Welcome back! In Lesson 13 (link here) we learned what it means to have an infinite list of numbers. Today we’ll start to talk about what it means to have an infinite sum of numbers. This will be the focus of the next few lessons.

Lesson 14: Infinite Series

Learning goals:

  • Explain the meaning of the sum of an infinite series.
  • Calculate the sum of a geometric series.
  • Evaluate the sum of a telescoping series.

Topic:

WeBWorK:

  • Series – Infinite Series

Motivating question

Here is an example of the question we’re trying to make precise so we can answer it: What is the value of the following infinite sum?

    \[\frac{1}{2} + \frac{1}{4}+ \frac{1}{8}+\frac{1}{16}+\frac{1}{32}+ \cdots\]

Warmup exercise 1

Determine whether the following sequence converges or diverges. If it converges, determine the value it converges to.

    \[\frac{1}{2}, \frac{3}{4},\frac{7}{8},\frac{15}{16},\frac{31}{32},\dots\]

Show answer 1

There are a few ways to approach this, but probably the easiest way would be to determine the general term. For reasons that won’t be clear until later, we’ll call the general term S_n (instead of the usual a_n). Then

  1. S_1 = \frac{1}{2}
  2. S_2 = \frac{3}{4}
  3. S_3 = \frac{7}{8}
  4. S_4 = \frac{15}{16}
  5. S_5 = \frac{31}{32}
  6. …and so on…

We need to figure out the pattern to determine what the general term S_n is.

Let’s look at the sequence of denominators of all these terms first: 2, 4, 8, 16, 32, \dots are all powers of 2. In fact, the powers match the index: 2^1, 2^2, 2^3, 2^4, \dots so this means the denominator of S_n is 2^n.

Now let’s look at the sequence of numerators: 1, 3, 4, 7, 15, 31, \dots. The pattern might not be as easy to detect in this case except we already know what pattern describes the denominators. The numerators are always one less than the denominators, so the numerator of S_n is 2^n-1.

This means that the general term is S_n = \frac{2^n-1}{2^n}. We still have to determine if this sequence converges. That is, we have to determine whether \lim_{n \to \infty} S_n exists.

To evaluate the limit, it might help to separate terms:

    \[\lim_{n \to \infty} S_n = \lim_{n \to \infty} \frac{2^n-1}{2^n}\]

    \[ = \lim_{n \to \infty} \left(\frac{2^n}{2^n} -\frac{1}{2^n} \right)\]

    \[ = \lim_{n \to \infty} \left( 1 - \frac{1}{2^n}\right) \]

    \[= 1-0\]

    \[=1\]

Introduction to series

If a sequence is a list of numbers: a_1, a_2, a_3, \dots then a series  is just the sum of the terms in the series: a_1+a_2+a_3+ \dots. Sequences and series don’t have to have infinitely many terms, but the ones we’re interested in will. (Language note: the plural of “series” is “series.”)

We’ll use sigma notation often when discussing series:

    \[ \sum_{n=1}^\infty a_n = a_1 + a_2 + a_3 + \cdots\]

If you need it, Video 1 below provides a refresher on Sigma notation.

Video 1

A good motivating example of a series to have in mind is \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+ \dots from today’s motivating question. This series is often referred to in relation to Xeno’s paradox. If you’re interested, you can learn more about Xeno’s paradox in the video linked here.

Here is a Desmos graph (link here) which shows that if we think of a sequence as a discrete version of a function, then the corresponding series is like a discrete version of a definite integral (area under a curve). Infinite series might be non-intuitive in the same way that improper integrals might be non-intuitive: something that seems big or unbounded in one sense is actually small or finite in another, hence Xeno’s paradox.

Partial sums and convergence of series

You might think that the infinite sum in our motivating example \frac{1}{2} + \frac{1}{4}+ \frac{1}{8}+\frac{1}{16}+\frac{1}{32}+ \cdots should be infinite. After all, we are adding up infinitely many positive numbers! However, there is a notion of convergence for infinite series, just like there is for improper integrals.

Video 2 below provides an introduction important definitions for discussing series: the sequence of partial sums of a series and what it means for a series to converge.

I should point out in the videos below you’ll see graphs of partial sums; those graphs and the Desmos graph linked above are related but not exactly the same.

Video 2

This video also describes the divergence test for series and shows some motivating examples of some series which converge or diverge. Officially, the divergence test isn’t covered until Lesson 15, but understanding it now will help us understand convergence better.

