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Show that for any $b>0$ and $n>0$, the alternating sum of the series

$ a_k=\prod_{i=1}^k\frac{1+b/(n+i)}{1+b/(n+1)} $

converges such that

$ \sum_{k=1}^{\infty} (-1)^k a_k \leq -\frac{1}{2} $

In fact

$ a_k=a_{k-1}\frac{(n+k+b)(n+1)}{(n+k)(n+1+b)} $

Can anybody help me?

  • 0
    The indexing in the summation is correct now. Thanks for the comment.2012-08-02

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