Question Video: Using the Integral Test to Determine the Convergence or Divergence of a Series Where the Summant Is a Rational Function Mathematics

Video Transcript

Use the integral test to determine whether the series the sum from 𝑛 equals one to ∞ of 𝑛 divided by three plus four 𝑛 squared converges or diverges.

In this question, we’re given an infinite series. And we’re asked to determine the convergence or divergence of this series by using the integral test. So we’ll start by recalling what the integral test tells us. This says, if 𝑓 of 𝑥 is a continuous, positive, decreasing function for all values of 𝑥 greater than or equal to some nonnegative integer 𝑘 and 𝑓 evaluated at 𝑛 is equal to our sequence 𝑎 𝑛 for all of our values of 𝑛, then the integral test tells us if the integral from 𝑘 to ∞ of 𝑓 of 𝑥 with respect to 𝑥 is convergent, then our series the sum from 𝑛 equals 𝑘 to ∞ of 𝑎 𝑛 must also be convergent.

However, if the integral from 𝑘 to ∞ of 𝑓 of 𝑥 with respect to 𝑥 is divergent, then our series the sum from 𝑛 equals 𝑘 to ∞ of 𝑎 𝑛 must also be divergent. So the integral test gives us a method of turning a question about the convergence or divergence of a series into a question about the convergence or divergence of an integral. So let’s try and apply the integral test to the series given to us in the question. First, we can see our series starts from 𝑛 is equal to one. So we’ll set 𝑘 equal to one. In fact, this means we can update our integral test for the value of 𝑘 equal to one.

Next, remember, we need 𝑓 evaluated at 𝑛 to be equal to our summand 𝑎 𝑛. So we’ll set 𝑓 of 𝑥 to be our summand with 𝑥 instead of 𝑛. This gives us 𝑓 of 𝑥 is equal to 𝑥 divided by three plus four 𝑥 squared. Now, to use the integral test, we need to check the following three properties are true for our function 𝑓 of 𝑥. First, we need to show that it’s continuous for all values of 𝑥 greater than or equal to one. In this case, our function 𝑓 of 𝑥 is the quotient of two polynomials. In other words, 𝑓 of 𝑥 is a rational function. And we know that all rational functions are continuous across their entire domain.

So to show that our function 𝑓 of 𝑥 is continuous for all values of 𝑥 greater than or equal to one, we just need to show it’s defined for all values of 𝑥 greater than or equal to one. And since this is a rational function, we know the only way this won’t be defined is if our numerator is equal to zero. However, if we were to solve three plus four 𝑥 squared equals zero, we would see it has no real solutions. Therefore, 𝑓 of 𝑥 is continuous for all real values of 𝑥. In particular, it will be continuous for 𝑥 greater than or equal to one.

Next, we need to show that 𝑓 of 𝑥 is positive for all values of 𝑥 greater than or equal to one. To do this, let’s take a closer look at our function 𝑓 of 𝑥 when 𝑥 is greater than or equal to one. First, we can see our numerator will be positive; it’s greater than or equal to one. Next, we can see in our denominator 𝑥 squared will be positive. We multiply this by four, which will be positive. And then we add three, which will also be positive. Therefore, 𝑓 of 𝑥 is the quotient of two positive numbers, which of course means that 𝑓 of 𝑥 is positive for all values of 𝑥 is greater than or equal to one.

The last thing we need to show is 𝑓 of 𝑥 is a decreasing function for all values of 𝑥 greater than or equal to one. And we could try and do this directly. However, since this is a rational function, we can always try and do this by checking the slope is negative when 𝑥 is greater than or equal to one by using the quotient rule. By using the quotient rule, we get that 𝑓 prime of 𝑥 is equal to the derivative of 𝑥 with respect to 𝑥 times three plus four 𝑥 squared minus the derivative of three plus four 𝑥 squared with respect to 𝑥 times 𝑥 all divided by three plus four 𝑥 squared all squared.

Now, we just need to simplify this expression. First, we’ll evaluate our derivatives by using the power rule for differentiation. We know the derivative of 𝑥 with respect to 𝑥 is equal to one. Next, the derivative of three plus four 𝑥 squared with respect to 𝑥 is equal to eight 𝑥. This gives us three plus four 𝑥 squared minus eight 𝑥 times 𝑥 all divided by three plus four 𝑥 squared all squared. And we can simplify this even further.

