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A Complicated-Looking Infinite Sum

Asked by Andrew Gill, student, State College Area High School on August 8, 1997:
I have been trying to find the sum of:
        ______                 _    2  q 
        \        (2q + 1 + pi\/n)(pi n) 
         \       _____________________
        /_____          (2q + 1)!

(where n is a positive integer), but have not been able to find the process for evaluating this power series. Any and all information on this problem would be greatly appreciated.

This summation can be broken up and rewritten. Splitting the fraction into two pieces by writing (2q + 1 + pi sqrt(n)) (pi^2 n)^q / (2q+1)! as the sum of (2q + 1) (pi^2 n)^q / (2q+1)! and (pi sqrt(n)) (pi^2 n)^q / (2q+1)!, then rewriting (pi^2 n)^q as (pi sqrt(n))^(2q) and simplifying, gives
        infinity                           infinity
        ______         _ 2q        _ 2q+1   ______         _ k
        \        (pi \/n)    (pi \/n)       \        (pi \/n)
         \      __________ + _______     =   \      _________
         /                                   /   
        /_____    (2q)!       (2q+1)!       /_____      k!
          q=0                                 k=0

(the first sum runs over all even integers k=2q, the second over all odd integers k=2q+1, so the combined sum runs over all integers k).

Now you should notice that the last summation is just the power series expansion of e^(pi sqrt(n)).

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