[jason_t]

[ QUOTE ]
My favorite summation trick using the geometric series is the following argument:

sum of x^n = (1 - x)^(-1)
differentiate to get
sum of n x^(n-1) = (1 - x)^(-2)
multiply by x to get
sum of n x^n = x * (1 - x)^(-2)

So, e.g.
sum of n / 2^n = 2.

[/ QUOTE ]

I like this one:

(d/dx) ln(1-x) = -1/(1-x) = -sum x^n

so

ln(1-x) = -sum x^(n+1) / (n+1)

hence

ln(2) = 1-1/2+1/3-1/4+1/5-1/6+....