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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


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    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


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    Staff: Mentor

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

    That was the idea I had in mind, indeed.
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