Alternating series test - Sequence and Series

Alternating Series Test

The alternating series test (also known as the Leibniz test), is type of series test used to determine the convergence of series that alternate. Keep in mind that the test does not tell whether the series diverges. In order to use this test, we first need to know what a converging series and a diverging series is.

What is a convergent series? What is a divergent series?

A convergent series is an infinite series which sums up to a finite number. For example, the famous convergent series:

Formula 1: Convergent Series
Formula 1: Convergent Series

This convergent series sum up to π26\frac{\pi^{2}}{6}! How do you get that? It is a long process that requires a lot of calculations, but usually it is sufficient enough to know that the series is convergent by the p-series test.

A divergent series is an infinite series where the sum is infinity. For example, the series

Formula 2: Divergent Series
Formula 2: Divergent Series

Adding up all the numbers will give you a sum of infinity. If that doesn’t convince you, take a look at this. Note that the series could be written as an NthN^{th} partial sum.

Formula 3: Power sum n^2
Formula 3: Power sum n^2

Now if we were to make it an infinite series, then we are going to take the limit as N goes to infinity of both sides. In other words,

Equation 1: Power sum n^2 Divergent
Equation 1: Power sum n^2 Divergent

Now that we know what a divergent and convergent series is, let’s take a look at the alternating series test.

Alternating Series Test

In order to use the alternating series test, the series must be alternating. In other words, the series are in the form:

Formula 4: Alternating Series with power n
Formula 4: Alternating Series with power n

or

Formula 5: Alternating Series with power n+1
Formula 5: Alternating Series with power n+1

where bn0b_{n} \leq 0. An alternating series is not limited to these two forms because the exponent on the (-1) can vary. Now the alternating series test states that if the two following conditions are met, then the alternating series is convergent:

Definition 6: Alternating Series Test
Definition 6: Alternating Series Test

For the second condition, bnb_{n} does not have to be strictly decreasing for all n1n \leq 1. As long as the sequence is decreasing at nn \infty, then that will be sufficient enough to show that it is decreasing. Now that we know what the alternating series test is, let us put it to use for the following examples.

Example 1: Show that the series

Equation 2: Alternating Series test pt.1
Equation 2: Alternating Series test pt.1

is convergent.

Before we want to use the alternating series test, we want to make sure that the series is actually alternating. In other words, turn this series into the form:

Equation 2: Alternating Series test pt.2
Equation 2: Alternating Series test pt.2

As you can see, we can turn our series into that form. Notice that:

Equation 2: Alternating Series test pt.3
Equation 2: Alternating Series test pt.3

And so we know that:

Equation 2: Alternating Series test pt.4
Equation 2: Alternating Series test pt.4

Hence, we can go ahead and use the alternating series test. Now remember the two conditions. First we have to show that

Equation 2: Alternating Series test pt.5
Equation 2: Alternating Series test pt.5

Since our bn=1n2b_{n} = \frac{1}{n^{2}}, then our limit is

Equation 2: Alternating Series test pt.6
Equation 2: Alternating Series test pt.6

Notice that this limit does to 0, hence

Equation 2: Alternating Series test pt.7
Equation 2: Alternating Series test pt.7

and so our first condition is fulfilled. Now let’s take a look at the second condition. We have to make sure that 1n2\frac{1}{n^{2}} is decreasing. How do we do this? There are a total of three ways to do this:

  • Method 1:

    Write out the first few terms We can write out the first few terms of bnb_{n} and then conclude if the sequence is decreasing. Notice that:
    Equation 2: Alternating Series test pt.8
    Equation 2: Alternating Series test pt.8
    Notice how the numerator never changes, but the denominator is getting bigger and bigger. As the denominator gets bigger, then the numbers itself get smaller. Hence we can conclude that the sequence is decreasing, and condition is fulfilled. Some teachers may not see this method as legitimate for more complicating questions (because it’s harder to compare). In that case, look at the other methods.
  • Method 2:

    Compare the nthn^{th} term and (n+1)th(n+1)^{th} term of the sequence bnb_{n}.

    Notice that the nthn^{th} term of the sequence bnb_{n} is:

    Equation 2: Alternating Series test pt.9
    Equation 2: Alternating Series test pt.9

    And the (n+1)th(n+1)^{th} term is

    Equation 2: Alternating Series test pt.10
    Equation 2: Alternating Series test pt.10

    Now comparing the two terms, you should notice that:

    Equation 2: Alternating Series test pt.11
    Equation 2: Alternating Series test pt.11

    The left side is bigger than the right side because the denominator in the right side is bigger, hence the term is actually smaller. So we just concluded that:

    Equation 2: Alternating Series test pt.12
    Equation 2: Alternating Series test pt.12

    for all nn>1. This means that the nthn^{th} term is always going to be bigger than the (n+1)th(n+1)^{th} term, which means the sequence is always decreasing. Hence again, the second condition is fulfilled. Again, sometimes it’s really hard to compare the two terms. In this case, look at method 3.

