Video: Partial Sums

In this video, we will learn how to find the 𝑛th partial sum of a series, and determine the convergence or divergence of the series from the limit of its partial sum.

15:18

Video Transcript

In this video, we’ll learn how to find the 𝑛th partial sum of a series and determine the convergence or divergence of the series from the limit of its partial sum. We will primarily consider something called telescoping series, but we will also look at geometric series and how we can establish convergence or divergence of these series.

A series is not a list of terms in a sequence, but the sum of the terms in that sequence. If a series has a finite number of terms, then it’s quite straightforward to add each term and evaluate the series. But when there are an infinite number of terms, it’s a little less straightforward. In fact, the series may not even have a finite sum. For example, the series one plus two plus three plus four plus all the way up to 𝑛 and so on won’t have a finite sum. Whereas the series given by a half plus a quarter plus an eighth plus a sixteenth and so on does because the 𝑛th term of this series gets ever smaller and smaller, eventually approaching zero as 𝑛 approaches ∞.

We define an infinite series as the sum of the terms π‘Ž one, π‘Ž two, π‘Ž three all the way up to π‘Ž 𝑛, and so on. And we denote it in short by the sum from 𝑛 equals one to ∞ of π‘Ž sub 𝑛. The thing is it doesn’t always make sense to think about the sum of infinitely many terms. And so, we define something called a partial sum. A partial sum of an infinite series is the sum of a finite number of consecutive terms beginning with the first. It could be helpful to examine the behaviour of partial sums of infinite series.

Formally, we say that given a series the sum from 𝑛 equals one to ∞ of π‘Ž sub 𝑛, 𝑆 sub 𝑛 is the 𝑛th partial sum. And that’s given by the sum from 𝑖 equals one to 𝑛 of π‘Ž sub 𝑖. This also leads us onto the definition of convergence of these series. We say that if the limit as 𝑛 approaches ∞ of the 𝑛th partial sum is some real number 𝑆, then the series the sum of π‘Ž 𝑛 is called convergent. This real number 𝑆 is called the sum of the series. Of course, if this is not true, we say that the series is called divergent. Now, often, partials sums can be computed with the sum function. Other times, we’ll need to spot patterns or apply formulae to look for evidence of convergence of these series. Let’s Look at an example.

Find the partial sum from the series the sum from 𝑛 equals one to ∞ of 𝑒 to the power of one over 𝑛 minus 𝑒 to the power of one over 𝑛 plus one. Is the series convergent or divergent?

Remember, the partial sum of our series is the sum of the first 𝑛 terms. In general, the 𝑛th partial sum of the series, the sum from 𝑛 equals one to ∞ of π‘Ž sub 𝑛, is the sum from 𝑖 equals one to 𝑛 of π‘Ž sub 𝑖. And that’s π‘Ž sub one plus π‘Ž sub two plus π‘Ž sub three all the way up to π‘Ž sub 𝑛. We generally define this as 𝑆 sub 𝑛. So, let’s find the 𝑛th partial sum for our series.

We know it’s going to be π‘Ž sub one plus π‘Ž sub two plus π‘Ž sub three all the way up to π‘Ž sub 𝑛. But here, π‘Ž sub 𝑛 is 𝑒 to the power of one over 𝑛 minus 𝑒 to the power of one over 𝑛 plus one. π‘Ž sub one is found by replacing 𝑛 with the number one. So, we get 𝑒 to the power of one over one minus 𝑒 to the power of one plus one, which is simply 𝑒 minus 𝑒 to the power of one-half. Then, π‘Ž sub two is 𝑒 to the power of one-half minus 𝑒 to the power of one over two plus one, or 𝑒 to the power of one-third.

In the same way, we can define the remaining terms as shown. This means 𝑆 sub 𝑛, the 𝑛th partial sum, is given by 𝑒 minus 𝑒 to the power of one-half. That’s π‘Ž sub one. Plus 𝑒 to the power of one-half minus 𝑒 to the power of one-third. Remember, that’s π‘Ž sub two. Plus 𝑒 to the power of one-third minus 𝑒 to the power of a quarter. That was π‘Ž sub three. And we continue adding these terms until we get to π‘Ž sub 𝑛, which is 𝑒 to the power of one over 𝑛 minus 𝑒 to the power of one over 𝑛 plus one.

