Lesson Video: Integral Test for Series Mathematics

Video Transcript

Integral test for series

In this video, we will learn how to use the integral test for series to determine whether a series containing nonnegative terms is convergent or divergent. We will be looking at a variety of examples of how the integral test can be used. Let’s start by covering what the integral test is. The integral test for series tells us that if we have some 𝑓 of 𝑥 which is a continuous, positive, decreasing function on the interval between 𝑘 and ∞ and that 𝑓 of 𝑛 is equal to 𝑎 𝑛, then, firstly, if the integral from 𝑘 to ∞ of 𝑓 of 𝑥 with respect to 𝑥 is convergent, so is the sum from 𝑛 equals 𝑘 to ∞ of 𝑎 𝑛. And secondly, if the integral from 𝑘 to ∞ of 𝑓 of 𝑥 with respect to 𝑥 is divergent, then so is the sum from 𝑛 equals one to ∞ of 𝑎 𝑛.

Now remember, if an integral or sum is convergent, then it is finite. And if it is divergent, then it is infinite. So essentially, what we’re doing with this test is using the convergence of an integral with an infinite bound to find the convergence of a series. Now, when we’re finding our function 𝑓 of 𝑥, it’s very important that these three conditions are satisfied. Now, the first two, continuous and positive, are usually quite easy to spot. However, sometimes we may need to do a bit of extra work to check that 𝑓 of 𝑥 is decreasing on the interval between 𝑘 and ∞. We can have a look at why this test works intuitively by considering the following graphs.

Let’s suppose we have the series which is the sum from 𝑛 equals 𝑘 to ∞ of 𝑎 𝑛 such that the function 𝑓 of 𝑛, which is equal to 𝑎 𝑛, is continuous, positive, and decreasing on the interval between 𝑘 and ∞. We can sketch a graph of 𝑓 of 𝑥, and it may look something like this. We can label 𝑘, 𝑘 plus one, 𝑘 plus two, and so on on our 𝑥-axis. Now, we have two possibilities for 𝑓 of 𝑥. Now, in one case, the integral from 𝑘 to ∞ of 𝑓 of 𝑥 with respect to 𝑥 is convergent. If this is the case, then we can draw bars on our graph to represent the terms in our series. Now, these bars that we’ve drawn here have a width of one and a height of 𝑓 of 𝑥. Therefore, the area of each of these bars is simply 𝑓 of 𝑥.

Now, for the first bar, 𝑓 of 𝑥 is equal to 𝑓 of 𝑘. And we have that 𝑓 of 𝑛 is equal to 𝑎 𝑛, and so the area of this bar is 𝑎 𝑘. Similarly, the area of the next bar is 𝑎 𝑘 plus one. And the area of the next bar is 𝑎 𝑘 plus two, and so on, and so on. Now, if we sum the area of each of these bars together all the way up to ∞, then what we’ll have is our series. But quite clearly, each of these bars is underneath our curve, 𝑓 of 𝑥. And the area under the curve is represented by this integral. So, we can see that our series must be smaller than our integral. Since our integral is convergent, this implies that our series must also be convergent.

Now, we consider the other case, which is, if our integral from 𝑘 to ∞ of 𝑓 of 𝑥 with respect to 𝑥 is divergent. If this is the case, then we can draw bars on our graph to represent our series as follows. Here, we’ve drawn the bars slightly differently. However, since the bars still have a width of one and a height of 𝑓 of 𝑥, the areas of the bars are still the terms in our series. And so, the sum of the area of these bars will be equal to our series.

Now, this time we have that the integral from 𝑘 to ∞ of 𝑓 of 𝑥 with respect to 𝑥 is divergent. And we can see that the area of the bars representing our series is at above the line 𝑓 of 𝑥. And so, they have a greater area than the integral from 𝑘 to ∞ of 𝑓 of 𝑥 with respect to 𝑥. Since this integral is divergent, this tells us that our series must also be divergent. Now that we’ve covered how the integral test for series works, let’s look at some examples.