  • Method 3:

    Take the derivative

    What were going to do is take the general term (bnb_{n}) and change all the nn’s to xx’s, and set it as f(x)f(x). In other words,

    Equation 2: Alternating Series test pt.13
    Equation 2: Alternating Series test pt.13

    Now we are going to take the derivative of this function. This will give us:

    Equation 2: Alternating Series test pt.14
    Equation 2: Alternating Series test pt.14

    Notice that for xx > 0, the denominator is going to be positive, and so the derivative f’(x) is negative. In other words, for increasing value of xx, the function is decreasing. Now let’s put that into the perspective of bnb_{n}. This means that for increasing values of n, the sequence bnb_{n} is always decreasing. Thus, we just fulfilled the second condition again.

    Since the two conditions are fulfilled, then we can conclude that the series

    Equation 2: Alternating Series test pt.15
    Equation 2: Alternating Series test pt.15

    converges. Now let’s take a look at a more interesting alternating series.

Example 2: Consider the alternating harmonic series:

Equation 3: Harmonic Alternating Series pt.1
Equation 3: Harmonic Alternating Series pt.1

Is it convergent? If it is, then what is the sum of this series?

  • What should we do here? You are probably thinking about one of the series convergence tests. You are most likely thinking about using the alternating series test. Again, we need to show that this is in fact an alternating series before we can apply the alternating series test. Recall that an alternating series could be of the form:
    Equation 3: Harmonic Alternating Series pt.2
    Equation 3: Harmonic Alternating Series pt.2
  • Notice that:
    Equation 3: Harmonic Alternating Series pt.3
    Equation 3: Harmonic Alternating Series pt.3
  • And so we know that
    Equation 3: Harmonic Alternating Series pt.4
    Equation 3: Harmonic Alternating Series pt.4
  • Since we know that it is an alternating series, then we can see if the two conditions are fulfilled. For the first condition, we see that:
    Equation 3: Harmonic Alternating Series pt.5
    Equation 3: Harmonic Alternating Series pt.5
  • So the first condition is fulfilled. Now the second condition states that bnb_{n} must be a decreasing sequence. Feel free to use any of the methods, but I will be using method two. See that the nthn^{th} term and (n+1)th(n+1)^{th} term are
    Equation 3: Harmonic Alternating Series pt.6
    Equation 3: Harmonic Alternating Series pt.6
  • Notice by comparing nthn^{th} term and (n+1)th(n+1)^{th} term of the sequence, we have:
    Equation 3: Harmonic Alternating Series pt.7
    Equation 3: Harmonic Alternating Series pt.7
  • Again, this is because the denominator on the right side is bigger, so the term is actually smaller. In other words, we just concluded that:
    Equation 3: Harmonic Alternating Series pt.8
    Equation 3: Harmonic Alternating Series pt.8
  • For nn>1. Hence, every term after is smaller than the one before it. So we can conclude that the sequence is decreasing. Since both of the conditions are fulfilled, then the series is convergent. But what is the sum of the series? Fortunately, there is an easy way to find this. First, we must recognize the Maclaurin series:
    Equation 3: Harmonic Alternating Series pt.9
    Equation 3: Harmonic Alternating Series pt.9
  • Now if we were to set xx = -1, then we will see that
    Equation 3: Harmonic Alternating Series pt.10
    Equation 3: Harmonic Alternating Series pt.10
  • In other words, we just concluded that:
    Equation 3: Harmonic Alternating Series pt.11
    Equation 3: Harmonic Alternating Series pt.11

So the sum of this alternating harmonic series is ln(2). If you want to take a look at more examples of using the alternating series test, click on this link.

http://tutorial.math.lamar.edu/Classes/CalcII/AlternatingSeries.aspx

Now here is an interesting question. If the alternating harmonic series is convergent, then what about the harmonic series itself? Is

Formula 6: Harmonic Series
Formula 6: Harmonic Series

convergent or divergent? You may realize that this isn’t an alternating series, so we are going to have to use another test. Why don’t we use the nthn^{th} term test?

nth term test for divergence

Recall that the nth term test (also known as the divergence test) states the following:

Definition 2: Nth term test
Definition 2: Nth term test

So if we take the limit and it is anything BUT 0, then we can say that the series diverge. However, if it does equal 0, then it does NOT mean the series converge. It just means the test has failed, and you would have to use something else to test the convergence.

