Does the Series (n^2+1)/(n^3+1) Converge or Diverge?

[dan38]

Homework Statement

series for (n^2+1) / (n^3+1)

Homework Equations

The Attempt at a Solution

I was under the impression that if the denominator is increasing at a larger rate than the numerator, then it will become smaller and smaller approaching zero? So it would converge

But I did the working and the comparison test dictates that it must diverge

Which one is right?

 

[vela]

Just because the terms of the series approach 0 doesn't mean the series will converge.

The converse, however, is true: If the series converges, the terms approach 0. The n-th term test is the contrapositive of this statement: If the terms don't approach 0, the series diverges. Note it says nothing about what happens when the terms do go to 0.

 

[dalcde]

Well, 1 + 1/2 + 1/3 + 1/4 + ... diverges but the terms approach 0 (see "Harmonic Series" in wikipedia)

 

[Infinitum]

The comparison test is right.

 

[Curious3141]

Expressing the general term as \frac{1}{n - \frac{n-1}{n^2+1}} and comparing to the harmonic series gives a quick answer.

 

[algebrat]

Expressing the general term as \frac{1}{n - \frac{n-1}{n^2+1}} and comparing to the harmonic series gives a quick answer.

Dividing top and bottom by n² (or even n³) might do what you intended a little nicer, me thinks. But like Curious said, it gives you a quick answer, and for instance is not a proof, that I know of.

 

[tt2348]

Use a comparison test, with n^2<n^2+1, and 1/(n^3+n^2)<1/(n^3+1), combining this will gives 1/(n+1)<Sn

 

[tt2348]

I think you're confusing series and sequences