Calc II - Alternating Series Test/Limits

In summary, the conversation discusses a homework question regarding convergence or divergence of an alternating series using the Alternating Series Test. The first condition is confirmed to be satisfied, but there is difficulty in confirming the second condition, which requires finding the limit of a function. The function is continuous, but as the variable approaches infinity, the function approaches 1, not 0. After some calculations and understanding of the properties of e, the concept is grasped and the limit is confirmed to be 1.
  • #1
jimbit
4
0
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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  • #2
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}##?
 
  • Like
Likes 1 person
  • #3
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?
 
  • #4
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.
 
  • #5
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..
 
  • #6
Oh, goodness,

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

Related to Calc II - Alternating Series Test/Limits

1. What is the Alternating Series Test and when should it be used?

The Alternating Series Test is a method used to determine the convergence or divergence of an alternating series. It can be used when the terms of the series alternate in sign (positive and negative) and decrease in absolute value as n increases. If these conditions are met, then the alternating series is said to converge.

2. How do you apply the Alternating Series Test?

To apply the Alternating Series Test, you first need to check if the series meets the necessary conditions (alternating signs and decreasing absolute value). Then, you can use the limit comparison test or the ratio test to determine if the series converges or diverges. If the limit of the absolute value of the terms is zero, then the series converges. If the limit is non-zero, then the series diverges.

3. Can the Alternating Series Test be used on series with positive terms?

No, the Alternating Series Test can only be used on series with alternating signs. If the series has all positive terms, then it is not an alternating series and the Alternating Series Test cannot be applied.

4. What is the difference between the Alternating Series Test and the Alternating Series Remainder Test?

The Alternating Series Test is used to determine the convergence or divergence of an alternating series, while the Alternating Series Remainder Test is used to estimate the error in the sum of a convergent alternating series. The remainder test is based on the fact that the sum of a convergent alternating series is equal to the sum of the first n terms plus the remainder term.

5. Are there any other methods for determining the convergence or divergence of an alternating series?

Yes, there are other methods such as the integral test, comparison test, and root test. However, the Alternating Series Test is specifically designed for alternating series and may be easier to apply in certain cases.

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