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Determine whether the series converges or diverges

  1. Dec 1, 2011 #1
    1) Determine whether the series converges or diverges: summation from n=1 to ∞ of (square root of (n+1) - square root of (n-1)) / n. clearly state which test you are using.

    2) Determine whether the series converges or diverges: summation from n=1 to ∞ of (1*3*5*... (2n-1)) / (2*5*8*... (3n-1)). clearly state which test you are using.

    For question #1, I tried multiplying the top and bottom by square root of (n+1) + square root of (n-1). On the top, the answer simplifies to 2 and on the bottom it simplifies to n multiplied by (square root of (n+1) + square root of (n-1)). I am thinking to divide the top and bottom by n so the limit as n approaches infinity is equal to 0. But by the nth term test for divergence, if the limit is equal to 0, then the series may converge or diverge. This is where I am stuck and can't think of anything else.

    For question #2, I am having trouble simplifying the problem. It can't be just (2n-1) / (3n-1) because that would change the whole series.
     
  2. jcsd
  3. Dec 1, 2011 #2

    jgens

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    For the first problem, note that (n-1)1/2 + (n+1)1/2 < 2(n-1)1/2. Use this to get a series you can apply the comparison test to. That should help you finish that one.

    For the second problem, we can write the series as:
    [tex]\sum_{n=1}^{\infty}\prod_{i=1}^n \frac{2n-1}{3n-1} = \sum_{n=1}^{\infty}\prod_{i=1}^n \left(\frac{2}{3} - \frac{1}{3(3n-1)}\right)[/tex]
    Now use an obvious approximation along with the comparison test to finish the problem.

    Edit: I can't seem to get the LaTeX tags to work properly :( Hopefully you can make some sense out of what I wrote.
    Edit 2: Fixed.
     
  4. Dec 1, 2011 #3
    For the first question, I don't understand how you got (n-1)^1/2 + (n+1)^1/2 < 2(n-1)^1/2. I know how to use the comparasion test but I am not sure what to compare the series with, since the numerator is complicated and the denominator is just n.
    Would the ratio test work on this series. a_n+1 / a_n . I tried it but it ends in a very complicated fraction.

    For the second question, I don't understand how you got the series that you wrote. If you divide (2n-1) and (3n-1) by n then the limit is equal to 2/3. Therefore, by the nth term test for divergence, the series will diverge? because the limit as n approaches infinity is equal to 2/3 which is not equal to zero. would this work?
     
  5. Dec 1, 2011 #4

    jgens

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    Problem 1:
    There is a typo in my first post. I should have written "(n-1)^1/2 + (n+1)^1/2 > 2(n-1)^1/2" instead. Hopefully this makes sense to you now. Sorry about any confusion.

    I think the comparison test is the best method to use for solving this problem. In particular, you have:
    [tex]\sum_{n=1}^{\infty}\frac{\sqrt{n+1}-\sqrt{n-1}}{n} = \sum_{n=1}^{\infty}\frac{2}{n(\sqrt{n+1}+\sqrt{n-1})} < \sum_{n=1}^{\infty}\frac{1}{n\sqrt{n-1}}[/tex]
    Apply the comparison test now.


    Problem 2:
    You method will not work since we are not summing the sequence (2n-1)/(3n-1). Anyway, I got that expression by noting that:
    [tex]\frac{2}{3} - \frac{1}{3(3n-1)} = \frac{2(3n-1) - 1}{3(3n-1)} = \frac{6n-3}{3(3n-1)} = \frac{2n-1}{3n-1}[/tex]
    If you think about how you can use the comparison test from here, the solution is fairly obvious.
     
    Last edited: Dec 1, 2011
  6. Dec 1, 2011 #5

    Mark44

    Staff: Mentor

    Mod note: moving to Calculus section
     
  7. Dec 1, 2011 #6
    So for the first question, if you plug in n=3, then 1/n*square root (n-1) is not greater than the other inequality but it is less than. So, I am thinking wouldn't the < be the other way around?
    Using the comparasion test, I could compare 2/... with 1/n*square root of (n-1). So therefore the series will converge?? Am I on the right track?

    For the second question, you got 2/3 - 1/3(3n-1). So if I divide the top and bottom of 1/3(3n-1) by n then the limit is equal to 2/3 - 0 = 2/3. This means that the series converges?
     
