Video: Harmonic and 𝑝-Series

In this video, we will learn how to find the condition for which the 𝑝-series converges, and we will prove the divergence of the harmonic series using the integral test.

12:05

Video Transcript

In this video, we’ll learn how to find the condition for which the 𝑝-series converges. And we’ll prove the divergence of the harmonic series using the integral test. We’ll see some examples of the application of the condition for the 𝑝-series to determine whether a series converges or diverges.

Let’s firstly remind ourselves of the difference between a convergent series and a divergent series. If we’re trying to determine whether a series is convergent, we’re trying to find out whether the sum of the terms gets closer and closer to a specific value as you increase the number of terms. Otherwise, we have a divergent series, which means the series does not have a finite limit. So this could be a series that tends to positive or negative ∞ when you add an infinite number of terms together.

There’s a really important series called the harmonic series, which is an infinite series of the form the sum from 𝑛 equals one to ∞ of one over 𝑛 which gives us one add one over two add one of the three add one over four, and so on. And this series actually got its name from its connection to music. A vibrating string produces wavelengths in proportion to one, one over two, one over three, one over four, and so on of the strings fundamental wavelength. But what we’re going to look at is whether the harmonic series is convergent or divergent. And we’re going to prove this using the integral test.

Recall, the integral test says that if we define a function 𝑓 from a series, which is a continuous positive and decreasing function on the interval between one and ∞, then if the improper integral of the function is convergent, then the series is convergent. If the improper integral is divergent, then the series is divergent. When the limits of this improper integral exist, the integral is convergent. And when the limit of the improper integral does not exist or is infinite, the integral is divergent. So let’s now apply the integral test to our series.

If we start by defining 𝑓 of π‘₯ to be one over π‘₯, then the integral we’re going to evaluate is the integral between one and ∞ of one over π‘₯ with respect to π‘₯. This is an improper integral because we have an infinite limit. And we deal with improper integrals by replacing the infinite limit with a variable and then taking the limits of the integral as that variable approaches ∞. And so we can rewrite this integral as the limit as 𝑑 approaches ∞ of the integral between one and 𝑑 of one over π‘₯ with respect to π‘₯. We know a general result that tells us that the integral of one over π‘₯ with respect to π‘₯ is the natural log of the absolute value of π‘₯ at a constant of integration.

And of course, when evaluating a definite integral from one to 𝑑, the constant will be subtracted from itself. So we get the limit as 𝑑 approaches ∞ of the natural log of the absolute value of π‘₯. This gives us the limit as 𝑑 approaches ∞ of the natural log of the absolute value of 𝑑 minus the natural log of the absolute value of one. But the natural log of one is zero. So we just need to evaluate the limit as 𝑑 approaches ∞ of the natural log of the absolute value of 𝑑. And we know this limit to be ∞. And so this improper integral is divergent. So by the integral test this series is divergent. So even though we’re going to be adding a smaller and smaller number each time, the partial sums are still growing without bound.

The harmonic series is actually a specific case of a more general series called a 𝑝-series. A 𝑝-series is a series of the form the sum from 𝑛 equals one to ∞ of one over 𝑛 to the 𝑝 power, such as a series the sum from 𝑛 equals one to ∞ of one over 𝑛 squared. And actually, the harmonic series is just a 𝑝-series with 𝑝 equal to one. We’re going to have a look at how we can tell if a 𝑝-series converges or diverges. We can do this using the test for divergence and the integral test. We’re going to break it down into different values for different cases of 𝑝. I’ll start by writing up the test for divergence.

Remember that we know when 𝑝 is equal to one, we have the harmonic series, which is divergent. And we’ll, firstly, consider the case when 𝑝 is equal to zero. Then the limit as 𝑛 approaches ∞ of one over 𝑛 to the 𝑝 power is just going to be one. As this is not equal to zero, by the test for divergence when 𝑝 is equal to zero, the series is divergent. What about the case when 𝑝 is strictly less than zero? Well, taking the limit as 𝑛 approaches ∞ of one over 𝑛 to the 𝑝 power is the same as taking the limit as 𝑛 approaches ∞ of 𝑛 to the negative 𝑝 power.

Remember here, that if 𝑝 is less than zero negative 𝑝 is greater than zero. And because 𝑛 represents the counting numbers for the series terms, 𝑛 is going to be an integer and so 𝑛 to the negative 𝑝 power will get larger and larger. So as 𝑛 approaches ∞, this limit approaches ∞. And so, by the test for divergence, when 𝑝 is less than zero, this series is divergent. Let’s now look at the case when 𝑝 is strictly greater than zero. To do this, we’ll apply the integral test. And here we’re assuming that 𝑝 is not equal to one because that would make it a harmonic series. And we already know that that diverges. So if we define a function one over π‘₯ to the 𝑝 power, then we evaluate the improper integral from one to ∞ of one over π‘₯ to the 𝑝 power with respect to π‘₯.

