CHAPTER
15

Infinite Series

In This Chapter

• Investigating infinite geometric series
• Studying tests of convergence: ratio, comparison, and integral
• Examining alternating series and absolute and conditional convergence
• Estimating sums of alternating series with a maximum error estimate

Whereas a sequence is a listing of numbers that come from a defined pattern, a series is the sum of the terms of the sequence. In the next two chapters, we will examine infinite series. We’ll lay down the basics for infinite series in this chapter and work further with some important series in Chapter 16.

Infinite Geometric Series

A geometric sequence is defined by the rule a_n = a₁r^{n – 1}. For example, the sequence a_n = 2(3)^{n – 1} is 2, 6, 18, … . Do you see why 2 must be the first term? a₁ = 2(3)^{1 – 1} = 2(3)⁰ = 2. The sum of the first six terms of this series is S₆ = 2 + 6 + 18 + 54 + 162 + 486. There are only a few terms in this series so we could add them up by hand. Imagine if we had to find the sum of the first 60 terms, you would want to use a formula. The process is called “eliminating the middle.”

1. We’ll multiply the equation for S₆ by the common ratio, 3, to get 3S₆ = 6 + 18 + 54 + 162 + 486 + 1458.

2. We subtract this from the equation for S₆. S_{6 –} 3S₆ = (2 + 6 + 18 + 54 + 162 + 486) – (6 + 18 + 54 + 162 + 486 + 1458) so that –2S₆ = 2 – 1458. (Do you see how all the middle terms are eliminated?)

3. Just so that we can get to a more general formula, we’ll rewrite 1458 as 2(3)⁶. (Grab your calculator to verify that is correct.)

4. So –2S₆ = 2 – 2(3)⁶ so that .

In general, if the first term of the geometric sequence is a₁ and the common ratio is r, the sum of the first n terms of the series is . (We’ll not compute the sum of the first 60 terms as it is in the neighborhood of 4 × 10²⁷.)

Example 1: Find the sum of the first 40 terms of the series 2400 + 1800 + 1350 + 1012.5 + ….

Solution: The first is 2400 and the common ratio is . Therefore, 9599.903.

You can see that as n gets larger, the value of a_n gets smaller. The question is “What happens to S_n?” The . As n gets larger, rⁿ → 0 because r < 1.

Example 2: Compute 16 + 8 + 4 + 2 + 1 + ….

Solution: The first term is 16 and the common ratio is , so the sum is .

Example 3: Compute 16 – 8 + 4 – 2 + 1 ….

Solution: The first term is 16 and the common ratio is , so the sum is .

Tests of Convergence

If all the terms in a series Σa_n are positive, the series is called a positive series. (As opposed to an alternating series that we will discuss later in this chapter.) There are a number of tests available to us to determine if a positive series is convergent.

We begin with a very simple, but important statement, called the Divergence Test.

Simply said, if the numbers being added to the series do not get infinitely small, the sum of the terms will continue to grow.

Example 4: Show that the series diverges.

Solution:

1. We’ll go back to the familiar limit theorem, .

2. If we replace x with becomes so that n → ∞ making .

3. Therefore, series diverges.

When the rule defining the series is a continuous function that can be integrated, you can apply improper integrals to test for the convergence or divergence of a series. This is called the Integral Test.

Example 5: Show that diverges.

Solution: . Because the integral diverges, the series also diverges.

Example 6: Show that converges.

Solution: . Because the integral converges, the series also converges.

The next test for us to look at is the Comparison Test. This is a good test to use when the unknown series looks like a known convergent or divergent series.

Example 7: Show that the series converges.

Solution:

1. This series looks strikingly similar to the series , which is a known convergent series.

2. We now have to show that there comes a point in which .

3. Both denominators are positive, so we can multiply the inequality by the common denominator without affecting the orientation of the inequality: n² + 5 > n².

4. This becomes 5 > 0, which is always true.

5. Because the unknown series is less than the known series for all values of n, the unknown series, converges.

Example 8: Show that the series diverges.

Solution:

1. It seems reasonable to compare this series to the divergent harmonic series. When is ? The answer is never.

2. For positive values of n, n is never greater than n + 1.

3. So what do we do? We come up with a different divergent series. Rather than use , use the series . When is ? Always!

4. Multiply by the common denominator to get 2n > n + 1 or that n > 1.

5. The unknown series is greater than the known divergent series so the series diverges.

(Did you notice that applying the Integral Test shows that the series diverges?)

