I have been trying to find the sum of: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, givesinfinity ______ _ 2 q \ (2q + 1 + pi\/n)(pi n) \ _____________________ / /_____ (2q + 1)! q=0(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.
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)).