Determine whether the series, which is the sum from 𝑛 equals zero to ∞ of 𝑒 to the negative 𝑛, converges or diverges.

We can try to find the convergence of this series using the integral test. The integral test tells us that if we have a function, 𝑓 of 𝑥, which is continuous, positive, and decreasing on the interval between 𝑘 and ∞ and that 𝑓 of 𝑛 is equal to 𝑎 𝑛. Then, if the integral from 𝑘 to ∞ of 𝑓 of 𝑥 with respect to 𝑥 is convergent, so is the sum from 𝑛 equals 𝑘 to ∞ of 𝑎 𝑛. And if the integral from 𝑘 to ∞ of 𝑓 of 𝑥 with respect to 𝑥 is divergent, so is the sum from 𝑛 equals 𝑘 to ∞ of 𝑎 𝑛. Now, the series we’re trying to find the convergence of is this sum from 𝑛 equals zero to ∞ of 𝑒 to the negative 𝑛. Therefore, 𝑎 𝑛 is equal to 𝑒 to the power of negative 𝑛. Since 𝑓 of 𝑛 is equal to 𝑎 𝑛, we obtain that 𝑓 of 𝑥 is equal to 𝑒 to the negative 𝑥. Since we’re taking the sum from 𝑛 is equal to zero, this means that 𝑘 is equal to zero.

And we now need to check whether 𝑓 is a continuous, positive, and decreasing function on the interval between zero and ∞. 𝑒 to the negative 𝑥 can also be written as one over 𝑒 to the 𝑥. Now, 𝑒 to the 𝑥 is positive for any value of 𝑥. Therefore, one over 𝑒 to the 𝑥 is also positive. So, we’ve satisfied the condition that 𝑓 of 𝑥 is positive on our interval. Now, one over 𝑒 to the 𝑥 is continuous since 𝑒 to the 𝑥 cannot be equal to zero at any value. Therefore, we’ve also satisfied this condition. And finally, since 𝑒 to the 𝑥 is an increasing function for all 𝑥, this means that one over 𝑒 to the 𝑥 is a decreasing function for all 𝑥. And so, we’ve satisfied the last condition on 𝑓 of 𝑥.

We’re therefore able to use the integral test. We need to find the convergence of the integral from zero to ∞ of 𝑒 to the negative 𝑥 with respect to 𝑥. We can use the fact that the integral from 𝑒 to the 𝑎𝑥 with respect to 𝑥 is equal to one over 𝑎 𝑒 to the 𝑎𝑥 plus 𝑐. In our case, 𝑎 is equal to negative one, and one over negative one is simply negative one. So, when we integrate 𝑒 to the negative 𝑥, we get negative 𝑒 to the negative 𝑥. And this is, of course, between our two bounds which is zero and ∞.

Since our upper bound is infinite, we need to find the limit as 𝑥 tends to ∞ of negative 𝑒 to the negative 𝑥. And we mustn’t forget to subtract negative 𝑒 to the negative zero. When trying to evaluate this limit, we can consider what happens as 𝑥 gets larger and larger and larger. As 𝑥 gets bigger, negative 𝑥 gets more and more negative. Therefore, 𝑒 to the power of negative 𝑥 gets closer and closer to zero.

And so, we can say that this limit is equal to zero. Then, we have 𝑒 to the power of negative zero, which is simply 𝑒 to the power of zero. Anything to the power of zero is one. And the two negative signs in front of the term cancel out with one another to give us a positive sign. From here, we can evaluate that the integral from zero to ∞ of 𝑒 to the negative 𝑥 with respect to 𝑥 is equal to one. Therefore, we found that our integral is convergent. And so, by the integral test, we can say that the sum from 𝑛 equals zero to ∞ of 𝑒 to the negative 𝑛 is convergent.

Let’s now move on to another example.

Use the integral test to determine whether the series, which is the sum from 𝑛 equals one to ∞ of one over 𝑛, converges or diverges.