Now using this test for the harmonic series we let:

Equation 4: Harmonic Series Divergence pt.1
Equation 4: Harmonic Series Divergence pt.1

Then you will see that:

Equation 4: Harmonic Series Divergence pt.2
Equation 4: Harmonic Series Divergence pt.2

Since we got 0, then the nth term test has failed and we would have to try something else. This leads to the question, maybe the harmonic series really converges?

Does 1/n converge

Let us assume that the harmonic series is convergent. Then that means the series must sum up to a finite number. Let’s call that sum S. So

Equation 4: Harmonic Series Divergence pt.3
Equation 4: Harmonic Series Divergence pt.3

Now we are going to play a little trick here. We can say that:

Equation 4: Harmonic Series Divergence pt.4
Equation 4: Harmonic Series Divergence pt.4

However on the right hand side of the inequality, see that

Equation 4: Harmonic Series Divergence pt.5
Equation 4: Harmonic Series Divergence pt.5

In other words, we are saying that

Equation 4: Harmonic Series Divergence pt.6
Equation 4: Harmonic Series Divergence pt.6

This is impossible, so we reached a contradiction. This situation causes us to get a mathematically illogical statement, so then the harmonic series must be divergent. A lot of people get confused by this method, so I have prepared another easier method to show that the harmonic series diverges. This method will require you to know the p-series rules.

Recall that p-series are in the form:

Equation 4: Harmonic Series Divergence pt.7
Equation 4: Harmonic Series Divergence pt.7

The p-series rule (or p-series test) states that if p>1, then the series converge. Otherwise, the series diverge. Notice that the series that we have is very similar to it.

Equation 4: Harmonic Series Divergence pt.8
Equation 4: Harmonic Series Divergence pt.8

In fact, our p=1 in this case. We also know that p=1\ngtr1, so then we know that the series diverge. Now we kind of went off topic here, but you must have realized that removing the (1)(n+1)(-1)^{(n+1)} from the series can actually change the convergence or divergence of a series.

Alternating Series Estimation Theorem

So we learned that it is possible to find the sum of an alternating harmonic series using a complicated formula that we were unfamiliar with. But what if we are dealing with another alternating series? How would we find the sum? Unfortunately there is no good way to find the exact sum of converging alternating series, but there is a way to estimate the sum. It is called the alternating series estimation theorem.

The alternating series estimation theorem states the following:

Suppose that the alternating series

Formula 5: Alternating Series with power n+1
Formula 5: Alternating Series with power n+1

Is convergent and converges to a finite number SS. Then

Definition 3: Alternating Series Estimation Theorem
Definition 3: Alternating Series Estimation Theorem

where:

SnS_{n} = the partial sum of n terms (sum of the first n terms)

RnR_{n} = the remainder (or error term) that we get from subtracting the actual value of the series with the sum of the first n terms. Sometimes it’s called the alternating series error.

bn+1b_{n+1} = the neglected term.

Note that this theorem only works if the series is alternating. You are probably very confused at what this theorem is saying, so let us use a series as an example.

Example 3: Let’s take a look at the alternating harmonic series we used earlier:

Equation 5: Harmonic Alternating Series Estimation pt.1
Equation 5: Harmonic Alternating Series Estimation pt.1
  • Let’s expand the series out. Doing so will give us:
    Equation 5: Harmonic Alternating Series Estimation pt.2
    Equation 5: Harmonic Alternating Series Estimation pt.2
  • Now let’s say that we want to estimate the sum of this series. What I’m going to do is estimate the sum of the series by summing the first 4 terms. Summing the first 4 terms gives us:
    Equation 5: Harmonic Alternating Series Estimation pt.3
    Equation 5: Harmonic Alternating Series Estimation pt.3
  • Since we only look at the first 4 terms, then the next term after it (the 5th5^{th} term) is the neglected term. Recall that the series can be rewritten as
    Equation 5: Harmonic Alternating Series Estimation pt.4
    Equation 5: Harmonic Alternating Series Estimation pt.4
  • So bn=1nb_{n} = \frac{1}{n}, and hence
    Equation 5: Harmonic Alternating Series Estimation pt.5
    Equation 5: Harmonic Alternating Series Estimation pt.5
  • Now what the theorem says is that
    Equation 5: Harmonic Alternating Series Estimation pt.6
    Equation 5: Harmonic Alternating Series Estimation pt.6
  • So using the 2 pieces of information that we have, then
    Equation 5: Harmonic Alternating Series Estimation pt.7
    Equation 5: Harmonic Alternating Series Estimation pt.7
  • Our remainder is kind of in the way of our equation, so we can get rid of it and have
    Equation 5: Harmonic Alternating Series Estimation pt.8
    Equation 5: Harmonic Alternating Series Estimation pt.8
  • Now instead of having the absolute value, we can rewrite our inequality to be:
    Equation 5: Harmonic Alternating Series Estimation pt.9
    Equation 5: Harmonic Alternating Series Estimation pt.9
  • Adding 0.58333 to all sides of the inequality gives us:
    Equation 5: Harmonic Alternating Series Estimation pt.10
    Equation 5: Harmonic Alternating Series Estimation pt.10
  • Hence, we just estimate what the sum of the series is! It must be between 0.38333 and 0.78333. Note that you can also do this by summing the first 5 term or the first 11 terms. Usually the questions you will be dealing with will tell you how many terms you need to sum. In fact, as you sum more and more terms and use this theorem, your estimation becomes more and more accurate.

