Series comparison test question

[cue928]

Problem:
Summation from n=1 to infinity: 1/(n^3 + n^2)

I understand, for example, another problem wherein it is 3/(4^n + 5) what you would compare that one to but how do you go about breaking this one up with two "n" terms? Are you supposed to, in general, pick the largest value of n? So perhaps compare it to 1/n^3 ?

Also, why does 1/n^2 from n=1 to inf. converge? Does it converge to zero?

 

[Dick]

1/n^2 does not converge to 0. It's a sum and all of the terms are positive. It can't possibly sum to zero. To prove it converges use an integral test. 1/n^3 is greater than 1/(n^3+n^2). That makes it a bad candidate for a comparison if you want to prove 1/(n^3+n^2) converges. Find something less than 1/(n^3+n^2) that you know converges.

 

[cue928]

But how do you find something a candidate for the test? That's what I don't see.

 

[Dick]

But how do you find something a candidate for the test? That's what I don't see.

Take everything in the denominator and replace it with something less than or equal to the term you are replacing but that still converges. That will give you a fraction that's greater, right? Sorry, I got the less and greater backwards the first time I posted this.

 

[Dick]

Ok, big hint. Which is larger 1/(n^3+n^2) or 1/(n^2+n^2)? Can you show the latter converges?