Let’s start by recalling the integral test. The integral test tells us that if 𝑓 is a continuous, positive, decreasing function on the interval between 𝑘 and ∞ and that 𝑓 of 𝑛 is equal to 𝑎 𝑛. Then, if the integral from 𝑘 to ∞ of 𝑓 of 𝑥 with respect to 𝑥 is convergent, so is the sum from 𝑛 equals 𝑘 to ∞ of 𝑎 𝑛. And secondly, if the integral from 𝑘 to ∞ of 𝑓 of 𝑥 with respect to 𝑥 is divergent, so is the sum from 𝑛 equals 𝑘 to ∞ of 𝑎 𝑛. Now, in our case, our series is the sum from 𝑛 equals one to ∞ of one over 𝑛. Therefore, 𝑎 𝑛 is equal to one over 𝑛. Using the fact that 𝑓 of 𝑛 is equal to 𝑎 𝑛, we can say that 𝑓 of 𝑥 is equal to one over 𝑥.

Now, we need to check that 𝑓 of 𝑥 is a continuous, positive, and decreasing function on the interval between 𝑘 and ∞. Noting that since our sum goes from 𝑛 equals one, 𝑘 must be equal to one. Let’s start by checking the continuity of our function over this interval. The only discontinuity of 𝑓 of 𝑥 occurs when 𝑥 is equal to zero. However, zero is not in our interval between one and ∞. Therefore, this function must be continuous between one and ∞. On this interval, 𝑥 is always positive. Therefore, one over 𝑥 must also be positive. And so, we’ve satisfied that condition too. Now, if 𝑥 starts at one and gets larger and larger and larger, then one over 𝑥 will get smaller and smaller and smaller. Therefore, we can say that this is a decreasing function on our interval.

Now that we’ve satisfied these three conditions, we’re able to use the integral test. We need to work out whether the integral from one to ∞ of one over 𝑥 with respect to 𝑥 is convergent or divergent. Now, we know that the differential of the natural logarithm of 𝑥 with respect to 𝑥 is equal to one over 𝑥. Therefore, the natural logarithm of 𝑥 is the antiderivative of one over 𝑥. Hence, we have that our integral is equal to the natural logarithm of 𝑥 between one and ∞. When using the bound of ∞, we need to take the limit as 𝑥 tends to ∞ of the natural logarithm of 𝑥. Then, for our lower bound of one, we simply subtract the natural logarithm of one.

Now, let’s evaluate this limit, the natural logarithm as an increasing function. Therefore, as 𝑥 gets larger, the natural logarithm of 𝑥 also gets larger. Hence, we can say that this limit is also equal to ∞. We also have that the natural logarithm of one is equal to zero. However, regardless of what constant this term is, this will not affect our answer of ∞. This tells us that our integral must be divergent. Hence, by the integral test, we can say that the sum from 𝑛 equals one to ∞ of one over 𝑛 is divergent.

Now, when we’re using the integral test, it’s not actually required for 𝑓 of 𝑥 to be decreasing for all 𝑥 in our interval between 𝑘 and ∞. All we require is for 𝑥 to be decreasing for all 𝑥 greater than or equal to some constant 𝑐, where 𝑐 is greater than or equal to 𝑘. We’ll see how this can be done in the next example.

Determine whether the series, which is the sum from 𝑛 equals one to ∞ of the square root of the natural logarithm of 𝑛 over 𝑛, converges or diverges.

We can use the integral test to test this convergence. The integral test tells us that if we have a function 𝑓 of 𝑥 which is continuous, positive, and decreasing on the interval between 𝑘 and ∞ and that 𝑓 of 𝑛 is equal to 𝑎 𝑛. Then, firstly, if the integral from 𝑘 to ∞ of 𝑓 of 𝑥 with respect to 𝑥 is convergent, so is the sum from 𝑛 equals 𝑘 to ∞ of 𝑎 𝑛. And secondly, if the integral from 𝑛 equals 𝑘 to ∞ of 𝑓 of 𝑥 with respect to 𝑥 is divergent, so is the sum from 𝑛 equals 𝑘 to ∞ of 𝑎 𝑛. Now, our series is the sum from 𝑛 equals one to ∞ of the square root of the natural logarithm of 𝑛 over 𝑛. And so, we can say that 𝑎 𝑛 is equal to the square root of the natural logarithm of 𝑛 over 𝑛. Since 𝑓 of 𝑛 is equal to 𝑎 𝑛, 𝑓 of 𝑥 is equal to the square root of the natural logarithm of 𝑥 over 𝑥.

