In this video I explore further into Infinite Sequences and Series and this time consider series that have involve the absolute values of a sequence. This involves the concept of “absolute convergence” which I show that it implies convergence in general; although the reverse is not always the case. In this case, the very interesting concept of “conditional convergence” arises which allows for the convergence of a series simply by re-arranging the terms such that the positive and negative terms cancel out; which is only possible if a series is not absolutely convergent. I also go over two specific tests to determine absolute convergence and they are the ratio and root test which are as their names describe. Half of this video is spent on 4 exercises at the end of the video and are a great way to reinforce the concepts introduced in this video.

The topics and their timestamps in the video are listed below:

1. @ 1:59 - Absolute Convergence
   • @ 2:43 - Definition 1
   • @ 3:46 - Example 1
   • @ 5:53 - Example 2
2. Conditional Convergence
   • @ 9:22 - Definition 2
   • @ 9:54 - Theorem 1
   • @ 14:34 - Example 3
3. @ 20:46 - The Ratio Test
   • @ 44:48 - Example 4
   • @ 50:37 - Note on Estimating Sums
   • @ 51:21 - Example 5
   • @ 58:51 - Note on the Ratio Test
4. @ 1:01:25 - The Root Test
   • @ 1:05:15 - Example 6
5. @ 1:08:46 - Rearrangements
6. Exercises
   • @ 1:24:52 - Exercise 1
   • @ 1:47:01 - Exercise 2: Proof of the Root Test
   • @ 2:12:30 - Exercise 3
   • @ 2:28:40 - Exercise 4

Please note the following list of corrections which I realized during the making of the video:

• @ 21:50 – Forgot to include the other condition of part (ii) of the Root Test: lim_n→∞ |a_n+1/a_n| = ∞. Note that in the video I corrected this @ 36:02 during the proof of part (ii).
• @ 2:42:42 - Should’ve written |s₉ – r| = |1 – 5| = 4 < |a₉| and also should've shown that we can ensure this to be the case regardless of the value of a₉.

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Infinite Sequences and Series: Absolute Convergence and the Ratio and Root Tests

Calculus Book Reference

Note that I mainly follow along the following calculus book:

• Calculus: Early Transcendentals Sixth Edition by James Stewart

Topics to Cover

1. Absolute Convergence
   • Definition 1
   • Example 1
   • Example 2
2. Conditional Convergence
   • Definition 2
   • Theorem 1
   • Example 3
3. The Ratio Test
   • Example 4
   • Note on Estimating Sums
   • Example 5
   • Note on the Ratio Test
4. The Root Test
   • Example 6
5. Rearrangements
6. Exercises
   • Exercise 1
   • Exercise 2: Proof of the Root Test
   • Exercise 3
   • Exercise 4

Absolute Convergence

Given any series ∑ a_n, we can consider the corresponding series:

whose terms are the absolute values of the terms of the original series.

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Definition 1

A series ∑ a_n is called absolutely convergent if the series of absolute values ∑ |a_n| is convergent.

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Notice that if ∑ a_n is a series with positive terms, then |a_n| = a_n and so absolute convergence is the same as convergence in this case.

Also note that in my earlier videos I went over convergence tests for series with positive terms and for alternating series.

But what if the signs of the terms switch back and forth irregularly?

We will see in Example 3 later in this video that the idea of absolute convergence sometimes helps in such cases.

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Example 1

The series:

is absolutely convergent because:

is a convergent p-series (p = 2).

Recall the p-series from my earlier video.

@mes/infinite-sequences-and-series-the-integral-test-and-estimate-of-sums

Retrieved: 26 August 2019
Archive: http://archive.fo/vapHa

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Example 2

We know, from my earlier video, that the alternating harmonic series:

is convergent.

Recall from my earlier video on alternating series.

@mes/infinite-sequences-and-series-alternating-tests

Retrieved: 26 August 2019
Archive: http://archive.fo/9neM8

Although the alternating harmonic series is convergent, it not absolutely convergent because the corresponding series of absolute values is:

which is the harmonic series (p-series with p = 1) and is therefore divergent.

