Infinite Sequences and Series: Alternating Tests

in mathematics •  3 months ago  (edited)

In this video I go over further into the wonderful world of Infinite Sequences and Series and this time look at Alternating Series; which are series that alternate from positive and negative terms. When dealing with alternating positive and negative terms it is important to develop specific theorems and methodologies in order to better analyze whether they converge or diverge. Often times it is best to view the odd and even terms of the series separately since they would have the same signs when grouped together. I go over this and much more throughout the video so make sure to follow along!

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

  1. @ 1:28 - Introduction to Alternating Series
  2. @ 6:06 - The Alternating Series Test
    • @ 14:00 - Proof of the Alternating Series Test
    • @ 27:50 - Example 1
    • @ 32:03 - Example 2
    • @ 36:28 - Example 3
  3. @ 53:43 - Estimating Sums
    • @ 55:05 - Alternating Series Estimation Theorem
    • @ 59:27 - Example 4
    • @ 1:11:26 - Important Note
  4. Exercises
    • @ 1:11:52 - Exercise 1
    • @ 1:48:22 - Exercise 2
      • @ 2:08:50 - Proof of the Multiplication Rule for Logarithms
    • @2:14:05 - Exercise 3

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Infinite Series and Sequences: Alternating Tests

Alternating Series.jpeg

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. Introduction to Alternating Series
  2. The Alternating Series Test
    • Proof of the Alternating Series Test
    • Example 1
    • Example 2
    • Example 3
  3. Estimating Sums
    • Alternating Series Estimation Theorem
    • Example 4
    • Important Note
  4. Exercises
    • Exercise 1
    • Exercise 2
      • Proof of the Multiplication Rule for Logarithms
    • Exercise 3

Introduction to Alternating Series

The convergence tests that we have looked at so far apply only to series with positive terms.

In this section and the next we learn how to deal with series whose terms are not necessarily positive.

Of particular importance are alternating series, whose terms alternate in sign.

An alternating series is a series whose terms are alternately positive and negative.

Here are two examples:

We see from these examples that that n-th term of an alternating series is of the form:

where bn is a positive number. (In fact, bn = |an|).

The following test says that if the terms of an alternating series decrease toward 0 in absolute value, then the series converges.


The Alternating Series Test

If the alternating series:

satisfies:

then the series is convergent.


Before giving the proof let's look at the following figure, which gives a picture of the idea behind the proof.

We first plot the partial sum s1 = b1 on a number line.

To find s2 we subtract b2, so s2 is to the left of s1.

Then to find s3 we add b3, so s3 is to the right of s2.

But, since b3 < b2, s3 is to the left of s1.

Continuing in this manner, we see that the partial sums oscillate back and for.

Since bn> → 0, the successive steps are becoming smaller and smaller.

The even partial sums s2, s4, s6, … are increasing and the odd partial sums s1, s3, s5, … are decreasing.

Thus it seems plausible that both are converging to some number s, which is the sum of the series.

Therefore we consider the even and odd partial sums separately in the following proof.


Proof of the Alternating Series Test

We first consider the even partial sums:

But we can also write:

Every term in brackets is positive, so s2n ≤ b1 for all n.

Therefore the sequence {s2n} of even partial sums is increasing and bounded above.

It is therefore convergent by the Monotonic Sequence Theorem.

Recall the Monotonic Sequence Theorem from my earlier video on Infinite Sequences.

https://steemit.com/mathematics/@mes/infinite-sequences-limits-squeeze-theorem-fibonacci-sequence-and-golden-ratio-more

Retrieved: 1 August 2019
Archive: http://archive.fo/i5ZhH

Essentially, if a sequence is always increasing (or decreasing), hence monotonic, and is bounded above (or below) then it will approach a number or limit L ≤ M (or L ≥ m if the sequence is decreasing).

Let's call its limit s, that is:

Now we compute the limit of the odd partial sums:

Since both the even and odd partial sums converge to s, we have limn→∞ sn = s (see Exercise 1a at the end of this video) and so the series is convergent.


