The series terms are,

\[{a_n} = \frac{{{n^2}}}{{{n^3} + 1}}\]

We can clearly see that for the range of \(n\) in the series the terms are positive and so that condition is met.

Show Step 3

In this case we need to be a little more careful with checking the decreasing condition. We can’t just plug in \(n\) + 1 into the series term as we’ve done in the first couple of problems in this section. Doing that would suggest that both the numerator and denominator will increase and so it’s not all that clear cut of a case that the terms will be decreasing.

Therefore, we’ll need to do a quick Calculus I increasing/decreasing analysis. Here the function for the series terms and its derivative.

\[f\left( x \right) = \frac{{{x^2}}}{{{x^3} + 1}}\hspace{0.25in}\hspace{0.25in}f'\left( x \right) = \frac{{2x - {x^4}}}{{{{\left( {{x^3} + 1} \right)}^2}}} = \frac{{x\left( {2 - {x^3}} \right)}}{{{{\left( {{x^3} + 1} \right)}^2}}}\]

With a quick number line or sign chart we can see that the function will increase for \(0 < x < \sqrt[3]{2} = 1.2599\) and will decrease for \(\sqrt[3]{2} = 1.2599 < x < \infty \). Because the function and series terms are the same we know that the series terms will have the same increasing/decreasing behavior.

So, from this analysis we can see that the series terms are not always decreasing but will be decreasing for \(n > \sqrt[3]{2}\) which is sufficient for us to use to say that this condition is also met.

Okay, we now know that both of the conditions required for us to use the Integral Test have been verified we can proceed with the Integral Test.

It is very important to always check the conditions for a particular series test prior to actually using the test. One of the biggest mistakes that many students make with the series test is using a test on a series that don’t meet the conditions for the test and getting the wrong answer because of that!

Show Step 4

Now, let’s compute the integral for the test.

\[\int_{0}^{\infty }{{\frac{{{x^2}}}{{{x^3} + 1}}\,dx}} = \mathop {\lim }\limits_{t \to \infty } \int_{0}^{t}{{\frac{{{x^2}}}{{{x^3} + 1}}\,dx}} = \mathop {\lim }\limits_{t \to \infty } \left. {\left( {\frac{1}{3}\ln \left| {{x^3} + 1} \right|} \right)} \right|_0^t = \mathop {\lim }\limits_{t \to \infty } \left( {\frac{1}{3}\ln \left| {{t^3} + 1} \right| - \ln \left( 1 \right)} \right) = \infty \] Show Step 5

Okay, the integral from the last step is a divergent integral and so by the Integral Test the series must also be a divergent series.