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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?

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Have you tried the ratio test for proving convergence? –  Timmy Turner Aug 2 '12 at 10:02
    
There are some indexing problems. Your sum starts with $k=0$, but definition of $a_k$ assumes that $k\geq 1$ –  no identity Aug 2 '12 at 10:04
    
I know the series converges. The issue is to find the limit of the sum. –  Alexander Aug 2 '12 at 10:08
    
The indexing in the summation is correct now. Thanks for the comment. –  Alexander Aug 2 '12 at 10:19

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