Example 1

The alternating harmonic series:

satisfies:

so the series is convergent by the Alternating Series Test.

The following figure illustrates Example 1 by showing the graphs of the terms an= (-1)n-1/n and the partial sums sn.

Notice how the values of sn zigzag across the limiting value, which appears to be about 0.7.

In fact, it can be proved that the exact sum of the series is ln 2 ≈ 0.693 (See Exercise 2 at the end of this video).


Example 2

The following series is alternating:

but:

so condition (ii) is not satisfied.

Instead, we look at the limit of the n-th term of the series:

This limit does not exist, so the series diverges by the Test for Divergence from my earlier video on Infinite Series:

https://steemit.com/mathematics/@mes/infinite-series-definition-examples-geometric-series-harmonics-series-telescoping-sum-more

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


Example 3

Test the following series for convergence or divergence:

Solution:

The given series is alternating so we try to verify conditions (i) and (ii) of the Alternating Series Test.

Unlike the situation in Example 1, it is not obvious that the sequence given by bn = n2/(n3 + 1) is decreasing.

However, if we consider the related function f(x) = x2/(x3 + 1), we find that:

Since we are considering only positive x, we see that:

Thus f is decreasing on the interval (21/3, ∞).

This means that:

Calculation Check: 2^(1/3) = 1.2599.

And therefore:

The inequality b2 < b1 can be verified directly but all that really matters is that the sequence {bn} is eventually decreasing.

Alternative Derivation:

Instead of verifying condition (i) of the Alternating Series Test by computing a derivative, we could verify that bn+1 < bn directly by using the technique of cross-multiplying as in Example 12 of my earlier video on Infinite Sequences, and shown below for reference:

Condition (ii) is readily verified:

Thus the given series is convergent by the Alternating Series Test.


Estimating Sums

A partial sum sn of any convergent series can be used as an approximation to the total sum s, but this is not of much use unless we can estimate the accuracy of the approximation.

The error involved in using s ≈ sn is the remainder Rn = s - sn.

The next theorem says that for series that satisfy the conditions of the Alternating Series Test, the size of the error is smaller than bn+1, which is the absolute value of the first neglected term.


Alternating Series Estimation Theorem

If s = ∑ (-1)n-1 bn is the sum of an alternating series that satisfies:

then:


Proof:

You can see geometrically why the Alternating Series Estimation Theorem is true by looking at the earlier figure I covered, and which I have shown below but from a screenshot of the exact figure from my Calculus book.

Notice that s - s4 < b5, |s - s5| < b6, and so on.

Notice also that s lies between any two consecutive partial sums, sn and sn+1.

It follows that:


Example 4

Find the sum of the following series correct to three decimal places.

Note that by definition: 0! = 1.

Solution:

We first observe that the series is convergent by the Alternating Series Test because:

To get a feel for how many terms we need to use in our approximation, let's write out the first few terms of the series:

Notice that:

Calculation Check: 1 - 1 + 1/2 - 1/6 + 1/24 - 1/120 + 1/720 = 0.368056.

By the Alternating Series Estimation Theorem we know that:

This error of less than 0.0002 does not affect the third decimal place, so we have s ≈ 0.368 correct to three decimal places.

In later videos, I will show that:

So what we have obtained in Example 4 is actually an approximation to the number e-1. #Amazing


IMPORTANT NOTE

The rule that the error (in using sn to approximate s) is smaller than the first neglected term is, in general, valid only for alternating series that satisfy the conditions of the Alternating Series Estimation Theorem.

The rule does not apply to other types of series.


Exercises

Here are 3 very in-depth exercises that explore further alternating series as well as sequences and series in general.

Exercise 1

(a) Show that if limn→∞ a2n = L and limn→∞ a2n+1 = L, then {an} is convergent and limn→∞ an= L.

(b) If a1 = 1 and:

find the first eight terms of the sequence {an}.

Then use part (a) to show that limn→∞ an= 21/2.