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Conditional Convergence

Definition 2

A series ∑ a_n is called conditionally convergent if it is convergent but not absolutely convergent.

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Example 2 shows that the alternating harmonic series is conditionally convergent.

Thus it is possible for a series to be convergent but not absolutely convergent.

However, the next theorem shows that absolute convergence implies convergence.

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Theorem 1

If a series ∑ a_n is absolutely convergent, then it is convergent.

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Proof:

Observe that the inequality:

is true because |a_n| is either a_n or -a_n.

If ∑ a_n is absolutely convergent then ∑ |a_n| is convergent, so ∑ 2|a_n| is convergent.

Therefore, by the Comparison Test, ∑ (a_n + |a_n|) is convergent.

Recall the Comparison Test from my earlier video.

@mes/infinite-sequences-and-series-the-comparison-tests

Retrieved: 26 August 2019
Archive: http://archive.fo/f0SM5

Then:

is the difference of two convergent series and is therefore convergent.

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Example 3

Determine whether the series:

is convergent or divergent.

Solution:

This series has both positive and negative terms, but it is not alternating.

The first term is positive, the next three are negative, and the following three are positive: The signs change irregularly.

Calculation Check:

• Note that my Calculus Book used Radians and my OneNote uses Degrees so I need to convert degrees to radians: half a circle is 180 degrees = π radians)
• cos(1*180/pi)/1^2 = 0.54030230586814
• cos(2*180/pi)/2^2 = -0.104036709136786
• cos(3*180/pi)/3^2 = -0.109999166288938
• cos(4*180/pi)/4^2 = -0.0408527263039757
• cos(5*180/pi)/5^2) = 0.011346487418529
• cos(6*180/pi) = 0.960170286650366
• cos(7*180/pi) = 0.753902254343305
• cos(8*180/pi) = -0.145500033808614

We can apply the Comparison Test to the series of absolute values:

Since |cos n| ≤ 1 for all n, we have:

We know that ∑ 1/n² is convergent (p-series with p = 2) and therefore ∑ |cos n|/n² is convergent by the Comparison Test.

Thus the given series ∑ (cos n)/n² is absolutely convergent and therefore convergent by Theorem 1.

The figure below shows the graphs of the terms a_n and partial sums s_n of the series in Example 3.

Notice that the series is not alternating but has positive and negative terms.

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The following test is very useful in determining whether a given series is absolutely convergent.

The Ratio Test

(i) If:

then the series ∑^∞_n=1 a_n is absolutely convergent (and therefore convergent).

(ii) If:

then the series ∑^∞_n=1 a_n is divergent.

(iii) If:

then the Ratio Test is inconclusive; that is no conclusion can be drawn about the convergence or divergence of ∑ a_n.

Proof:

(i) The idea is to compare the given series with a convergent geometric series.

Recall the geometric series from my earlier video on Infinite Series.

@mes/infinite-series-definition-examples-geometric-series-harmonics-series-telescoping-sum-more

Retrieved: 27 July 2019
Archive: http://archive.fo/FGAyJ

Since L < 1, we can choose a number r such that L < r < 1.

Since:

the ratio |a_n+1/ a_n| will eventually be less than r; that is, there exists an integer N such that:

Or, equivalently,

Putting n successively equal to N, N + 1, N + 2, … in the above inequality, we obtain:

Now the series:

is convergent because it is a geometric series with 0 < r < 1.

So the general inequality above, together with the Comparison Test, shows that the series:

is also convergent.

It follows that the series ∑^∞_n=1 |a_n| is convergent.

Recall that a finite number of terms doesn't affect convergence.

Therefore ∑ a_n is absolutely convergent (and therefore convergent).

(ii) If:

then the ratio |a_n+1/ a_n| will eventually be greater than 1; that is, there exists an integer N such that:

This means that:

and so:

Therefore ∑ a_n diverges by the Test for Divergence.

Recall the Test for Divergence also from my earlier video on Infinite Series.

@mes/infinite-series-definition-examples-geometric-series-harmonics-series-telescoping-sum-more

Note: Part (iii) of the Ratio Test says that if:

then the test gives no information.

