# Determine if integral is convergent or divergent?

1. Mar 18, 2012

### theBEAST

1. The problem statement, all variables and given/known data
I attached the problem to this post.

3. The attempt at a solution
I was wondering if I could use the limit comparison test for this integral. My professor taught us this test that can be used for series but could it work for improper integrals as well?

So what I would do is an = 1/(2ex-x) and bn = 1/2ex.

Then take the limit approaching infinity of an/bn which yields 1. Since bn converges so does an. Does that make sense?

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2. Mar 18, 2012

### zooxanthellae

I don't quite understand how you're applying the limit comparison test from series to integrals. However, my instinct is that you are making a jump that can't be made. You might check this yourself by seeing if you can find a counterexample to your method.

However, there does exist a technique that is similar to the limit comparison test. The idea is to find an integral that you know converges/diverges on the interval and use that. So if we had 1/x^2 + 1, we know that on [0,infinity) this integral is < the integral of 1/x^2 on [0,infinity) and therefore converges since the integral of 1/x^2 converges (why? this one is easily computed manually). Can you find such a function here?

3. Mar 18, 2012

### SammyS

Staff Emeritus

This is the equivalent comparison test for integrals.

If $\displaystyle |g(x)|\ge|f(x)|\text{ for all }x \ge a\,,$ and if $\displaystyle \int_a^\infty\ |g(x)|\ dx$ converges, then so does $\displaystyle \int_a^\infty\ |f(x)|\ dx$ converge.

4. Mar 18, 2012

### theBEAST

Yup I can use 1/e^x but why wouldn't the limit comparison test work for integrals as well?

5. Mar 19, 2012

### zooxanthellae

I initially interpreted your post as using the a_n, a_n+1 limit test. What you've actually described is very close to what SammyS has posted, which is what the Integral Limit Comparison Test usually describes. I'm reasonably sure that your method works, but the method SammyS has posted is, I think, usually easier.