We also have that 𝑘 is equal to one. We now need to check whether 𝑓 of 𝑥 is continuous, positive, and decreasing on the interval between one and ∞. Firstly, for continuity, we need the inside of the square root, so that’s the natural logarithm of 𝑥, to be nonnegative. If the natural logarithm of 𝑥 is greater than or equal to zero, this means that 𝑥 must be greater than or equal to one. Since our value of 𝑘 is one, we are looking at the interval between one and ∞. Therefore, 𝑥 is always greater than or equal to one. And so, the numerator of our function is continuous. The only other discontinuity may occur when the denominator of our fraction is equal to zero. However, 𝑥 is always greater than or equal to one, and so 𝑥 cannot be equal to zero. Therefore, our function is continuous on the interval.

Next, we need to check whether our function is positive over this interval. The square root function in the numerator is always positive, since it’s a positive square root. And the denominator is always positive, since 𝑥 is greater than or equal to one. Therefore, we’ve satisfied this condition. Now, we need to check that our function is decreasing between one and ∞. Now, it’s quite difficult to check this without performing a calculation. So, we can find the values of 𝑥 for which 𝑓 of 𝑥 is decreasing by differentiating 𝑓 with respect to 𝑥. This will give us 𝑓 prime of 𝑥. Now, 𝑓 of 𝑥 is a quotient. Therefore, we can use the quotient rule. The quotient rule tells us that the differential of 𝑢 over 𝑣 is equal to 𝑣 multiplied by the differential of 𝑢 minus 𝑢 multiplied by the differential of 𝑣 all over 𝑣 squared.

Now, in our case, 𝑢 is equal to the square root of the natural logarithm of 𝑥 and 𝑣 is equal to 𝑥. Finding d𝑣 by d𝑥 is fairly straightforward. We differentiate 𝑥 and simply get one. Now, when we differentiate the square root of the natural logarithm of 𝑥, we need to use the chain rule. Now, we may find it easier to rewrite 𝑢. And we can rewrite it as the natural logarithm of 𝑥 to the power of one-half.

To find d𝑢 by d𝑥, we multiply by the power, decrease the power by one, and finally multiply by the differential of the inside of the function. So, that’s the differential of the natural logarithm of 𝑥. And this is simply equal to one over 𝑥. Now, d𝑢 by d𝑥 and d𝑣 by d𝑥 is simply 𝑢 prime and 𝑣 prime. And these can be substituted into the formula for the quotient rule. We obtain that 𝑓 prime of 𝑥 is equal to 𝑥 multiplied by one-half multiplied by one over 𝑥 times the natural logarithm of 𝑥 to the power of negative one-half minus the square root of the natural logarithm of 𝑥 all over 𝑥 squared.

Now, we’re looking for the values of 𝑥 such that 𝑓 is decreasing. So, that means that 𝑓 prime of 𝑥 is negative. Now, the denominator of 𝑓 prime of 𝑥, that’s 𝑥 squared, is always positive since the square is always positive. Therefore, the values of 𝑥 for which 𝑓 prime is negative occur when the numerator of the fraction is negative. So, this is when the square root of the natural logarithm of 𝑥 is greater than 𝑥 multiplied by one-half multiplied by one over 𝑥 multiplied by the natural logarithm of 𝑥 to the power of negative one-half. Here, we can cancel out the one over 𝑥 and the 𝑥.