This gives the continued fraction expansion:

Solution:

Note that this exercise was originally part of an early section but it is useful to include it in this video for the derivation of the Alternating Series Test.

(a)

Let ε > 0.

Since limn→∞ a2n = L, there exists N1 such that:

Similarly, since limn→∞ a2n+1 = L, there exists N2 such that:

If n is even, then n = 2m where m > N1, so:

If n is odd, then n = 2m + 1, where m > N2, so:

Thus, effectively all we are doing is varying slightly the terms of the definition of a limit which depends on being arbitrarily close to the limit L as the number of terms n is made arbitrarily large.

Calculation Check:

  • a1 = 1
  • a2 = 1 + 1/(1+1) = 1.5
    • 3/2 = 1.5
  • a3 = 1 + 1/(1+3/2) = 1.4
    • 7/5 = 1.4
  • a4 = 1 + 1/(1+7/5) = 1.4167
    • 17/12 = 1.4167
  • a5 = 1 + 1/(1+17/12) = 1.4138
    • 41/29 = 1.4138
  • a6 = 1 + 1/(1+41/29) = 1.4143
    • 99/70 = 1.4143
  • a7 = 1 + 1/(1+99/70) = 1.4142
    • 239/169 = 1.414201
  • a 8 = 1 + 1/(239/169) = 1.7071
    • 577/408 = 1.414216

Notice that:

It appears that the odd terms are increasing and the even terms are decreasing.

Let's use Mathematical Induction to show that:

Recall from my earlier video on Mathematical Induction:

https://youtu.be/WdIr_onvUtE

Retrieved: 2 July 2019
Retrieved: http://archive.fo/4lslc

Suppose that:

We have thus shown, by induction, that the odd terms are increasing and the even terms are decreasing.

Also all the terms lie between 1 and 2 (i.e. we are adding a positive fraction less than 1 to 1), so both {an} (odds) and {bn} (evens) are bounded monotonic sequences are therefore convergent by the Monotonic Sequence Theorem.

Exercise 2

Use the following steps to show that:

Let hn and sn be partial sums of the harmonic and alternating harmonic series.

(a) Show that s2n = h2n - hn.
(b) From Exercise 3 later in this video, we have:

And therefore:

Use these facts together with part (a) to show that:

Solution:

We will prove this also by Mathematical Induction.

Let P(n) be the proposition that s2n = h2n - hn.

P(1) is the statement s2 = h2 - h1, which is true since:

So suppose that P(n) is true.

We will show that P(n+1) must be true as a consequence:

Proof of the Multiplication Rule for Logarithms


Exercise 3

Use the following steps to show that the sequence has a limit:

Note that the value of the limit is denoted by γ and is called Euler's constant.

(a) Draw a picture like the figure below from my earlier video on the Integral Test with f(x) = 1/x and interpret tn as an area (or use the inequality also shown in my earlier video and shown below) to show that tn > 0 for all n.

https://steemit.com/mathematics/@mes/infinite-sequences-and-series-the-integral-test-and-estimate-of-sums

Retrieved: 28 July 2019
Retrieved: http://archive.fo/3YMxr

(b) Interpret:

as a difference of areas to show that tn - tn+1 > 0.

Therefore, {tn} is a decreasing sequence.

(c) Use the Monotonic Sequence Theorem to show that {tn} is convergent.

Solution:

Note that this is also an exercise from an earlier section in my calculus book but it is fitting to include it in this video.

The sum of the areas of the n rectangles in the graph below is:

And the integral or area below the curve is less than the area of rectangles because the rectangles extend above the curve y = 1/x, so:

(b) Note that:

The area under f(x) = 1/x between x = n and x = n + 1 is:

Thus clearly the area below the curve is greater than the inscribed rectangle in the figure above (which is 1/(n+1)), so:

Thus tn > tn+1 and so {tn} is a decreasing sequence.

(c) We have shown that {tn} is decreasing and tn > 0 for all n.

Thus, 0 < tn ≤ t1 = 1, so {tn} is a bounded monotonic sequence, and hence converges by the Monotonic Sequence Theorem.

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