Solution 2 Tues., Dec. 19, 2006

$\sum 4 \frac{1}{(4k)!} = \sum 2 (1 + (-1)^k) \frac{1}{(2k)!}$

$= 2 \sum \frac{1}{(2k)!} + 2 \sum \frac{(-1)^k}{(2k)!}$

$= 2 \cosh(1) + 2 \cos(1) \,$

Retrieved from "http://www.efnet-math.org/w/Solution_2_Tues.%2C_Dec._19%2C_2006"