Absolute and Conditional Convergence Tests for Series | Homework Equations

In summary: I'm supposed to use the definition for all of them (i.e. divergence test, integral test, etc.) and the only one i remember off the top of my head is the limit comparison test, but i don't think i can use that for the above problem.i stated what you did above about the denominators and their reciprocals, but i have no other proof than that. also, I'm not sure what i would say for my other problems. I'm supposed to use the definition for all of them (i.e. divergence test, integral test, etc.) and the only one i remember off the top of my head is the limit comparison test, but i don't think i can use that for the above problem.
  • #1
whatlifeforme
219
0

Homework Statement


Absolute, Conditional, - convergence, or Divergence.

Homework Equations


[itex]\displaystyle \sum^{∞}_{n=1} (-1)^n e^{-n} [/itex]

The Attempt at a Solution


1. Alternating Series Test
2. Ratio Test for ABsolute Convergence

1. [itex]\displaystyle (-1)^n (1/e)^n [/itex]
an > 0 for n=1,2,3,4 - YES
decreasing - YES
limit (n->inf) = 0 - YES

Converges.

2.[itex]\displaystyle (1/e)^{n+1} * (e^n) = 1/e[/itex]
limit (n->inf) = 1/e

1/e < 1

Thus, Absolute Convergence.
 
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  • #2
whatlifeforme said:

Homework Statement


Absolute, Conditional, - convergence, or Divergence.


Homework Equations


[itex]\displaystyle \sum^{∞}_{n=1} (-1)^n e^{-n} [/itex]


The Attempt at a Solution


1. Alternating Series Test
2. Ratio Test for ABsolute Convergence

1. [itex]\displaystyle (-1)^n (1/e)^n [/itex]
an > 0 for n=1,2,3,4 - YES
decreasing - YES
limit (n->inf) = 0 - YES

Converges.

2.[itex]\displaystyle (1/e)^{n+1} * (e^n) = 1/e[/itex]
limit (n->inf) = 1/e

1/e < 1

Thus, Absolute Convergence.

Yes, the series converges, and converges absolutely. Your work is a little sketchy in parts, so if your instructor is picky, you might lose points. For example, in the alt. series test you need to show that an > 0 for all n ≥ 1, not just n = 1, 2, 3, and 4. And you don't show any work that justifies your saying that the sequence an is decreasing or that the limit of the sequence is zero.
 
Last edited:
  • #3
i did not show the work here, but i did show the work for decreasing by setting f(x) = series and taking derivative.
for the limit i have lim(x->inf) (1/e)^n = 0 (not sure how to show that this is zero)?

also, how do i show that an > 0 for all n >= 1 ?
 
  • #4
whatlifeforme said:
i did not show the work here, but i did show the work for decreasing by setting f(x) = series and taking derivative.
for the limit i have lim(x->inf) (1/e)^n = 0 (not sure how to show that this is zero)?
The limit as you wrote it doesn't make much sense, since you have x increasing, but the expression you're taking the limit of doesn't involve x.

One way to show that this limit is zero is to show that (1/e)n can be made arbitrarily close to zero. IOW, using the definition of the limit, which in this case involves N and ##\epsilon##. I don't know if actual limit definitions have been presented in your class just yet.
whatlifeforme said:
also, how do i show that an > 0 for all n >= 1 ?
Here, an = 1/en. The only way for a rational expression to be equal to zero is for its numerator to equal zero. Obviously that can't happen here. The denominators in this sequence, e1, e2, ..., en, ... are an increasing sequence of positive numbers, so it must be true that their reciprocals are decreasing and positive. An induction proof, which would be easy, would be convincing here.
 
  • #5
everytime we take a limit, i don't think the instructor wants us to use the definition of the proof. also, it's been a bit since i took a number theory class (discrete mathematics) which involved induction proofs.i stated what you did above about the denominators and their reciprocals, but i have no other proof than that. also, I'm not sure what i would say for my other problems.
 

1. What is the difference between absolute and conditional convergence?

Absolute convergence refers to a series where the sum of the absolute values of the terms converges, while conditional convergence refers to a series where the sum of the terms only converges when certain conditions are met.

2. What are some commonly used tests for absolute and conditional convergence?

Some commonly used tests for absolute convergence include the Ratio Test, the Root Test, and the Comparison Test. For conditional convergence, the Alternating Series Test and the Leibniz Test are frequently used.

3. Can a series be both absolutely and conditionally convergent?

No, a series can only be either absolutely or conditionally convergent. If a series is absolutely convergent, it is also conditionally convergent, but the reverse is not always true.

4. How do these convergence tests help in determining the convergence of a series?

Convergence tests provide a way to determine whether a series will converge or diverge without having to find the actual sum. By applying these tests to a series, we can determine if the series converges absolutely, converges conditionally, or diverges.

5. Are there any other factors besides convergence tests that can affect the convergence of a series?

Yes, there are other factors that can affect the convergence of a series, such as the behavior of the terms in the series, the order of the terms, and the presence of any special patterns or cancellations within the series.

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