Trig substitution into integrals

[tadf2]

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?

 

[Mark44]

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.

What does "fit inside the sine" mean?

Is it still possible to make the substitution? Or is there a restriction when this happens?

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.

 

[tadf2]

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?

 

[gopher_p]

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.