give!


Is the function ##f(w) = e^w## continuous in the variable ##w##? For the case of ##w = 2/x## what is the limiting value of ##w## as ##x \to \infty##? So, what is the limit of ##e^{2/x}##?
 
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f(w) = e^w is continuous.
For 2/x, it is not continuous at 0. But it is continuous from 1 to infinity.
So, the limit is 1?
 
jimbit said:
f(w) = e^w is continuous.
For 2/x, it is not continuous at 0. But it is continuous from 1 to infinity.
So, the limit is 1?

You tell me.
 
Well, as 2/x becomes infinitely small, e goes to 1. I see that now, after doing some calculations.
However, I'm just not sure why it doesn't go to zero instead. I seem to be missing some crucial fact about e..
 
Oh, goodness,

I understand now, as 2/x becomes increasingly small, it basically becomes 0.
e^0 = 1.
 

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