But if we look really carefully at our 𝑛th partial sum, we should see that some of the terms will cancel. We have negative 𝑒 to the power of one-half plus 𝑒 to the power of one-half. Well, that’s zero. We then have negative 𝑒 to the power of one-third plus 𝑒 to the power of one-third, which is also zero. But let’s add in π‘Ž sub 𝑛 minus one. And when we do, we see that this process repeats all the way up to negative 𝑒 to the power of one over 𝑛 plus 𝑒 to the power of one over 𝑛. And so, what this means is all we’re left with to describe our 𝑛th partial sum is 𝑒 minus 𝑒 to the power of one over 𝑛 plus one. And so, the partial sum for our series is 𝑒 minus 𝑒 to the power of one over 𝑛 plus one.

The next part of this question asks us whether the series is convergent or divergent. Well, remember, we say that if the limit as 𝑛 approaches ∞ of the 𝑛th partial sum is equal to some real number 𝑆, then that means the series, the sum of π‘Ž 𝑛, is convergent. So, let’s evaluate the limit as 𝑛 approaches ∞ of 𝑆 sub 𝑛.

In that case, that’s the limit as 𝑛 approaches ∞ of 𝑒 minus 𝑒 to the power of one over 𝑛 plus one. Now, actually, 𝑒 itself is completely independent of 𝑛. And we should notice that as 𝑛 grows larger, one over 𝑛 plus one grows smaller. As 𝑛 approaches ∞, one over 𝑛 plus one approaches zero. So, the limit as 𝑛 approaches ∞ of the 𝑛th partial sum is 𝑒 minus 𝑒 to the power of zero. But of course, 𝑒 to the power of zero β€” in fact, anything to the power of zero β€” is equal to one. So, our limit is equal to 𝑒 minus one. This is indeed a real number. And so, we can say that our series is convergent.

Now, notice how the partial sum of our series eventually had just a few terms after cancelling. This is called the method of differences. And the series is called a telescoping series. This is a series whose partial sums eventually only have a fixed number of terms after cancelling. Let’s have another look at a series of this form.

Find the partial sum of the series the sum from 𝑛 equals one to ∞ of one over two 𝑛 plus one times two 𝑛 minus one. Is the series convergent or divergent?

It might not instantly be obvious how we’re going to find the 𝑛th partial sum of this series, but π‘Ž sub 𝑛 here is a fraction. We’re going to manipulate the expression for the 𝑛th term of our sequence by writing it in partial fraction form. Remember, this is a way of simplifying the fraction to the sum of two or more less complicated rational functions.

In this case, we write one over two 𝑛 plus one times two 𝑛 minus one as 𝐴 over two 𝑛 plus one plus 𝐡 over two 𝑛 minus one. We want to make the expression on the right-hand side look like that on the left. And so, we create a common denominator. And we achieve this by multiplying the numerator and denominator of our first fraction by two 𝑛 minus one and of our second fraction by two 𝑛 plus one. Once we’ve done this, we can simply add the numerators. And so, on the right-hand side, we get 𝐴 times two 𝑛 minus one plus 𝐡 times two 𝑛 plus one all over two 𝑛 plus one times two 𝑛 minus one.

Notice now that the denominators on the right- and left-hand side of our equation are equal. This, in turn, means that for these two fractions to be equal, their numerators must themselves be equal. That is to say, one equals 𝐴 times two 𝑛 minus one plus 𝐡 times two 𝑛 plus one. And then, we need to work out the values of 𝐴 and 𝐡. And we have a couple of ways that we can do this. We could distribute the parentheses and equate coefficients on the left- and right-hand side of our equation. Alternatively, we can substitute the zeros of two 𝑛 plus one times two 𝑛 minus one into the entire equation. That is, let 𝑛 equal one-half, see what happens, and let 𝑛 equal negative one-half and see what happens.

By letting 𝑛 be equal to one-half, we get one equals 𝐴 times one minus one plus 𝐡 times one plus one. But of course, 𝐴 times one minus one is 𝐴 times zero, which is zero. And that’s the whole purpose of substituting these zeros in; it leaves us with an equation purely in terms of one of our unknowns. In this case, this simplifies to one is equal to two 𝐡. And if we divide through by two, we find 𝐡 is equal to one-half.

