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Homework Help: Sum of an infinite series

  1. Nov 3, 2016 #1
    1. The problem statement, all variables and given/known data

    Find the sum of the given infinite series.
    $$S = {1\over 1\times 3} + {2\over 1\times 3\times 5}+{3\over 1\times 3\times 5\times 7} \cdots $$

    2. Relevant equations

    3. The attempt at a solution
    I try to reduce the denominator to closed form by converting it to a factorial.

    $$\sum_{k \ge 1} {k\over {\prod^{k + 1}_{a = 1} 2a -1 }}$$
    $$\sum_{k \ge 1} {k\times \prod^{k + 1}_{a = 1} 2a\over \prod^{k + 1}_{a = 1} 2a -1 \times \prod^{k + 1}_{a = 1} 2a}$$
    $$\sum_{k \ge 1} {k\times 2^{k+1} \times (k+1)! \over (2(k+1))!}$$

    I hit the dead end here. Although i can simply this a bit more but i still can't find a series that i can sum easily.
    Please provide some hints as to how can i proceed further.
    Last edited: Nov 3, 2016
  2. jcsd
  3. Nov 3, 2016 #2


    Staff: Mentor

    What are you supposed to show in this problem? Your problem description is lacking this information.
  4. Nov 3, 2016 #3
    The trick is to write it as a telescopic sum:
    \frac{n}{(2n+1)!!} = \frac{1}{2(2n-1)!!} - \frac{1}{2(2n+1)!!}
    (in case you haven't seen it before, !! is the double factorial)
  5. Nov 3, 2016 #4
    I am really sorry Mark. I have edited the question. Please see if my edit is sufficient enough.
  6. Nov 3, 2016 #5


    Staff: Mentor

    Much better. I wasn't sure whether you were supposed to determine if the series converged or diverged, or if it converged, find the sum.
  7. Nov 3, 2016 #6
    I don't know what double factorial means. Sorry.
    Does double factorial means ##n!! = n(n-2)(n-4)\times\cdots \times3 \times 1## or ## (n!)!##?

    Also, can you tell how you know that this is a telescopic series ?
    Have you done this question before ?
  8. Nov 3, 2016 #7
    This one.
    Nope, haven't done it before - but its a standard technique for evaluating infinite summation of series. It probably helps to examine the structure of the first few terms. This can lead you to a good guess / ansatz that you can then prove in general for an arbitrary term.
  9. Nov 3, 2016 #8
    Thank you very much. You are genius for sure.
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