1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    AlephZero

    User Avatar
    Science Advisor
    Homework Helper

    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

    AlephZero

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted