I was testing for convergence of a series: ∑[itex]\frac{1}{n^2 -1}[/itex] 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 the substitution? Or is there a restriction when this happens?
What does "fit inside the sine" mean? Sure, you can make the substitution. The integral will be from 3 to, say b, and you take the limit as b → ∞. Not that you asked, but it's probably simpler and quicker to break up 1/(n^{2} - 1) using partial fractions.
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?
If you're looking for an appropriate trig substitution for the definite integral (and not just one that gets you a correct antidierivative), then ##\sec u## is the way to go. But like Mark44 said, partial fractions is really the "right" technique of integration for this particular integral.