1. ## challenging factorial problem

Simplify

2/1! - 3/2! + 4/3! - 5/4! + ... + 2010/2009! - 2011/2010!,

where k! denotes the product of all integers from 1 up to k,

that is: 1! = 1, 2! =1 * 2 = 2, 3! = 3*2*1 = 6 and so on...

2. $Tn = ((n+1)/n!)(-1)^{n+1}
=(-1)^{n+1}(1/(n-1)! + 1/n!)
$

n varies from 1 to 2010, therefore, the series reduces to,
$1+1 -1 -1/2!+1/2!+1/3!-1/3!-1/4!+............-1/2010!$

All the intermediate terms vanish, and what survives is
$S=1-1/2010!$

Thus the value of this expression is $1-1/2010!$

3. you might want to try the expansion of

$xe^x$

Originally Posted by the undertaker
Simplify

2/1! - 3/2! + 4/3! - 5/4! + ... + 2010/2009! - 2011/2010!,

where k! denotes the product of all integers from 1 up to k,

that is: 1! = 1, 2! =1 * 2 = 2, 3! = 3*2*1 = 6 and so on...