And let’s, firstly, rewrite this as a negative exponents π‘₯ to the negative 𝑝 power. And we take our usual approach to evaluating improper integrals by finding the limit as 𝑑 approaches ∞ of the integral between one and 𝑑 of π‘₯ to the negative 𝑝 power with respect to π‘₯. And using the general rule that the integral of π‘₯ to the power 𝑛 with respect to π‘₯ is π‘₯ to the power of 𝑛 add one over 𝑛 add one plus a constant of integration. So we find our integral to be the limit as 𝑑 approaches ∞ of π‘₯ to the negative 𝑝 add one power over negative 𝑝 add one. And remember, this is okay because we’re assuming that 𝑝 is not equal to one. And then we substitute in our limits for integration and then take out one over one minus 𝑝 as a factor.

From here, we know that one raised to the power of negative 𝑝 add one is going to give us one. And also we can write 𝑑 to the power of negative 𝑝 add one and its fractional form as one over 𝑑 to the power of 𝑝 minus one. And now there’s two cases we need to consider here, if 𝑝 is greater than one and if 𝑝 is less than one. Well, if 𝑝 is greater than one, then 𝑝 minus one is going to be positive. So as 𝑑 approaches ∞, 𝑑 to the 𝑝 minus one power also approaches ∞. So one over 𝑑 to the 𝑝 minus one power approaches zero. So if this term here approaches zero, then what we have left is one over one minus 𝑝 multiplied by negative one, which is negative one over one minus 𝑝 which is the same as one over 𝑝 minus one. And because this limit exists, then the improper integral is convergent. And so, by the integral test, the series is convergent for 𝑝 greater than one.

And now, we look at the case when 𝑝 is less than one. If 𝑝 is less than one, then 𝑝 minus one is less than zero, so one over 𝑑 to the 𝑝 minus one power, which is the same as 𝑑 to the negative 𝑝 minus one power in a negative exponents form. Note that because 𝑝 minus one is negative, then negative 𝑝 minus one is positive. And so 𝑑 to the negative 𝑝 minus one must approach ∞ as 𝑑 approaches ∞. Because this is an infinite limit, the improper integral, the integral from one to ∞ of π‘₯ to the negative 𝑝 power with respect to π‘₯, must diverge when 𝑝 is less than one. And so by the integral test, this series is divergent for 𝑝 less than one.

So we found this series converges only when 𝑝 is greater than one. So I’m going to clear some space to write our conclusion. So we conclude that the 𝑝-series the sum from 𝑛 equals one to ∞ of one over 𝑛 to the 𝑝 power, is convergent if 𝑝 is greater than one and divergent if 𝑝 is less than or equal to one. Note that this is less than or equal because, remember, we already found the harmonic series when 𝑝 is equal to one to be divergent. Let’s now have a look at some examples of how we can apply this condition to some 𝑝-series.

Determine whether the series the sum from 𝑛 equals one to ∞ of one over 𝑛 to the fourth power converges or diverges.

Well, we recognise this sum to be the form of a 𝑝-series. So we can start by writing out the condition for convergence of a 𝑝-series. The 𝑝-series the sum from 𝑛 equals one to ∞ of one over 𝑛 to the 𝑝 power is convergent if 𝑝 is greater than one and divergent if 𝑝 is less than or equal to one. So for this series, 𝑝 is equal to four, which is greater than one. So we can conclude that this series converges.

Determine whether the series the sum from 𝑛 equals one to ∞ of one over 𝑛 to the power of two-fifths converges or diverges.

We start by recognising that this sum is in the form of a 𝑝-series, which is a series of the form the sum from 𝑛 equals one to ∞ of one over 𝑛 to the 𝑝 power. So let’s write out the condition for convergence for a 𝑝-series. That is the 𝑝-series the sum from 𝑛 equals one to ∞ of one over 𝑛 to the 𝑝 power is convergent if 𝑝 is greater than one and divergent if 𝑝 is less than or equal to one. And so for this series, 𝑝 equals two-fifths which is less than one. So we can say that this series diverges.

Determine whether the series the sum from 𝑛 equals one to ∞ of one over the square root of 𝑛 cubed converges or diverges.

If we start by rewriting this sum, using the fact that we can write the square root of π‘Ž as π‘Ž to the half power, we can say that this sum is equivalent to the sum from 𝑛 equals one to ∞ of one over 𝑛 cubed raised to the half power. We can then use the fact that π‘Ž to the π‘₯ power, then raised to the 𝑦 power, equals π‘Ž to the π‘₯ multiplied by 𝑦 power. So we can write our sum as the sum from 𝑛 equals one to ∞ of one over 𝑛 to the three over two power. And then we can recognise this to be a 𝑝-series, which is a series of the form the sum from 𝑛 equals one to ∞ of one over 𝑛 to the 𝑝 power.

So let’s write out the conditions for convergence for a 𝑝-series. The 𝑝-series the sum from 𝑛 equals one to ∞ of one over 𝑛 to the 𝑝 power is convergent if 𝑝 is greater than one and divergent if 𝑝 is less than or equal to one. So for this series 𝑝 is equal to three over two; this is the same as 1.5, which is greater than one. So this series converges.

So let’s summarise the main points. a 𝑝-series is a series of the form. the sum from 𝑛 equals one to ∞ of one over 𝑛 to the 𝑝 power. And this is convergent if 𝑝 is greater than one and divergent if 𝑝 is less than or equal to one. For 𝑝 equals one, this series is called the harmonic series. And this series is divergent.

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