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Determine whether the following series converges

  1. Jun 4, 2009 #1
    1. The problem statement, all variables and given/known data

    Determine whether the following series converges:

    [tex]
    \sum \frac{(n-1)^3}{\sqrt{n^8+n+2}}
    [/tex]


    2. Relevant equations

    Definition of convergence:

    Let [tex]\sum a_{n}[/tex] be a series.

    If the sequence of (sn) partial sums converges to L (finite). Then we say the series converges to L or has sum L. If (sn) diverges we say [tex]\sum a_{n}[/tex] diverges.


    3. The attempt at a solution

    With some manipulation i can see the sequence acts like 1/n, thus the series [tex]\sum a_{n}[/tex] would diverge. However i can't use 1/n as a comparison since we would not have an > bn. So I havent been able to find a suitable divergent sequence.

    So basically we need a sequence bn such:

    [tex]\frac{(n-1)^3}{\sqrt{n^8+n+2}}}>b_{n}[/tex]

    This doesnt need to hold for all natural numbers n we can have it for some n>=a.

    And bn must of course be divergent
     
    Last edited: Jun 4, 2009
  2. jcsd
  3. Jun 4, 2009 #2
    Re: Convergence

    Next question, if someone can tell me why my LaTex isn't working that would be fantastic ???
     
  4. Jun 4, 2009 #3

    Mark44

    Staff: Mentor

    Re: Convergence

    Fixed your LaTeX. See below for what I did.
    Some of your beginning tex tags had a '\' character in them, and all of you ending tex tags had \ instead of /.

    Regarding your problem, do you know the Limit Comparison Test? If you can show that the arbitrary term in your series is > 1/n, the Limit Comparison Test might be helpful.
     
  5. Jun 4, 2009 #4
    Re: Convergence

    Cheers, and crikey i made loads of mistakes in there - fixing :)
     
  6. Jun 4, 2009 #5
    Re: Convergence

    I think the arbitrary term in my series is less than 1/n except for very small values of n below zero, so 1/n isn't suitable.

    With some manipulation we can express the nth term as as:

    [tex]
    \frac{(1-1/n)^3}{n\sqrt{1+n^{-7}+2n^{-8}}}
    [/tex]
     
    Last edited: Jun 4, 2009
  7. Jun 4, 2009 #6
    Re: Convergence

    I'm not thinking clearly at the moment, need a break, but I think 1/4n would work though for n>=3.
     
  8. Jun 4, 2009 #7
    Re: Convergence

    You mean 1/n isn't suitable for the Limit Comparison Test?
     
  9. Jun 4, 2009 #8

    Mark44

    Staff: Mentor

    Re: Convergence

    If the arbitrary term in your series is less than 1/n, then the Comparison Test is no help, but try the Limit Comparison Test, if you know it. There is also the Ratio Test, which might be useful.

    Also, you probably aren't interested in negative values of n. The index for series usually starts at 0 or 1. Not always, but usually.
     
  10. Jun 4, 2009 #9
    Re: Convergence

    I do know the limit comparison however i just get the sequence tending to 0 rather than a finite L. So doesn't seem to be of use

    Comparison test should work for 1/4n, for n>=2

    [tex]\frac{(1-1/n)^{3}}{n\sqrt{1+n^{-7}+2n^{-8}}}\geq\frac{1}{4n}\][/tex]

    So by comparison test the series diverges?
     
  11. Jun 4, 2009 #10
    Re: Convergence

    No i meant for the comparison test, however if i use 1/n in the limit comparison i get an/bn tending to 0
     
  12. Jun 4, 2009 #11
    Re: Convergence

    You'd have to prove or show how you got that inequality in order to use it for the comparison test.

    If you use 1/n for the limit comparison test, you should not get 0 if you start with your manipulated series.
     
  13. Jun 4, 2009 #12

    Mark44

    Staff: Mentor

    Re: Convergence

    I don't see your work using the limit comparison test, so can't comment on the correctness of your result. You shouldn't get a sequence that tends to zero -- maybe you meant the limit was zero.

    That test works this way, with an being the general term in your series and bn being the one in 1/n.

    If [tex]\lim_{n \rightarrow \infty} \frac{a_n}{b_n} = L \neq 0[/tex]
    and L is a finite number, then both series converge, or they both diverge. The wikipedia article (http://en.wikipedia.org/wiki/Limit_comparison_test) is somewhat misleading about when you can use this test. It's not just for convergent series. The external reference at the end of the wikipedia article I cited has a more complete presentation.

    I haven't calculated this limit, so I can't tell if this test would have been conclusive. If you can actually prove the inequality you have in your last post, the comparison test tells you that your series diverges.
     
  14. Jun 4, 2009 #13
    Re: Convergence

    Ah no i made a mistake! (again!!!)

    Limit comparison works, tends to 1 when i divide by 1/n.

    Thanks!
     
  15. Jun 4, 2009 #14
    Re: Convergence

    So to conclude

    By the standard comparison test:

    since (1-1/n)^3 >= 1/2 for n>=2

    and [tex]n\sqrt{1+n^{-7}+2n^{-8}[/tex] <= 2n

    we must have
    [tex]
    \frac{(1-1/n)^{3}}{n\sqrt{1+n^{-7}+2n^{-8}}}\geq\frac{1/2}{2n}=\frac{1}{4n}\]
    [/tex]

    More simply by the limit comparison test

    [tex]\frac{(1-1/n)^{3}/n\sqrt{1+n^{-7}+2n^{-8}}}{1/n}\rightarrow\frac{1/n}{1/n}=1\][/tex]
     
  16. Jun 4, 2009 #15
    Re: Convergence

    That's not true, try it for n=2 and n=3. Are you forgetting the ^3?
     
  17. Jun 4, 2009 #16
    Re: Convergence

    I meant n>=5 :)
     
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