For instance, for the convergent series ∑ 1/n² we have:

whereas for the divergent series ∑ 1/n we have:

Therefore, if lim_n→∞ |a_n+1/ a_n| = 1 the series ∑ a_n might converge or it might diverge.

In this case the Ratio Test fails and we must use some other test.

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Example 4

Test the following series for absolute convergence:

Solution:

We use the Ratio Test with:

Thus, by the Ratio Test, the given series is absolutely convergent and therefore convergent.

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Note on Estimating Sums

In my earlier videos we used various methods for estimating the sum of a series - the method depended on which test was used to prove convergence.

What about series for which the Ratio Test works?

There are two possibilities:

If the series happens to be an alternating series, as in Example 4, then it is best to use the methods of my earlier video on alternating series.

If the terms are all positive, then we can use the special methods explained in Exercise 1 later in this video.

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Example 5

Test the convergence of the series:

Solution:

Since the terms a_n = n^{n / n! are positive, we don't need the absolute value signs.}

Recall from my earlier video that this is in fact the number e written as a limit.

Retrieved: 26 August 2019
Archive: http://archive.fo/sJyuL

The Theorem 2.5.8 mentioned in the above proof is shown in my other earlier video. #EarlierVideos

Retrieved: 26 August 2019
Archive: http://archive.fo/035sm

Since e > 1, the given series is divergent by the Ratio Test.

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Note on the Ratio Test

Although the Ratio Test works in Example 5, an easier method is to use the Test for Divergence.

Since:

It follows that a_n does not approach 0 as n → ∞.

Therefore the given series is divergent by the Test for Divergence.

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The following test is convenient to apply when n-th powers occur.

Its proof is similar to the proof of the Ratio Test and shown in Exercise 2 later in this video.

The Root Test

(i) If:

then the series ∑^∞_n=1 a_n is absolutely convergent (and therefore convergent).

(ii) If:

then the series ∑^∞_n=1 a_n is divergent.

(iii) If:

then the Root Test is inconclusive.

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If lim_n→∞ |a_n|1/n then part (iii) of the Root Test says that the test gives no information.

The series ∑ a_n could converge or diverge.

If L = 1 in the Ratio Test, don't try the Root Test because L will again be 1.

And if L = 1 in the Root Test, don't try the Ratio Test because it will fail too.

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Example 6

Test the convergence of the series:

Solution:

Thus the given series converges by the Root Test.

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Rearrangements

The question of whether a given convergent series is absolutely convergent or conditionally convergent has a bearing on the question of whether infinite sums behave like finite sums.

If we rearrange the order of the terms in a finite sum, then of course the value remains unchanged.

But this is not always the case for an infinite series.

By a rearrangement of an infinite series ∑ a_n we mean a series obtained by simply changing the order of the terms.

For instance, rearrangement of ∑ a_n could start as follows:

It turns out that:

If ∑ a_n is an absolutely convergent series with sum s, then any rearrangement of ∑ a_n has the same sum s.

However, any conditionally convergent series can be rearranged to give a different sum.

To illustrate this fact let's consider the alternating harmonic series:

Recall that this summation is proved in my earlier video on alternating tests.

@mes/infinite-sequences-and-series-alternating-tests

If we multiply this series by 1/2, we get:

Inserting zeros between the terms of the series, we have:

Note that adding these zeroes does not affect the sum of the series; each term in the sequence of partial sums is repeated, but the limit is the same.

Now we add the series 1 and 2 using the theorem also from my earlier video on infinite series.

@mes/infinite-series-definition-examples-geometric-series-harmonics-series-telescoping-sum-more

Thus we have:

Notice that the above series contains the terms as in series 1, but rearranged so that one negative term occurs after each pair of positive terms.

The sums of these series, however, are different.

In fact, Riemann proved that:

If ∑ a_n is a conditionally convergent series and r is any real number whatsoever, then there is a rearrangement of ∑ a_n that has a sum equal to r.

A proof of this fact is outline in Exercise 4.