We can also rewrite the natural logarithm of 𝑥 to the power of negative one-half as one over the square root of the natural logarithm of 𝑥. Therefore, the right-hand side of our inequality becomes one over two times the square root of the natural logarithm of 𝑥. We can multiply both sides by the square root of the natural logarithm of 𝑥. We obtain that 𝑓 of 𝑥 is decreasing for values of 𝑥, such that the natural logarithm of 𝑥 is greater than one-half. Using a calculator, we find that this is values of 𝑥 which are greater than 1.648 and so on. Therefore, we found that our function is not decreasing over the entirety of our interval.

However, we are able to rewrite our series. We can take the first term out of our series so that our series is equal to the square root of the natural logarithm of one over one plus the sum from 𝑛 equals two to ∞ of the square root of the natural logarithm of 𝑛 over 𝑛. Since the natural logarithm of one is equal to zero, this term on its own is simply zero. And so, this new sum that we found is in fact equal to the original sum. However, we are summing from 𝑛 is equal to two. Therefore, our value of 𝑘 changes to two. Now, we have that 𝑓 of 𝑥 is decreasing for 𝑥 is greater than 1.648. Therefore, we’ve satisfied the last condition to use the integral test and we’re able to use the integral test. So, let’s do that now.

We need to find out whether the integral from two to ∞ of the square root of the natural logarithm of 𝑥 over 𝑥 with respect to 𝑥 is convergent or divergent. In order to solve this integral, we can in fact use a 𝑢 substitution. We can let 𝑢 be equal to the square root of the natural logarithm of 𝑥. Rearranging this, we have that 𝑥 is equal to 𝑒 to the power of 𝑢 squared. We can differentiate 𝑥 with respect to 𝑢 to find that d𝑥 by d𝑢 is equal to two 𝑢𝑒 to the 𝑢 squared. An equivalent statement is d𝑥 is equal to two 𝑢𝑒 to the 𝑢 squared d𝑢. When 𝑥 is equal to two, 𝑢 is equal to the square root of the natural logarithm of two. And when 𝑥 is equal to ∞, 𝑢 is also equal to ∞. We’re now ready to perform our substitution.

We can start by substituting in the bounds of our integral, which will be the square root of the natural logarithm of two and ∞. Then, we have that the square root of the natural algorithm of 𝑥 is 𝑢. And 𝑥 is 𝑒 to the 𝑢 squared. Finally, we need to substitute in d𝑥. Now, we have completed our 𝑢 substitution. We’re able to cancel out the 𝑒 to the 𝑢 squared in the denominator with the other 𝑒 to the 𝑢 squared. And our integral becomes the integral from the square root of the natural logarithm of two to ∞ of two 𝑢 squared d𝑢.

We can integrate this by raising the power of 𝑢 by one and dividing by the new power. So, now we have two-thirds 𝑢 cubed between our two bounds. Substituting in our bounds, we obtain this. And even though the second term here is constant, the limit as 𝑢 tends to ∞ of two-thirds 𝑢 cubed is in fact infinite. Therefore, we have that this integral is divergent. So, by the integral test, our series from 𝑛 equals two to ∞ diverges. Since this series is equal to the one in the question, we have found our solution, which is that our series is divergent.

We’ve now covered a variety of examples. Let’s recap some key points of the video.

Key points

The integral test tells us that if 𝑓 is a continuous, positive, decreasing function on the interval between 𝑘 and ∞ and that 𝑓 of 𝑛 is equal to 𝑎 𝑛. Then, firstly, if the integral from 𝑘 to ∞ of 𝑓 of 𝑥 with respect to 𝑥 is convergent, so is the sum from 𝑛 equals 𝑘 to ∞ of 𝑎 𝑛. And secondly, if the integral from 𝑘 to ∞ of 𝑓 of 𝑥 with respect to 𝑥 is divergent, so is the sum from 𝑛 equals 𝑘 to ∞ of 𝑎 𝑛. Sometimes our function may not be decreasing on the interval between 𝑘 and ∞, but we’re still able to use the integral test as long as 𝑓 of 𝑥 is decreasing for all values of 𝑥 greater than or equal to 𝑐, where 𝑐 is a constant such that 𝑐 is greater than or equal to 𝑘.