Example 9: Determine whether the series converges or diverges.

Solution:

1. Once again, the series will make for a good comparison.

2. When is ?

3. Solve: n² + 5n – 1250 > n² becomes 5n – 1250 > 0 or n > 250.

4. The inequality takes 250 terms before it holds true, but eventually it is true that .

5. The series converges.

The Limit Comparison Test also takes advantage of known convergent and divergent series in determining the nature of the convergence of an unknown series.

This test has the advantage that is saves us from doing some ugly algebra.

Example 10: Show that diverges.

Solution: Use the Limit Comparison Test with the divergent series . . Therefore, both series diverge.

Example 11: Show that the series converges. (Sorry, I got tired of using n as the variable.)

Solution: Compare this series to . Therefore, both series converge.

Alternating Series

Now that we’ve developed ground rules for how to deal with series containing only positive terms, we take on the issue of series with alternating signs. This will be the basis for the work we do with power series in the next chapter.

This statement is called the Alternating Series Test.

Example 12: Show that the alternating harmonic series converges.

Solution: The series meets the condition that decreases (use the derivative if you want; it’s always negative) and . Therefore, converges.

Example 13: Determine if the series is convergent or divergent.

2. This does not tell us that the series is divergent, but it does tell us we need to try something else. At this point, the only something else we have is the Divergence Test. If does not equal 0, then the series diverges.

3. As you can see, will oscillate between –1 and 1 for very large values of n.

4. The limit fails to exist, so the limit does not equal 0.

5. We can use the Divergence Test to conclude that is divergent.

So we now know that the harmonic series is divergent but the alternating harmonic series is convergent. This leads us to consider the other series that are related by their absolute values.

Therefore, the alternating harmonic series is conditionally convergent.

Example 14: Is the series absolutely convergent, conditionally convergent, or neither?

Solution: The series is convergent based on the p-series. Therefore, is absolutely convergent.

Example 15: Is the series absolutely convergent, conditionally convergent, or neither?

2. However, if we consider that –1 ≤ cos(n) ≤ 1, then .

3. Use the Comparison Test to note that converges conditionally then so does .

The Ratio Test is a major player in the study of alternating series. It is especially useful for series involving nth powers and factorials.

Example 16: Determine if the series is absolutely convergent, conditionally convergent, or neither.

Solution: Using the Ratio Test, . By the terms of the Ratio Test, is absolutely convergent.

Example 17: Determine if the series is absolutely convergent, conditionally convergent, or neither.

Solution: . With e > 0, the series diverges.

The Root Test is the last of the tests for convergence/divergence that we need to consider.

A useful fact to have when using the Root Test is .

Example 18: Determine if the series is absolutely convergent, conditionally convergent, or neither.

Solution: Applying the Root Test, . The series is absolutely convergent.

Estimating the Sum of Alternating Series

Let’s reinforce two items about alternating series that will help with the work we are going to do on estimating sums.

• If we define the alternating series as , the terms a_n are positive and they are decreasing.
• It is the factor (–1)ⁿ that causes the signs to alternate.

The sum of the first n terms of the series, called the partial sum S_n, estimates the true sum. However, there is a difference between partial sum and the true sum. The maximum value of this difference is the next term in the series, a_{n + 1}.

Example 19: Estimate the maximum error when is estimated by S¹⁵.

Solution: so the maximum error is .

Example 20: How many terms are needed to estimate the sum so that the maximum error is less than 0.0001?

Solution:

1. The maximum error is .

2. becomes 10000 < (k + 1)!.

3. Use your calculator to determine that k + 1 = 8 so k = 7.

4. Use seven terms to insure that the maximum error is less than 0.0001.

5. The sum , which is approximately –0.632121. is –0.632143.

6. The first four decimal places are exactly the same.

The Least You Need to Know

• Look to see if is equal to 0. If it’s not, the series diverges by the Divergence Test.
• A geometric series converges if |r| < 1; the series diverges if |r| > 1.
• Use the Integral Test when the series has the form of a function that can be integrated.
• Use the Comparison Test when the series looks like a p-series.
• When the series involves the ratio of polynomials, use the Comparison Test or Limit Comparison Test after checking that all the terms of the series are positive.
• Use the Ratio Test when the series involve factorials or constants raised to powers.
• Use the Alternating Series Test when the series have the form (-1)ⁿa_n.