Back to the motivating example

From Video 1, we learned that a series \{a_n\} converges if its corresponding sequence of partial sums \{S_n\} converges, so we can use this to analyze our motivating example

    \[\frac{1}{2} + \frac{1}{4}+ \frac{1}{8}+\frac{1}{16}+\frac{1}{32}+ \cdots\]

. Let’s compute the first few partial sums for this series:

  1. S_1 = a_1 = \frac{1}{2}
  2. S_2 = a_1+a_2 = \frac{1}{2}+\frac{1}{4} = \frac{3}{4}
  3. S_3 = a_1+a_2 +a_3= \frac{1}{2}+\frac{1}{4} + \frac{1}{8} = \frac{7}{8}
  4. S_4 = a_1+a_2 +a_3+a_4= \frac{1}{2}+\frac{1}{4} + \frac{1}{8} + \frac{1}{16} = \frac{1}{16}.
  5. …and so on…

So the sequence of partial sums is precisely the sequence \{S_N\} from the Warmup exercise 1! Since we showed in Warmup exercise 1 that the sequence \{S_n\} converges to 1, this means that the infinite sum \frac{1}{2} + \frac{1}{4}+ \frac{1}{8}+\frac{1}{16}+\frac{1}{32}+ \cdots converges to 1. In sigma notation:

    \[ \sum_{n=1}^\infty \left(\frac{1}{2^n}\right)^n = 1\]

Geometric Series

The series from the motivating example is an example of a geometric series (which you might remember from your precalculus class). Geometric series are your friends! They’re among the nicest types of series you might hope to meet. One reason that they’re so nice is that if a series is geometric, it’s easy to tell if it converges or diverges. If a geometric series converges, it’s also easy to see what it converges to.

Over the next few lessons, we’ll be assembling a new toolbox full of tests you can try to apply to see if a series converges. Some of the tests in the toolbox might look complicated at first, but we’ll point out when a particular test is saying, “Well…this series isn’t actually geometric, but it’s practically eventually almost geometric, and that’s good enough.” We’ll come back to this property in Lesson 18 (link here).

Video 3 below tells you everything you need to know about infinite geometric series. Notice that the motivating example \sum_{n=1}^\infty \left(\frac{1}{2}\right)^n is a geometric series. We already know by looking at the sequence of partial sums that this series converges to 1, but we could also use the tools in Video 2 to determine this.

Video 3

Theorem: Let \sum_{n=1}^\infty ar^{n-1} be a geometric series.

  • If |r| < 1 then the series converges to \frac{a}{1-r}.
  • If |r| \geq 1 then the series diverges.

Telescoping Series

Telescoping series are another nice type of series. Think of an old-fashioned telescope like this one. If you are a person who needs to carry around a telescope, it’s not practical to carry a big long tube around with you, so you might want one that collapses into a smaller tube (really a bunch of tubes nested inside one another). Telescoping series work much the same way. When you write out the individual terms of the summation, it looks like some big long thing, but if you arrange the terms differently, you’ll see a lot of cancellation, which is like the collapsing of a telescope.

Video 4 below takes us through telescoping series.

Video 4

Sum of the first n integers

One formula that you might need to know is the sum of the first n positive integers:

\sum_{i=1}^n i= 1 + 2 + \dots + (n-1) + n = \frac{n(n+1)}{2}.

The story goes that 18th century German mathematician Gauss was 10 years old when he discovered this formula. He was a kind of annoying kid and his teacher tried to give him a problem to keep him quiet for a while and asked him to determine the sum of the first 100 positive integers. But Gauss was too clever! If you’re interested in how he solved this problem very quickly, check out the video linked here (there’s a small error between 0:40 and 0:45 in the video; can you spot it???). This sum is an example of a finite arithmetic series, so you may have seen this formula when you were studying them in your precalculus class.

Be careful!

We’ll see the divergence test in more detail in Lesson 15, but for now there is one easy mistake to make that you should avoid making.

In all of the examples we’ve seen of convergent series, the individual terms form a sequence that converges to zero. The divergence test tells us that this must happen if our series has any hope of converging. But just because the sequence of individual terms of a series converges to zero, this does not guarantee that the series converges. Indeed, over the next few lessons we’ll see several divergent series whose terms converge to zero. Keep this in the back of your mind at all times: if the individual terms approach zero, the series itself may converge or diverge. The point of the next several lessons is to fill up our toolbox with convergence tests so we can tell whether such series converge or diverge.

Exit ticket

Determine whether the following series converges or diverges. If it converges, determine the value it converges to.

    \[\sum_{n=1}^\infty \frac{5}{10^n}\]