First, in the second term in our numerator, 𝑥 multiplied by 𝑥 can be simplified to give us 𝑥 squared. Then the last piece of simplification we’ll do in our numerator is four 𝑥 squared minus eight 𝑥 squared is equal to negative four 𝑥 squared. So we get 𝑓 prime of 𝑥 is equal to three minus four 𝑥 squared all divided by three plus four 𝑥 squared all squared. Remember, 𝑓 will be decreasing whenever its slope is negative. So we want to check the slope of our function for values of 𝑥 greater than or equal to one. So let’s check the numerator and the denominator separately.

Let’s start with our denominator. We can see this is the square of a positive number, so this will be positive. But we get a different story in our numerator. Consider what happens to 𝑥 squared. If 𝑥 is greater than or equal to one, we know 𝑥 squared will also be greater than or equal to one. We can then multiply both sides of this inequality through by negative four to get that negative four 𝑥 squared will be less than or equal to negative four. Finally, we can add three to both sides of this inequality to get three minus four 𝑥 squared will be less than or equal to negative one. But this is exactly what we have in the numerator of this expression.

So our numerator is negative. However, our denominator is positive. So this is the quotient of a negative number and a positive number, which means it’s negative. So we’ve shown when 𝑥 is greater than or equal to one. The slope of 𝑓 of 𝑥 is negative, which means it’s a decreasing function. So we’ve now shown all of the prerequisites for the integral test. So we can try and use this to determine the convergence or divergence of our series. So let’s clear some space and try to determine the convergence or divergence of this integral.

By using what we know about improper integrals, we know the convergence or divergence of the integral from one to ∞ of 𝑥 divided by three plus four 𝑥 squared with respect to 𝑥 will be the same as the convergence or divergence of the limit as 𝑡 approaches ∞ of the integral from one to 𝑡 of 𝑥 divided by three plus four 𝑥 squared with respect to 𝑥. So now our integrand is some finite value of 𝑡. And we’ve already shown that our integrand is continuous for all values of 𝑥 greater than or equal to one. So now, we can use any of our tools for integration to help us evaluate this integral.

There’s a few different ways we could try evaluating this integral. However, the easiest is to recall the following rule of integration. The integral of 𝑓 prime of 𝑥 divided by 𝑓 of 𝑥 with respect to 𝑥 is equal to the natural logarithm of the absolute value of 𝑓 of 𝑥 plus a constant of integration 𝐶. This relies on us noticing that if we differentiate our denominator, we get eight 𝑥. This is a linear multiple of our numerator. In fact, we could just multiply our numerator by eight and divide our entire integral by eight. And now, our integral is in exactly this form. In fact, we don’t need to add our constant of integration 𝐶 since we’re using this in a definite integral.

So by using our integral rule, we can rewrite our limits as the limit as 𝑡 approaches ∞ of one-eighth times the natural logarithm of the absolute value of three plus four 𝑥 squared evaluated at the limits of integration 𝑥 is equal to one and 𝑥 is equal to 𝑡. Now, all we need to do is evaluate this at the limits of integration. Doing this gives us the limit as 𝑡 approaches ∞ of one-eighth times the natural logarithm of the absolute value of three plus four 𝑡 squared minus the natural logarithm of the absolute value of three plus four times one squared.

And now, we can start simplifying. First, three plus four times one squared is equal to seven. Next, we can take the one-eighth outside of our limit. And this gives us the following limit, which we can evaluate directly. We see 𝑡 is approaching ∞. As 𝑡 approaches ∞, three plus four 𝑡 squared is approaching ∞. And if this is approaching ∞, then the natural logarithm of this is also approaching ∞. And of course, the second term in this limit is constant; it doesn’t change if the value of 𝑡 changes. So our limit is growing without bound and therefore is divergent.

But remember, this limit being divergent tells us that our integral is divergent. And our integral being divergent tells us by using the integral test that our series must be divergent. Therefore, by using the integral test, we were able to determine the sum from 𝑛 equals one to ∞ of 𝑛 divided by three plus four 𝑛 squared is divergent.