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Homework Help: Convergence by Comparison Test

  1. Feb 2, 2014 #1
    1. The problem statement, all variables and given/known data
    Use $$\sum\limits_{n=1}^∞ \frac{1}{n^2}$$ to prove by the comparison test that $$\sum\limits_{n=1}^∞ \frac{n+1}{n^3} $$ converges.

    2. Relevant equations
    $$\sum\limits_{n=1}^∞ \frac{n+1}{n^3} \equiv \sum\limits_{n=1}^∞ \frac{1}{n^2} + \sum\limits_{n=1}^∞ \frac{1}{n^3} $$

    3. The attempt at a solution
    $$ \frac{n+1}{n^3} < \frac{1}{n^2}$$
    $$ n + 1 < n $$
    $$ 1 < 0 $$ which is false.
    I cannot prove the convergence of the series by direct comparison with 1/n2, correct? The series is a sum of 2 p-series (p > 1) so it is clearly a convergent series and can be shown by limit comparison with 1/n2, but I cannot see how I can prove this by direct comparison because there is no way that 1/n2 + 1/n3 is less than 1/n2 if n > 0...
  2. jcsd
  3. Feb 2, 2014 #2


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    True, but can you show 1/n2 + 1/n3 is less than some constant times 1/n2?
  4. Feb 2, 2014 #3
    Looks like direct comparison test doesn't work, but it's possible that "comparison test" in the problem could also refer to the limit comparison test -- see if that works instead.
  5. Feb 2, 2014 #4
    The problem was referring to direct comparison; limit comparison does in fact work as I stated in my original post. I'm aware that I'm dealing with a convergent series, but I just wanted to confirm whether or not direct comparison with 1/n2 is inconclusive.
  6. Feb 2, 2014 #5
    Yes, that seems feasible.
    ##\frac{n+1}{n^3} < c(\frac{1}{n^2}) ##
    ##\frac{n+1}{n^3} < \frac{c}{n^2} ##
    ##1 + \frac{1}{n} < c##
    ##\frac{1}{n} < c-1 ## if c > 2.

    Let c = 3
    ## \frac{n+1}{n^3} < 3(\frac{1}{n^2}) ##
    ## n^3 + n^2 < 3n^3 ##
    ## 1 + \frac{1}{n} < 2 ##
    ## \frac{1}{n} < 1 ## for all n > 1.

    Therefore, ##\frac{n+1}{n^3} < \frac{c}{n^2} ## where c > 2,
    but I'm still kinda iffy about this, because I'm asked to prove the convergence with 1/n2.
    Last edited: Feb 2, 2014
  7. Feb 2, 2014 #6
    assume the sum 1/n^2 converges. for each positive integer n, (n+1)/n^3 = 1/n^2 + 1/n^3 which is less than or equal to 2/n^2. and the sum of 2/n^2 converges since each term in the sequence of partial sums equals 2 times the corresponding term in the sequence of partial sums for 1/n^2. (there is a theorem that states that if (xn) is a sequence of real numbers that converges to L, then the sequence (a * xn) where a is a real number converges to a * L).
  8. Feb 2, 2014 #7


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    Well, we don't know how pedantic your math lecturer is.

    But I would have thought that if you are told that ##\sum\frac{1}{n^2}## converges, it seems perfectly reasonable to use the fact that ##\sum\frac{2}{n^2}## also converges.
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