Prove that converges.
here is a solution that my friend gave me.
using the hint we have
the first expression converges to . for the third expression we can change the lower limit to 1, since this will not effect the convergence. so we have
To handle the second expression we apply integration by parts once again and get.
to evaluate the first term we can either use l'hopitals rule or taylor series, your pick, either way you find that it does converge. For the last 2 we use fact that
I don't see why this integral shouldn't coverge, though obviously my argument above is hardly a proof.
If, the continous functions
And, converges then so does (Of course I mean the improper intergral).
But your inequality does not satisfy the one above, namely because it can be negative.
However, I have a way to fix this error.
Using the integral series test we need to test,
and see whether it converges. It does since its absolutely convergent, since .