# Trig substitution into integrals

5
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 the substitution? Or is there a restriction when this happens?

### Staff: Mentor

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/(n2 - 1) using partial fractions.