Converge/Diverge: Show Work & Tests Used

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In summary, the first sum converges due to comparison with a convergent sum, and the second sum converges due to the Leibniz rule for alternating series.
  • #1
sdg612
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Determine whether the following converge or diverge (show all work and state all tests and theormes used!)

a)[tex]\infty[/tex] (n=1)[tex]\sum[/tex]cos[tex]\hat{}[/tex]2 n/n [tex]\sqrt{}[/tex]n

b)[tex]\infty[/tex] (n=1)[tex]\sum[/tex]4(-1)[tex]\hat{}[/tex]n-1/n[tex]\hat{}[/tex]5+n[tex]\hat{}[/tex]3

i need help answering these 2 problems! any help wld be appreciated! Thnx!
 
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  • #2
Firstly, you need to show your work; we're not here to do your homework for you! Secondly, your expressions are not clear. Here, I'll rewrite them for you; click on the images to see the code:

[tex]\sum_{n=1}^{\infty}\frac{\cos^2n}{n\sqrt n}[/tex]

[tex]\sum_{n=1}^{\infty}\frac{4(-1)^{n-1}}{n^5+n^3} [/tex]

Are they the questions you are asking?

Finally, please be sure to post any further questions in the homework help forums.

Welcome to PF, BTW.
 
  • #3
thank you! yes those are the questions! it took me forever just to get them looking like that! Thank you again I'm still trying to get used to this forum...
 
  • #4
Hi, i suposse i would procceed in the following way:
Note that due to the fact that cos(x) is always smaller or equal than 1,
cos^2(x) has the same behavior. So, the sum that you are asking:
[tex]\sum_{n=1}^{\infty}\frac{\cos^2n}{n\sqrt n}[/tex] is smaller than
[tex]\sum_{n=1}^{\infty}\frac{\1}{n\sqrt n}[/tex] and this last converges
because it's a "p-type" sum with p=3/2>1. So by comparission, the first sum
must converge.
In the second one, you may see that it's an alternant series (the 4 doesn't matter)
and also that the sequence formed by the terms is a positive and decreasing one, so
due to the Leibniz rule (that applies only in this case) the sum is convergent.

Sory for the english i still don't know it very well
 
  • #5
there was supposed to be a 1 in the numerator of the second sum...
 

1. What does it mean for a series to converge or diverge?

Convergence and divergence refer to the behavior of a series, which is a sequence of numbers added together. If the terms of the series approach a specific value as the number of terms increases, the series is said to converge. If the terms of the series do not approach a specific value, the series is said to diverge.

2. How can I determine if a series converges or diverges?

There are several tests that can be used to determine convergence or divergence, such as the ratio test, the integral test, or the comparison test. The specific test used will depend on the form of the series and the information given. It is important to carefully consider the conditions of each test and use the appropriate one for the given series.

3. What is the purpose of showing work when determining convergence or divergence?

Show work is important because it allows for a clear understanding of the steps taken to determine convergence or divergence. It also helps to identify any errors that may have been made and allows for easier communication and collaboration with other scientists or mathematicians.

4. Are there any real-world applications for determining convergence or divergence?

Yes, there are many real-world applications for determining convergence or divergence. For example, in economics, convergence and divergence can be used to analyze the behavior of financial markets or the growth of a company. In physics, convergence and divergence are important in understanding the behavior of infinite series in the study of electromagnetic fields and wave equations.

5. Can a series both converge and diverge?

No, a series can only either converge or diverge. It cannot do both simultaneously. However, some series may have some parts that converge and other parts that diverge, in which case the overall behavior of the series can be determined by evaluating each part separately.

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