I have to use the comparison test although I'm bad at coming up with what to compare it to
What if I compare it to 1/sqrt(x)
The Attempt at a Solution
I take the integral of 1/sqrt(x) from [0,PI] I get -2sqrt(x) and then I use limits and evaluate
I get 2sqrt(PI)
Can I do this? So since this integral is larger then the one in question my integral converges?