Let’s repeat this process for 𝑛 is equal to negative one-half. Our equation becomes one equals 𝐴 times negative one minus one plus 𝐡 times negative one plus one. 𝐡 times negative one plus one is 𝐡 times zero, which is zero. And so, this time, we have an equation purely in terms of 𝐴. Its simplifies to one equals negative two 𝐴. And if we divide through by negative two, we find 𝐴 is equal to negative one-half. And so, we replace 𝐴 with negative one-half and 𝐡 with one-half. And we see that our 𝑛th term, one over two 𝑛 plus one times two 𝑛 minus one, can be written as negative one-half over two 𝑛 plus one plus a half over two 𝑛 minus one.

Let’s clear some space and simplify this a little. We can write a half over two 𝑛 minus one as one over two times two 𝑛 minus one. Similarly, we can write negative a half over two 𝑛 plus one as negative one over two times two 𝑛 plus one. To find the 𝑛th partial sum, we’re going to list out the first few terms of our series. π‘Ž sub one is found by replacing 𝑛 with one. This is the first term in our series. We get one over two times two minus one minus one over two times two plus one. This simplifies to a half minus one-sixth.

Then, π‘Ž sub two is found by replacing 𝑛 with two. And we get one over two times four minus one minus one over two times four plus one. That’s a sixth minus a tenth. In the same way, we find π‘Ž sub three to be equal to a tenth minus fourteenth, π‘Ž sub four to be equal to a fourteenth minus an eighteenth, and so on. And so, we find the 𝑛 partial sum to be the sum of all of these all the way up to π‘Ž sub 𝑛. So, that’s a half minus a sixth plus a sixth minus a tenth plus a tenth minus fourteenth plus a fourteenth minus an eighteenth all the way up to one over two times two 𝑛 minus one minus one over two times two 𝑛 plus one.

But now look what happens. Negative a sixth plus one-sixth is zero. Negative a tenth plus one-tenth is zero. Negative one fourteenth plus one fourteenth is zero. And this continues all the way up to one over two times two 𝑛 minus one. And so, we’re actually only left with two terms in our 𝑛th partial sum. They are a half and negative one over two times two 𝑛 plus one. We are almost done. But we’re going to add these fractions once again by creating a common denominator.

This time, we achieve this simply by multiplying the numerator and denominator of our first fraction by two 𝑛 plus one. Which gives us two 𝑛 plus one minus one over two times two 𝑛 plus one. And one minus one is zero. Then, we see that these twos cancel. And we have our expression for the 𝑛th partial sum of our series. It’s 𝑛 over two 𝑛 plus one.

The second part of this question asks us to establish whether the series is convergent or divergent. And so, we recall that if the limit as 𝑛 approaches ∞ of 𝑆 sub 𝑛 exists as some real number 𝑆, then the sum of π‘Ž 𝑛, this series, is convergent. Well, in our case, 𝑆 sub 𝑛 is 𝑛 over two 𝑛 plus one. So, we need to find the limit as 𝑛 approaches ∞ of 𝑛 over two 𝑛 plus one. We can’t use direct substitution because if we were substitute in 𝑛 equals ∞, we get ∞ over ∞, which we know to be undefined.

What we can do, though, is manipulate the fraction a little. We divide both the numerator and denominator by the highest power of 𝑛 in our denominator. In this case, we divide by 𝑛. So, on the numerator, we get 𝑛 over 𝑛, which is equal to one. And on the denominator, the first part is two 𝑛 over 𝑛, which is two. And then, we add one over 𝑛.

And now, we only have one term involving an 𝑛. It’s this, one over 𝑛. Now, as 𝑛 grows larger, one over 𝑛 grows smaller, eventually approaching zero. And so, we find the limit as 𝑛 approaches ∞ of 𝑆 sub 𝑛 of our 𝑛th partial sum simplifies to one over two plus zero, which is just one-half. Since this exists as a real number, we can say that our series is convergent.

So far, we’ve looked at telescoping series, but there are a handful of series whose partial sums can be calculated using formulae. One of the key ones of these is the geometric series. Now, it’s outside of the scope of the video to prove these results, but we can quote them.

In a geometric series, each turn is obtained from the preceding one by multiplying it by the common ratio π‘Ÿ. We say that the 𝑛th partial sum of a geometric series is given by 𝑆 sub 𝑛 equals π‘Ž times one minus π‘Ÿ to the 𝑛th power over one minus π‘Ÿ. Importantly, we say that the series is convergent if the absolute value of π‘Ÿ is less than one and its sum is then given by π‘Ž over one minus π‘Ÿ. We also say that if the absolute value of the common ratio π‘Ÿ is greater than or equal to one, the series is divergent. We’ll now have a look at an example of this form.

