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.

 

[CellCoree]

Yes, it is still possible to make the substitution. The substitution of n as 2sin(u) is a valid substitution for this integral, regardless of the value of the upper bound. The upper bound being infinity does not affect the validity of the substitution. However, it is important to keep in mind that when evaluating the integral, the upper bound must also be substituted as infinity. This may require some additional algebraic manipulation to simplify the integral before evaluating it at infinity.