So if I compare it to 1/n that new comparison diverges.
But 1/n is bigger than the original function so I cant conclude that the original diverges too.
If I divide the original by the 1/n and take the limit as n-> infinity I keep going in loops using lhopitals.
How do I solve this?
Another possibility is the integral test. We can compare the sum to the function . Since the function is always decreasing when , a left-hand endpoint estimate of the integral will be an over-estimate. Notice that when we choose to have each of our subintervals be 1 unit in length, then the integral can be approximated by . But since this is an overestimate, we can say
Since the sum is greater than this divergent integral, the sum must also diverge.