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    • Thread Starter

    \sum_{n = M}^N \alpha^n = \frac{\alpha^M - \alpha^{N-1}}{(1 - \alpha)}

    What if \alpha=1????

    For what values of M and N does this formula hold? Can they both be negative? Does N need to be greater than M?
    • PS Helper

    PS Helper
    If \alpha = 1 then \alpha^n = 1 for each value of n, so your sum is 1 + 1 + \cdots + 1. You just have to work out how many 1s you're adding together.

    And to work out what values of M and N it holds for, think about how the formula is derived.

    We can fix M and for each N \ge M write S_N = \displaystyle \sum_{n=M}^N \alpha^n. Then:
    S_N = \alpha^M + \alpha^{M+1} + \cdots + \alpha^{N-1} + \alpha^N

    and so:
    \alpha S_N = \alpha^{M+1} + \alpha^{M+2} + \cdots + \alpha^{N} + \alpha^{N+1}

    You can take one of these away from the other (so loads of terms on the RHS cancel, leaving just two). Rearranging gives you the formula you have. Does this place any condition on M,N? I'll let you decide that for yourself.

    [EDIT: Just to clarify a point, if N<M then by definition the sum is zero (e.g. if you're summing the integers from 3 "up to" 1 it doesn't make much sense).]
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Updated: September 29, 2010
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