Recent content by tadf2

Cool, I get it now. You answered my question. Thanks Zondrina!

\frac{A}{z} this part is not necessarily needed for partial fractions because it doesn't require any expansions..

Homework Statement f = \frac{1}{z(z-1)(z-2)} Homework Equations Partial fraction The Attempt at a Solution R1 = 0 < z < 1 R2 = 1 < z < 2 R3 = z > 2 f = \frac{1}{z(z-1)(z-2)} = \frac{1}{z} * (\frac{A}{z-1} + \frac{B}{z-2}) Where A = -1 , B = 1. f = \frac{1}{z} *...

Inside the sine meaning, the argument of the 'arcsine' would only range from -1 to 1. So I'm guessing you can't make the substitution because arcsin(infinity) = error?

I was testing for convergence of a series: ∑\frac{1}{n^2 -1} from n=3 to infinity I used the integral test, substituting n as 2sin(u) so here's the question: when using the trig substitution, I realized the upperbound, infinity, would fit inside the sine. Is it still possible to make...