Now let’s take a look at a question where we don’t know how many terms we need to sum, but we know that our remainder (or error) has to be less than a certain number.

Example 4: Determine the number of terms required to approximate the sum of the series

Equation 6: Harmonic Alternating Series Error pt.1
Equation 6: Harmonic Alternating Series Error pt.1

with an error of less than 0.0001.

  • Recall that the theorem states that:
    Equation 6: Harmonic Alternating Series Error pt.2
    Equation 6: Harmonic Alternating Series Error pt.2
  • If the error must be less than 0.0001, then basically we are saying that
    Equation 6: Harmonic Alternating Series Error pt.3
    Equation 6: Harmonic Alternating Series Error pt.3
  • or in a more simple manner,
    Equation 6: Harmonic Alternating Series Error pt.4
    Equation 6: Harmonic Alternating Series Error pt.4
  • Now we can’t really do anything with this inequality because there are too many unknowns. So we need to think of a better way to do this. This is where it gets a little tricky because we actually have to think about this.
  • Instead of saying that 0.0001 is less than the neglected term (0.0001bn+1)(0.0001 \leq b_{n+1}), why don’t we say that the neglected term is less than 0.0001. In other words, let’s say that
    Equation 6: Harmonic Alternating Series Error pt.5
    Equation 6: Harmonic Alternating Series Error pt.5
  • Why? Think about it. We know that
    Equation 6: Harmonic Alternating Series Error pt.6
    Equation 6: Harmonic Alternating Series Error pt.6
  • The error term is less than the neglected term. So if the neglected term is less than 0.0001, then it must be true that the error term is less than 0.0001. Basically we are saying that:
    Equation 6: Harmonic Alternating Series Error pt.7
    Equation 6: Harmonic Alternating Series Error pt.7
  • So this works! Now going back to our inequality
    Equation 6: Harmonic Alternating Series Error pt.8
    Equation 6: Harmonic Alternating Series Error pt.8
  • We know that
    Equation 6: Harmonic Alternating Series Error pt.9
    Equation 6: Harmonic Alternating Series Error pt.9
  • So
    Equation 6: Harmonic Alternating Series Error pt.10
    Equation 6: Harmonic Alternating Series Error pt.10
  • That means
    Equation 6: Harmonic Alternating Series Error pt.11
    Equation 6: Harmonic Alternating Series Error pt.11
  • We can rewrite this inequality to be:
    Equation 6: Harmonic Alternating Series Error pt.12
    Equation 6: Harmonic Alternating Series Error pt.12
  • That means as long as nn > 9999, then the error (or remainder) will be less than 0.0001. Let’s pick nn=10000. This means that summing the first 10000 terms will guarantee the error to be less than 0.0001. That is a lot of terms to sum up! If you want to look at more examples, then take a look at this link:

    http://mathonline.wikidot.com/error-estimation-for-approximating-alternating-series

Alternating series test

In this section we will learn about what alternating series are. It is very easy to see if the series is alternating by expanding it out. If the terms go from positive to negative and negative to positive, then it is alternating. We will also examine the convergence of alternating series by using a method called the alternating series test. The test requires two conditions, which is listed below. Keep in mind that if you cannot fulfill these conditions, that does not mean the alternating series is divergent. There is still a possibility that it is convergent.

Lessons

Notes:
An Alternating series is in the form:
(1)nbn\sum(-1)^nb_n
or
(1)n+1bn\sum(-1)^{n+1}b_n
Where bn0b^n \geq0 An alternating series is not limited to these two forms because the exponent on the (-1) can vary.
The Alternating Series Test states that if the two following conditions are met, then the alternating series is convergent:

1. lim\limn →\infty bn=0b_n=0
2. The sequence bnb_n is a decreasing sequence.

For the second condition, bnb_n does not have to be strictly decreasing for all n1n\geq 1. As long as the sequence is decreasing for nn\infty, then that will be sufficient enough.
  • 2.
    Convergence of the Alternating Series Test
    Show that the following series converge:
Teacher pug

Alternating series test

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