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Determine whether the series is convergent or divergent

  1. Mar 17, 2009 #1
    1. The problem statement, all variables and given/known data
    Determine whether the series is convergent or divergent.

    1 + 1/8 + 1/27 + 1/64 + 1/125 ....

    2. Relevant equations

    3. The attempt at a solution

    I know this is convergent but not sure how to prove this mathematically.
  2. jcsd
  3. Mar 17, 2009 #2
    It might help to write the series in it's summation form...

    Do you know any of the tests for convergence?
  4. Mar 17, 2009 #3
    Well this chapter is about the integral test so can I apply that method?
  5. Mar 17, 2009 #4
    Yeah you can apply that test, I suppose -- How does the integral test work?
  6. Mar 17, 2009 #5
    i don't think the integral test can apply to this because this cannot be represented as a function.

    however, there is another test, the remainder test,

    Remainder test states that :

    Suppose f(k) = ak, where f is a continuous,positive,decreasing function for x>=n
    and [tex]\sum a_n [/tex] is convergent. If Rn= s-sn,

    then [tex]\int f(x) dx[/tex] [tex]\leq R_n[/tex] [tex]\leq \int f(x) dx[/tex], where the limit is from n to infinity
  7. Mar 17, 2009 #6
    I think you're misunderstanding the integral test

    It says that [tex]\sum_{n=1}^{infinity} f(n)[/tex] converges if [tex]\int_{n}^{infinity} f(x) dx[/tex] converges.
  8. Mar 17, 2009 #7

    Through some induction it can be seen that the summation is: [tex]$\displaystyle\sum_{n=1}^\infty \frac{1}{n^3}.[/tex]

    Which is in the form of a p-series:

    [tex]$\displaystyle\sum_{n=1}^\infty \frac{1}{n^p}.[/tex]

    So if you were to know the rules of the p-series I assume it would be pretty straightforward
  9. Mar 17, 2009 #8
    Ok, I get that, but how can we represent the problem above as a sum of a series, so
    we can use the integral test?
  10. Mar 17, 2009 #9


    Staff: Mentor

    Do you know what the general term (i.e., nth term) in your series looks like?

    1 + 1/8 + 1/27 + 1/64 + 1/125 ... + ??? + ...

    There's a definite pattern going on here.
  11. Mar 17, 2009 #10
    Ya thats the problem, i can't seem to recognize any patterns.
  12. Mar 17, 2009 #11
    EDIT: In helping you recognize the pattern, you should first look at the numerator...

    The numerator is always 1, so you can sort of look beyond that and concentrate on the denominator...

    Since the first value of the series is 1, you can surmise that the summation probably begins at [tex]n=1[/tex] or has a number raised to the [tex]n[/tex] value (which if n began at 0 would yield 1). In this case however, beginning at 0 doesn't make sense as the following terms don't correspond with a series beginning at an index of 0.

    Now that you can concentrate on the denominator, you can't do much more than use induction to realize what each of the denominators has in common. In this case, each of the 5 terms' denominators, (1, 8, 27, 64, 125) is a perfect cube number. Once you "see" this, try to plug in your index and the succeeding values of the series to see if they would match up with [tex]n^3[/tex].

    In this case they do! Thus this series can be expressed as simply: [tex]$\displaystyle\sum_{n=1}^\infty \frac{1}{n^3}.[/tex]

    I hope this helps!

    If you insist on using the Integral test on this just take the integral of the summation we've previously noted:
    [tex]$\displaystyle\sum_{n=1}^\infty \frac{1}{n^3}.[/tex]

    For example:

    [tex]\displaystyle\int^\infty_1 n^{-3}dn[/tex]

    This maintains the convergence you've already established.
    Last edited: Mar 17, 2009
  13. Mar 17, 2009 #12
    Oh man good eye, I just got off spring break and have been trying hard to get back into
    the mode of thinking. Thanks man!.
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