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Quick question about a telescoping series

  1. Feb 27, 2005 #1
    [tex] \sum_{j=k}^{\infty}\left\{{\frac{1}{b_{j}}-\frac{1}{b_{j+1}}}\right\}=\frac{1}{b_{k}}[/tex]

    This series holds for all real monotone sequences of [tex] b_j [/tex].

    So if I were to carry this series out to say n I end up with a partial sum that looks like:

    [tex] S_n=\frac{1}{b_k}-\frac{1}{b_{k+(n+1)}} [/tex]

    Now as n goes to infinity we are left with just [tex] b_k [/tex]. This of course implies that [tex]\frac{1}{b_{k+(n+1)}}[/tex] goes to zero as n goes to infinity. So does this mean that the monotone sequence [tex]b_j[/tex] must equal {1,2,3,4,5,...,j} ? If not what exactly are the constraints on [tex]b_j[/tex] to make that series an identity?

    Thanks for the help everyone.

    JTB
     
  2. jcsd
  3. Feb 27, 2005 #2
    Never mind...I figured it out. As long as [tex]b_j[/tex] is strictly increasing then the identity holds. The amount of jump between any two terms is irrelevant.

    If there are any further comments please feel free to make them other wise I will let this thread die peacefully.

    JTB
     
  4. Feb 28, 2005 #3

    NateTG

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    Actually, it's a bit cleaner to talk about it as:
    [tex]\sum_{i=k}^{\infty} \left(a_i-a_{i+1}\right)[/tex]
    Then any partial sum can easily be evaluated:
    [tex]\sum_{i=k}^{n} \left(a_i-a_{i+1}\right) = a_k-a_{n+1}[/tex]
    so we have
    [tex]\lim_{n \rightarrow \infty} \sum_{i=k}^{n} \left(a_i-a_{i+1}\right) = \lim_{n \rightarrow \infty} a_k-a_{n+1}=a_k - \lim_{n \rightarrow \infty} a_{n+1}[/tex]

    There's no need to restrict the series to being monotone or real.
     
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