  8. Dec 1, 2011 #7

    jgens

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    I am pretty sure that the direction of the '<' signs are right now. And yes. You just use the obvious comparison test to show that the series will converge.

    As to the second question, you cannot just ignore the product sign. You might not want to work with it, but you have to deal with it. And even if you did, the method you mentioned does not show convergence. Instead, think about an obvious function you could use the comparison test on.
     
  9. Dec 1, 2011 #8
    for the first question, I am starting to understand what you are getting at. But for the comparasion test, is it important to show whether the series I am comparing the original to converges or diverges. Since, 1/n square root (n-1) is greater than 2/..., the series will converge, but I am having trouble figuring out how 1/n square root (n-1) converges.

    For the second question, I could compare the series with 1/3n-1? I am not quite understanding this one.
     
  10. Dec 1, 2011 #9

    jgens

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    It is a well-known fact that the series Ʃ(n-1)-3/2 converges. You can prove this using the integral test. There are other ways to prove this, but I think the tests used are a little more complicated. That should help you finish the first one.

    For the second question, think about the product. How can you find something that you can guarantee will be greater than the product? There is a very obvious choice. The biggest hint I will give you is that
    [tex]0 < \frac{2}{3} - \frac{1}{3(3n-1)} < \frac{2}{3}[/tex]
    whenever n > 0.
     
  11. Dec 1, 2011 #10
    First, I don't understand where you got the (n-1)^3/2 from. I want to compare the original series with 1/n(n-1)^1/2. First I have to prove that 1/n(n-1)^1/2 converges/diverges and then use the comparasion test to show that 1/n(n-1)^1/2 is greater than the orginial series which means that the series converges overall. The problem is that I am not sure how to determine convergence or divergence of 1/n(n-1)^1/2. Once I solve this problem, then I've got my answer.

    For the second question, are you trying to say that:
    2/3 - ( (2/3) / 3(3n-1) ) is greater than 2/3 - ( 1/3(3n-1)) ?? which is true for n>0. This way I can use the comparasion test. Is this enough to prove that the series will converge overall or do I have to show the convergence/divergence of 2/3 - ( (2/3) / 3(3n-1) ) too? I think the series with converge overall.
     
  12. Dec 1, 2011 #11

    jgens

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    I wrote (n-1)-3/2. The negative sign is important. And you need to show that the series Ʃ(n-1)-3/2 converges. You can show this using the integral test. There are some other tests that show it as well, but I think they are more complicated.

    Now, you need to figure out why we are using Ʃ(n-1)-3/2 is the series we're using for the comparison test. This is really easy to figure out.

    I am trying to set things up so you can use the comparison test, but you keep ignoring the product. I will not give you any more hints on this one until you take into account that you are summing ∏(2/3 -1/[3(3n-1)]).


    The series does converge. But not for any of the reasons you have stated.
     
  13. Dec 1, 2011 #12
    Yes sorry about the negative sign. I meant (n-1)^3/2. I used the integral test and this series converges. I compared this series to the original and the series will converge overall. This problem is resolved.

    Sorry, but can you just specify what you mean by "the product"? That is what is confusing me the most and the reason why I am not able to understand the hints you are giving. Please clarify that problem for me and I hope I will understand it better when I realize what you meant by the product.

    Thank you.
     
  14. Dec 1, 2011 #13

    jgens

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    The series is:
    [tex]\sum_{n=1}^{\infty}\prod_{i=1}^n\left(\frac{2}{3}-\frac{1}{3(3i-1)}\right)[/tex]
    So it is insufficient to just consider the nth term in the product as you keep doing. That is, we cannot just consider 2/3-1/[(3(3n-1)]; instead, we have to consider ∏(2/3 - 1/[3(3i-1)]). Once you do that, it's pretty easy.
     
  15. Dec 1, 2011 #14
    Okay I understand what you mean. However, i've never used that product symbol (it looks like the pi symbol when you type it) in calculus so you please explain briefly by what that indicates? I have indeed used i=1 though.
     
  16. Dec 1, 2011 #15
    okay i finally understand what you meant. The product symbol, okay I figured it out. So if you work out the series, you get 1/2*3/5*5/8*7/9 ... (2i-1)/(3i-1). You end up with the original question. So, now what?
     
  17. Dec 2, 2011 #16

    jgens

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    Now you use the following fact:
    [tex]0 < \frac{2}{3} - \frac{1}{3(3i-1)} < \frac{2}{3}[/tex]
    for all i > 0.
     
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