### Video Transcript

Use the integral test to determine whether the series the sum from π equals one to β of one divided by three π plus seven all squared is convergent or divergent.

In this question, weβre given an infinite series. And we need to determine the convergence or divergence of this series by using the integral test. So the first thing weβre going to need to do is recall the integral test. This tells us if π of π₯ is a continuous, decreasing, positive function for all values of π₯ greater than or equal to some nonnegative integer π and we have that π evaluated at π is equal to our sequence ππ for all of our values of π, then we know the following by using the integral test.

If the integral from π to β of π of π₯ with respect to π₯ is convergent, then we know our series the sum from π equals π to β of ππ will also be convergent. However, if the integral from π to β of π of π₯ with respect to π₯ is divergent, then our series the sum from π equals π to β of ππ will also be divergent. So the integral test gives us a way of turning a question about the convergence or divergence of a series into a question about the convergence or divergence of an integral. The first thing weβll need to do is check that we can use the integral test on the series given to us in the question.

First, we see that our series starts from π is equal to one. So weβll start by trying to set our value of π equal to one. And in fact, we can use this to rewrite our integral test. Next, remember, π evaluated at π needs to be our summand. So weβll set π of π₯ equal to our summand with π equal to π₯. In other words, π of π₯ will be equal to one over three π₯ plus seven all squared. Now, to use the integral test, we need to check the following three conditions are true for our function π of π₯. It needs to be continuous, it needs to be decreasing, and it needs to be a positive function for all values of π₯ greater than or equal to one.

Weβll start with the first condition, checking that π of π₯ is continuous for all values of π₯ greater than or equal to one. To do this, we notice that our function π of π₯ is a rational function. Itβs the quotient of two polynomials. And we know that all rational functions are continuous across their entire domain. And we also know the domain of a rational function will be everywhere except when its denominator is equal to zero. And for our function π of π₯, its denominator will be equal to zero when π₯ is equal to negative seven over three. So our function π of π₯ is continuous everywhere except when π₯ is equal to negative seven over three.

And weβre only interested in values of π₯ greater than or equal to one. So we know that π of π₯ is continuous on this interval. Next, we want to know if π of π₯ is a decreasing function on this interval. To do this, letβs take a closer look at our function π of π₯. First, we see the numerator is a constant, one, which is positive, and our denominator is a square, which is greater than or equal to zero. In fact, we know on this interval, it will be positive. In fact, this also tells us our third condition is true. However, weβll be needing this to show that this is a decreasing function.

Now that we know π of π₯ is the quotient of two positive numbers on this interval, we can show that itβs a decreasing function. To do this, think what happens in our denominator inside of the parentheses when we increase our value of π₯. We know that three π₯ plus seven is a positive, sloping, linear function. We know this means itβs an increasing function. So taking a larger value of π₯ will increase the value of three π₯ plus seven. But we then square this number and take one divided by this number. In other words, increasing our value of π₯ will make our denominator a larger positive number.

So increasing the value of π₯ means weβre decreasing the value of π of π₯. In other words, this is a decreasing function. Therefore, weβve shown for values of π₯ greater than or equal to one, π of π₯ is a decreasing function. This means weβve now shown all of our prerequisites are true for using the integral test. This means instead of checking the convergence or divergence of our series, we can instead look at the convergence or divergence of the integral. So we want to evaluate the convergence or divergence of the integral from one to β of one divided by three π₯ plus seven all squared with respect to π₯.

To do this, we need to notice this is an improper integral. So we need to add a variable π‘ in our upper limit and then tend this value of π‘ to β. In other words, the convergence or divergence of this integral is the same as the convergence or divergence of the limit as π‘ approaches β of the integral from one to π‘ of one over three π₯ plus seven all squared with respect to π₯. And now, π‘ is just some constant inside of our integral. This means our integrand is continuous across the entire interval of integration. And we can use any of our integral rules to help us evaluate this integral.

In particular, to help us use this one, weβll use a substitution. Weβll substitute π’ is equal to three π₯ plus seven. So weβll need to differentiate both sides of our substitution π’ is equal to three π₯ plus seven with respect to π₯. This gives us dπ’ by dπ₯ is equal to three. And remember, we know that dπ’ by dπ₯ is not a fraction. However, when weβre using integration by substitution, we can treat it a little bit like a fraction. Using this, we can rearrange this equation to give us the equivalent statement in terms of differentials. One-third dπ’ is equal to dπ₯.

Weβre now almost ready to evaluate this integral by using our substitution. However, remember, this is a definite integral. So weβre going to need to find our new limits. To find our new upper and lower limits of integration, weβre going to need to substitute π₯ is equal to one and π₯ is equal to π‘ into our substitution. This gives us a new upper limit of π’ is equal to three π‘ plus seven and a new lower limit of π’ is equal to three times one plus seven, which is of course equal to 10.

So by using the substitution π’ is equal to three π₯ plus seven and the fact that one-third dπ’ will be equal to dπ₯, we now have the limit as π‘ approaches β of the integral from 10 to three π‘ plus seven of one over three times π’ squared with respect to π’. And now, we can just evaluate this integral by using the power rule for integration. It might help us to rewrite our integrand as one-third times π’ to the power of negative two. So we need to add one to our exponent of π’ and then divide by this new exponent. This gives us the limit as π‘ approaches β of negative π’ to the power of negative one divided by three evaluated at the limits of integration π’ is equal to 10 and π’ is equal to three π‘ plus seven.

And before we evaluate this at the limits of integration, weβll rewrite π’ to the power of negative one over three as one over three π’. Now, we just need to evaluate this at the limits of integration. We get the limit as π‘ approaches β of negative one divided by three times three π‘ plus seven plus one over three times 10. And we could start simplifying. However, at this point, we can just evaluate this limit. Our limit is as π‘ is approaching β. In our first term, our numerator remains constant. However, we can see our denominator contains a factor of π‘. Itβs growing without bound.

Therefore, if our numerator remains constant but our denominator is growing without bound, our first term is tending to zero. Of course, we can see that our second term doesnβt change as the value of π‘ changes. Itβs constant. This means that its limit will just be equal to itself. So our limit was equal to one over 30. And since this is convergent, that means that our integrand is convergent.

And remember, by the integral test, if our integral is convergent, then our series must also be convergent. Therefore, by using the integral test, we were able to determine the sum from π equals one to β of one divided by three π plus seven all squared is convergent.