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 of alternating series

  1. Jan 20, 2016 #1
    1. The problem statement, all variables and given/known data
    Do the following series converge or diverge?
    ## \sum_{n=2}^\infty \frac{1}{\sqrt{n} +(-1)^nn}## and
    ##\sum_{n=2}^\infty \frac{1}{1+(-1)^n\sqrt{n}}##.

    2. Relevant equations
    Leibniz convergence criteria:
    If ##\{a_n\}_{k=1}^\infty## is positive, decreasing and ##a_n \to 0##, the alternating series ##\sum_{n=1}^\infty (-1)^{n-1}a_n## is convergent.

    3. The attempt at a solution
    I suspect the first series converges and the second diverges but I need to show that.
    Starting with the first series it can be rewritten
    ##\sum_{n=2}^\infty \frac{1}{\sqrt{n}}\frac{(-1)^n}{(-1)^n+\sqrt{n}}##
    At this point I had hopes that there exists an ##N## for which ##\forall n \ge N## ##a_n## is decreasing. That is show that
    ##\frac{1}{\sqrt{n+1}}\frac{1}{-1+\sqrt{n+1}} \le \frac{1}{\sqrt{n}}\frac{1}{1+\sqrt{n}}##. Sadly based on some numerical experiments this doesn't seem to be true so I need another approach.

    Another idea was to be able to pair together the coefficients
    ##\frac{1}{\sqrt{n+1}}\frac{1}{-1+\sqrt{n+1}}+ \frac{1}{\sqrt{n}}\frac{1}{1+\sqrt{n}}## to show that the partial sum ##\lim_{n\to \infty} S_{2n}## exists but I see no useful way to show this either.
     
  2. jcsd
  3. Jan 20, 2016 #2

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    You should be able to show the following.

    ##\displaystyle \frac{1}{\sqrt{n}}\frac{1}{(-1+\sqrt{n})} \ge \frac{1}{\sqrt{n}}\frac{1}{(1+\sqrt{n})}##

    then ...

    Does ##\displaystyle \lim_{{n\to\infty}} \left(\frac{1}{\sqrt{n}}\frac{1}{(-1+\sqrt{n})}\right) = 0 \ ?##
     
  4. Jan 21, 2016 #3
    Both of these statements are obviously true (the first since ##(-1+\sqrt{n}) < (1+\sqrt{n})##). But I don't see how either one helps me ?
    I still have that alternating sign that doesn't allow me to use the comparison test. For example I can't say that ##\sum_{n=2}^\infty \frac{(-1)^n}{\sqrt{n}((-1)^n+\sqrt{n})} \le \sum_{n=2}^\infty \frac{(-1)^n}{\sqrt{n}(-1+\sqrt{n})}##and neither series converges absolutely either.

    I'm not sure I should use the Leibniz criteria either, these exercises are taken from a section covering that but I suspect they're meant to show cases where I can't apply that criteria directly or perhaps not at all and have to find other methods.
     
  5. Jan 21, 2016 #4
    I think I solved it now. Summing up the ##2n## and ##2n+1## terms seem to work.
    ##\frac{1}{\sqrt{2n}+2n}+ \frac{1}{\sqrt{2n+1}-2n-1} = \frac{\sqrt{2n+1}+\sqrt{2n}-1}{-4n^2+\dots}##
    which is of order ##\frac{1}{n^{3/2}}## and hence converges.

    While for the second one doing the same thing
    ##\frac{1}{1+\sqrt{2n}} + \frac{1}{1-\sqrt{2n+1}} = \frac{\sqrt{2n}-\sqrt{2n+1}+2}{2n}## which is of order ##\frac{1}{\sqrt{n}}## and hence diverges if you do compare it against the harmonic series.
     
  6. Jan 22, 2016 #5

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    What I intended with my suggestion was to suggest a way for you to use the Leibniz criteria.

    You have for the first series: ##\displaystyle \ \sum_{n=2}^\infty \frac{1}{\sqrt{n}}\frac{(-1)^n}{(-1)^n+\sqrt{n}} = \sum_{n=2}^\infty (-1)^n a_n \ ##

    Thus ##\displaystyle \ a_n = \frac{1}{\sqrt{n}}\frac{1}{(-1)^n+\sqrt{n}} \ . ## We see that an is positive. for all n ≥ 2, so the series is indeed alternating.

    To see that the sequence, [ an ] converges, compare it to the sequence, [ bn ], where
    ##\displaystyle b_n = \frac{1}{\sqrt{n}}\frac{1}{(-1+\sqrt{n})} \ . ##​

    When n is odd, an = bn .

    When n is even, an < bn , which is consistent with the first inequality I gave in post #2, and is:
    ##\displaystyle \frac{1}{\sqrt{n}}\frac{1}{(1+\sqrt{n})}<\frac{1}{\sqrt{n}}\frac{1}{(-1+\sqrt{n})}##​

    I think you'll find that the sequence, [bn] converges to 0, therefore, [an] converges to zero
     
  7. Jan 22, 2016 #6
    Right that's a good way to show that ##a_n## converges. I never thought about that part since it seemed quite obvious that the sequence converges with the ##\sqrt{n}## factor in the denominator. But this still wouldn't allow us to use the Leibniz criteria right since we don't have a decreasing sequence?

    So let's see if I understand this right, your post would be a way to start if Leibniz criteria would be applicable, but in this case it turned out to not be? I included the criteria in the post since most problem in the same section seemed to be using that criteria so I thought it would be of use.
     
  8. Jan 22, 2016 #7

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Oh! Of course.

    I missed the word 'decreasing'.

    DUH !
     
  9. Jan 22, 2016 #8
    Then we're on the same page! It didn't help that I thought I should use that criteria for a problem it didn't apply to at the start.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Convergence of alternating series
Loading...