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Exercises

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Exercise 1

Let ∑ a_n be a series with positive terms and let r^{n = a_n+1/ a_n.}

Suppose that lim_n→∞ r^{n = L < 1, so ∑ a_n converges by the Ratio Test.}

As usual, we let Rn be the remainder after n terms, that is:

(a) If {r^{n} is a decreasing sequence and rn+1 < 1, show, by summing a geometric series, that:}

(b) If {r^{n} is an increasing sequence, show that:}

Solution:

(a) The remainder is:

Since {r^{n} is decreasing we have:}

Since r^{n+1 < 1 thus we have that Rn is less than a convergent Geometric Series:}

(b) Note that since {r^{n} is increasing and given that:}

So, starting with Equation 1 from (a), we have:

And since L < 1 we have that the Remainder again is less than a Geometric Series:

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Exercise 2: Proof of the Root Test

Prove the Root Test.

Hint for part (i): Take any number r such that L < r < 1 and use the fact that there is an integer N such that:

Solution:

Recall the Root Test from earlier in this video:

(i) Following the hint, we get that:

Since 0 ≤ L < 1 and L < r < 1, we have 0 < r < 1 and thus the following geometric series converges:

Since |a_n| < r^{n for n ≥ N, by the Comparison Test:}

And since convergence does not depend on a finite number of terms, we have:

And thus:

(ii) Since:

then there is an integer N such that:

Thus we have:

(iii) Consider the following two series:

Divergent harmonic p-series (p = 1):

If this limit exists then we have:

Similarly for the convergent p-series (p = 2):

For each, sum:

so the Root Test is inconclusive.

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Exercise 3

Given any series ∑ a_n, we define a series ∑ a⁺_n whose terms are all the positive terms of ∑ a_n and a series ∑ a^-_n whose terms are all the negative terms of ∑ a_n.

To be specific, we let:

Notice that if a_n > 0, then a⁺_n = a_n and a^-_n = 0, whereas if a_n < 0, then a^-_n = a_n and a⁺_n = 0. (Pretty ingenious when you think about it.)

(a) If ∑ a_n is absolutely convergent, show that both of the series ∑ a⁺_n and ∑ a^-_n are convergent.

(b) If ∑ a_n is conditionally convergent, show that both of the series ∑ a⁺_n and ∑ a^-_n are divergent.

Solution:

(a) Since ∑ a_n is absolutely convergent, and since:

we thus conclude by the Comparison Test that both ∑ a⁺_n and ∑ a^-_n must be absolutely convergent.

Note: We can also solve Part (a) as follows:

(b) We will show by contradiction that both ∑ a⁺_n and ∑ a^-_n must diverge.

For suppose that ∑ a⁺_n converged.

Then so would the following series converge:

Note that ∑ |a_n| diverges because ∑ a_n is only conditionally convergent.

Hence, ∑ a⁺_n can't converge.

Similarly, neither can ∑ a^-_n.

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Exercise 4

Prove that if ∑ a_n is a conditionally convergent series and r is any real number, then there is a rearrangement of ∑ a_n whose sum is r.

Hints:

(1) Use the notation of Exercise 3.

(2) Take just enough positive terms a⁺_n so that their sum is greater than r.

(3) Then add just enough negative terms a^-_n so that the cumulative sum is less than r.

(4) Continue in this manner and use the following theorem also from my earlier video:

@mes/infinite-series-definition-examples-geometric-series-harmonics-series-telescoping-sum-more

Solution:

Let ∑ b_n be the rearranged series constructed in the hint.

Note that this series can be constructed this way by virtue of the result from Exercise 3b which states that if ∑ a_n is conditionally convergent than both ∑ a⁺_n and ∑ a^-_n are divergent; hence we can keep adding their terms to obtain as large of a number we want since there is no limit.

This series ∑ b_n will have partial sums s_n that oscillate in value back and forth across r.

Since ∑ a_n is convergent, we can apply the given theorem from my earlier video:

We can ensure that our construction of ∑ b_n has oscillations |s_n - r| that are always less than |a_n|, for example:

Thus we have that: