- #1
tadf2
- 5
- 0
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?
∑[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?