Are these the best tests for convergence of the following series?

In summary, the given series (a), (b), (c), (d), (e), and (f) all converge. Series (a) and (c) converge by the alternating series test, series (b) and (e) converge by the basic comparison test, series (d) diverges by the comparison test with 1/n, and series (f) can be evaluated using the power series expansion of e^x.
  • #1
end3r7
171
0

Homework Statement


(a) [tex]\displaystyle{\sum_{n=1}^{\infty}\Bigg(\frac{(-1)^{n}}{n^{1 + \frac{1}{n}}}\Bigg)[/tex]
(b) [tex]\displaystyle{\sum_{n=1}^{\infty}\Bigg(\frac{e^{\frac{1}{n}}}{n^{2}}\Bigg)[/tex]
(c) [tex]\displaystyle{\sum_{n=1}^{\infty}\Bigg(\frac{(-1)^{n}n!}{n^{n}}\Bigg)[/tex]
(d) [tex]\displaystyle{\sum_{n=1}^{\infty}\Bigg(\frac{1}{n^{1 + \frac{1}{n}}}\Bigg)[/tex]
(e) [tex]\displaystyle{\sum_{n=1}^{\infty}\Bigg(\frac{ln(x)}{n^{\frac{3}{2}}}\Bigg)[/tex]
(f) [tex]\displaystyle{\sum_{n=1}^{\infty}\Bigg(1 - e^{\frac{-1}{n}}\Bigg)[/tex]

Homework Equations


The test that we have messed with are:
Telescoping, Geometric, P-Series, Ratio, Root, Simple Comparison, Limit Comparison, Absolute Convergence, Alternating Series, Dirichlet, Integral, Gauss

The Attempt at a Solution



First, I got they all converge:

(a) (d)
I worked 'd' first. I did a limit comparison test with 1/n
[tex]\frac{\frac{1}{n}}{\frac{1}{n^{1 + \frac{1}{n}}}} = n^{n}[/tex]
That limit is 1. And since its absolute value converges (a) converges.

(b) Basic Comparion test wtih <= [tex]\frac{e}{n^{2}}[/tex]

(c) Alternating series test

(e) Integral test

(f) I am not sure which test to apply
 
Last edited:
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  • #2
(d) is wrong,a dn so is (a), not sure how to prove if they are convergent (if they are).

Is (d) divergent?
 
  • #3
(a) is convergent by alternating series test.

Did I end up proving (d) divergent or is my proof wrong?

I'm also quite lost on (f) since I can't integrate that function nicely
 
  • #4
Too many questions at once! d) is divergent. It's the same as 1/(n*n^(1/n)). For n^(1/n) show that the log of that approaches zero. So n^(1/n) approaches 1. So you can do a comparison with say 1/(2n). For f) expand e^(-1/n) in a power series using e^x=1+x+x^2/2!+etc and keep only the terms that matter.
 

Related to Are these the best tests for convergence of the following series?

1. What is convergence of a series?

Convergence of a series refers to the behavior of a infinite sequence of numbers. A series is said to converge if its terms approach a finite limit as we take more and more terms. If the terms do not approach a finite limit, the series is said to diverge.

2. How do I determine if a series is convergent?

There are several tests that can be used to check for convergence of a series, such as the comparison test, the ratio test, and the integral test. These tests involve comparing the given series to simpler series or using mathematical techniques to analyze the behavior of the series.

3. What is the best test for convergence of a series?

The best test for convergence of a series depends on the specific series being evaluated. Some series may be easily evaluated using a particular test, while others may require a combination of tests. It is important to try different tests and use the one that is most appropriate for the series at hand.

4. Are these tests for convergence always accurate?

No, these tests for convergence are not always accurate. In some cases, a series may appear to converge according to a certain test, but may actually diverge. This is why it is important to understand and use multiple tests to confirm the convergence or divergence of a series.

5. Can a series be both convergent and divergent?

No, a series cannot be both convergent and divergent. A series can only have one of these two behaviors. If a series does not converge, it is said to diverge. However, it is possible for a series to have different convergence or divergence behavior depending on the specific test used.

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