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Ricky
2006-03-06 17:29:54

And g(x) is 1/x

So f(x) <= g(x)

But g(x) is divergent.  Thus, but the comparison test, this says absolutely nothing about convergence.

Basically, the covergent thm just says that if you have something less than a finite number (i.e. convergent), it is also a finite number.  If you have something greater than an infinite number (divergent), it is also infinite.

But what if you have something greater than a finite number?  You can't really tell if that's finite or infinite.  And if you have something less than an infinite number?  You can't really say if that is still finite or infinite.

Alk
2006-03-06 17:23:45

I don't think I understand the actual process of determining this solution. So far I have
f(x) >= g(x) >= 0

f(x) = dx/(x + e^2x)

Would someone be willing to explain the steps in solving this?

Ricky
2006-03-06 16:47:54

The comparison test is as follows:

If f(x) < g(x) and g(x) is convergent, then f(x) must be convergent

If f(x) > g(x) and g(x) is divergent, then f(x) must be divergent.

So you can't use < for divergent like you did.

Alk
2006-03-06 15:51:50

The problem I am working on is:

Use the comparison theorem to determine whether the integral is convergent of divergent. The integral is dx/(x + e^2x) from 1 to infinite.

My question is why does this not work.
For x >= 1, 1/(x+ e^2x)   <    1/x, therefore 1/x+e^2x is divergent.