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 series

  1. May 5, 2013 #1
    1. The problem statement, all variables and given/known data

    Let [itex](a_n)[/itex] be a sequence.
    (i) Prove that if [itex]\sum\limits_{n = 1}^\infty {{a_n}} [/itex] converges, then [itex]\sum\limits_{n = 1}^\infty {\left( {{a_{2n - 1}} + {a_{2n}}} \right)} [/itex] also converges.

    (ii) Prove that if [itex]\sum\limits_{n = 1}^\infty {\left( {{a_{2n - 1}} + {a_{2n}}} \right)} [/itex] converges and [itex]a_n \to 0[/itex], then [itex]\sum\limits_{n = 1}^\infty {{a_n}} [/itex] converges.

    2. Relevant equations



    3. The attempt at a solution

    (i) Let [itex]{R_n} = \sum\limits_{k = 1}^n {{a_k}} [/itex], [itex]{S_n} = \sum\limits_{k = 1}^n {{a_{2k - 1}}} [/itex], and [itex]{T_n} = \sum\limits_{k = 1}^n {{a_{2k}}} [/itex]

    Then [itex]{R_{2n}}={S_n}+{T_n}[/itex], since [itex]\sum\limits_{n = 1}^\infty {{a_n}} [/itex] converges, the sequence [itex](R_n)[/itex] converges, and so is the subsequence [itex](R_{2n})[/itex]. It follows that [itex]\sum\limits_{n = 1}^\infty {\left( {{a_{2n - 1}} + {a_{2n}}} \right)} [/itex].

    (ii) Ok so I'm stuck on this part. I already have [itex](R_{2n})[/itex] converges and [itex](a_n)[/itex] is bounded, how can I go about proving that [itex](R_{n})[/itex] converges as well? Thank you!
     
  2. jcsd
  3. May 5, 2013 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    The convergence of Rn does not guarantee the convergence of Sn or Tn, you cannot write it as sum like that.
    The idea to use a sequence and subsequences is good, however.

    (ii) how can R2n+1 deviate from R2n, if an->0?
     
  4. May 8, 2013 #3
    I'm sorry I wasn't able to reply earlier.

    For (i), I agree with you that the convergence of [itex]R_n[/itex] does not imply the convergence of [itex]S_n[/itex] or [itex]T_n[/itex] but it can imply that ([itex]S_n[/itex]+[itex]S_n[/itex]) converges right?

    Btw, I think I may get your point for part (ii), let me just say what I'm thinking: Since [itex]R_{2n}[/itex] = [itex]R_{2n-1}[/itex]+[itex]a_{2n}[/itex], it can be deduced that [itex]R_{2n-1}[/itex] converges to the same limit as [itex]R_{2n}[/itex], as a result, [itex]R_{n}[/itex] converges.
     
  5. May 8, 2013 #4

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    If you replace the second S by a T (typo?), yes.

    That was the idea I had in mind, indeed.
     
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 series
  1. Series Convergence (Replies: 3)

  2. Convergence of series (Replies: 2)

  3. Series convergence (Replies: 26)

  4. Convergence of a series (Replies: 24)

  5. Series Convergence (Replies: 15)

Loading...