Find the partial sum for the series the sum from 𝑛 equals one to ∞ of two times one-half to the power of 𝑛 minus one. Is the series convergent or divergent?

We could begin by listing out the first few terms of this series. The 𝑛th partial sum is the sum of the first 𝑛 terms. So, it’s π‘Ž one plus π‘Ž two plus π‘Ž three plus π‘Ž four all the way up to π‘Ž sub 𝑛, where here π‘Ž sub 𝑛 is two times a half to the power of 𝑛 minus one. And this means that π‘Ž sub one is found by replacing 𝑛 with one. We get two times a half to the power of one minus one, which is two times a half to the power of zero, or just two.

The second term, π‘Ž sub two, is two times a half to the power of two minus one, which is two times one-half to the power of one, or just two times a half. The third term is two times a half to the power of three minus one, which is two times one-half squared. And we continue this process all the way up to the term two times a half to the power of 𝑛 minus one.

Now, at first glance, it might look like this is a really complicated partial sum. But, in fact, we have a special type of series. It’s a geometric series. Each term is obtained from the preceding one by multiplying it by the common ratio π‘Ÿ. Now, there is a formula we can quote to find the partial sum of a geometric series. It’s given by π‘Ž times one minus π‘Ÿ to the 𝑛th power over one minus π‘Ÿ. So, if we can define π‘Ž and π‘Ÿ from our series, we’ll be able to quite easily find the 𝑛th partial sum.

Well, we know that the first term in our series is two times a half to the power of zero. Well, a half to the power of zero is one. So, that’s two times one. And we can, therefore, say that π‘Ž is equal to two. And then, we can see that π‘Ÿ is equal to one-half. Each time, our term is being multiplied by one-half. And so, substituting these values of π‘Ž and π‘Ÿ into the formula for the 𝑛th partial sum, and we get two times one minus a half to the 𝑛th power over one minus a half.

Now, in fact, the denominator of this fraction, one minus a half, becomes one-half. And, of course, dividing by one-half is actually the same as multiplying by two. So, we can multiply the numerator of our expression by two. And we’re left with the 𝑛th partial sum. It’s four times one minus a half to the 𝑛th power.

The second part of this question asks us to decide whether this series is convergent or divergent. Well, in fact, we could find the limit as 𝑛 approaches ∞ of 𝑆 sub 𝑛. Alternatively, we can quote a general result. We say that a geometric series with a common ratio of π‘Ÿ is convergent if the absolute value of π‘Ÿ is less than one. And it’s divergent if the absolute value of π‘Ÿ is greater than or equal to one.

Well, here, π‘Ÿ is equal to one-half. And the absolute value of one-half is one-half, which is less than one. And so, in this case, we can say that this series is convergent. In fact, we’re also able to generalise this and say that if the series is convergent, it’s sum is given by π‘Ž over one minus π‘Ÿ. In this case, that’s two over one minus one-half, which is equal to four.

In this video, we’ve learned that a partial sum of an infinite series is a sum of a finite number of consecutive terms beginning with the first. Given a series the sum from 𝑛 equals one to ∞ of π‘Ž sub 𝑛, its 𝑛th partial sum is defined at 𝑆 sub 𝑛. Which is π‘Ž one plus π‘Ž two plus π‘Ž three all the way up to π‘Ž sub 𝑛. We say that if the limit as 𝑛 approaches ∞ of 𝑆 sub 𝑛 is equal to some real member 𝑆, then the series is called convergent. This number 𝑆 is called the sum of the series. And if this is not true, the series is called divergent.

We looked primarily at telescoping series. And those are a series whose partial sums eventually only have a fixed number of terms after cancelling. And this cancellation process is often known as the method of differences. Finally, we looked at geometric series. And in a geometric series, each term is obtained from the preceding one by multiplying it by the common ratio π‘Ÿ.

The partial sum of a geometric series is given by π‘Ž times one minus π‘Ÿ to the 𝑛th power over one minus π‘Ÿ. We also saw that our geometric series are convergent if the absolute value of π‘Ÿ is less than one. And if the absolute value of π‘Ÿ is greater than or equal to one, the geometric series is divergent.

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