26th CanMO 1994 Problem 1 Find -3/1! + 7/2! - 13/3! + 21/4! - 31/5! + ... + (1994 2 + 1994 + 1)/1994!
∑ (-1) n (n/n! + 1/n!) = -1 + 1/1994! (the series telescopes). Similarly, ∑ (-1) n n 2 /n! = ∑ (-1) n n/(n-1)! = ∑ (-1) n (1/(n-2)! + 1/(n-1)! ) = 1/1993! . So expression given is -1 + 1/1994! + 1/1993! = -1 + 1995/1994! . 26th CanMO 1994 © John Scholes jscholes@kalva.demon.co.uk 18 Aug 2002 Last corrected/updated 28 Nov 03