Re: Limit Comparison Series

Quote:

Originally Posted by

**minneola24** 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?

Thanks.

which is a harmonic series, and is divergent. Therefore the original series is divergent.

Re: Limit Comparison Series

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.

Re: Limit Comparison Series

Quote:

Originally Posted by

**minneola24** 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.

When using the limit comparison test it is best to do direct computations.

Let

Now

Thus we do not need to 'worry' with direct comparison.