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Math Help - Problem involving series

  1. #1
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    Problem involving series

    The following problem is in a section on convergence tests for series (comparison, ratio, root, integral, p-series, etc.) I've been stuck on it for about a day, so any help would be much appreciated.

    Suppose that \Sigma a_{n} is a convergent series with positive terms. Let \{S_{n}\} be the sequence of partial sums for \Sigma a_{n}. Let S = lim \ S_{n} and let \rho_{k} = \frac{a_{k+1}}{a_{k}}. Suppose there is a number N for which \rho_{k} \ge \rho_{k+1} if k \ge N and \rho_{N} < 1. Show that S - S_{N} \le \frac{a_{N+1}}{1 - \rho_{N}}.

    So far all I've been able to do is show that S - S_{N} = a_{N+1}(1 + \rho_{N+1} + \rho_{N+1}\rho_{N+2} + \rho_{N+1}\rho_{N+2}\rho_{N+3} + ...) which means I need to show that 1 + \rho_{N+1} + \rho_{N+1}\rho_{N+2} + \rho_{N+1}\rho_{N+2}\rho_{N+3} + ... \le \frac{1}{1 - \rho_{N}}, but I'm not sure where to take it from here, or if this is even a good approach.
    Last edited by spoon737; May 9th 2012 at 02:37 PM.
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  2. #2
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    Re: Problem involving series

    So you know 0 < \rho_N < 1. What can you say about \Sigma (\rho_N)^n? Does it converge? Where to? You also know that \rho_N \geq \rho_{N+1}. So then what can you say about the sum that you found?
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  3. #3
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    Re: Problem involving series

    Ah, I see. Term for term, my series is less than or equal to a geometric series converging to \frac{1}{1 - \rho_{N}}. Thanks a billion!
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