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INFINITE SEQUENCES AND SERIES The convergence tests that we have looked at so far apply only to series with positive terms.
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11.5 Alternating Series In this section, we will learn: How to deal with series whose terms alternate in sign. INFINITE SEQUENCES AND SERIES
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ALTERNATING SERIES An alternating series is a series whose terms are alternately positive and negative. Here are two examples:
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ALTERNATING SERIES From these examples, we see that the nth term of an alternating series is of the form a n = (–1) n – 1 b n or a n = (–1) n b n where b n is a positive number. In fact, b n = |a n |
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ALTERNATING SERIES The following test states that, if the terms of an alternating series decrease toward 0 in absolute value, the series converges.
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ALTERNATING SERIES TEST If the alternating series with b n > 0, satisfies i. b n+1 b n for all n ii. then the series is convergent.
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ALTERNATING SERIES Before giving the proof, let us look at this figure, which gives a picture of the idea behind the proof.
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ALTERNATING SERIES First, we plot s 1 = b 1 on a number line. To find s 2,we subtract b 2. So, s 2 is to the left of s 1.
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ALTERNATING SERIES Then, to find s 3, we add b 3. So, s 3 is to the right of s 2. However, since b 3 < b 2, s 3 is to the left of s 1.
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ALTERNATING SERIES Continuing in this manner, we see that the partial sums oscillate back and forth. Since b n → 0, the successive steps are becoming smaller and smaller.
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ALTERNATING SERIES The even partial sums s 2, s 4, s 6,... are increasing. The odd partial sums s 1, s 3, s 5,... are decreasing.
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ALTERNATING SERIES Thus, it seems plausible that both are converging to some number s, which is the sum of the series. So, we consider the even and odd partial sums separately in the following proof.
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ALTERNATING SERIES TEST - PROOF First, we consider the even partial sums: s 2 = b 1 – b 2 0 since b 2 b 1 s 4 = s 2 + (b 3 – b 4 ) s 2 since b 4 b 3
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ALTERNATING SERIES TEST - PROOF In general, s 2n = s 2n – 2 + (b 2n – 1 – b 2n ) s 2n – 2 since b 2n b 2n – 1 Thus, 0 s 2 s 4 s 6 … s 2n …
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ALTERNATING SERIES TEST - PROOF However, we can also write: s 2n = b 1 – (b 2 – b 3 ) – (b 4 – b 5 ) – … – (b 2n – 2 – b 2n – 1 ) – b 2n Every term in brackets is positive. So, s 2n b 1 for all n.
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ALTERNATING SERIES TEST - PROOF Thus, the sequence {s 2n } of even partial sums is increasing and bounded above. Therefore, it is convergent by the Monotonic Sequence Theorem.
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ALTERNATING SERIES TEST - PROOF Let us call its limit s, that is, Now, we compute the limit of the odd partial sums:
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ALTERNATING SERIES TEST - PROOF As both the even and odd partial sums converge to s, we have See Exercise 80(a) in Section 11.1 Thus, the series is convergent.
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ALTERNATING SERIES The alternating harmonic series satisfies i. b n+1 < b n because ii. It is convergent by the Alternating Series Test. Example 1
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ALTERNATING SERIES The series is alternating. However, So, condition ii is not satisfied. Example 2
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ALTERNATING SERIES Instead, we look at the limit of the nth term of the series: This limit does not exist. So, the series diverges by the Test for Divergence. Example 2
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ALTERNATING SERIES Test the series for convergence or divergence. The given series is alternating. So, we try to verify conditions i and ii of the Alternating Series Test. Example 3
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ALTERNATING SERIES Unlike the situation in Example 1, it is not obvious that the sequence given by b n = n 2 /(n 3 + 1) is decreasing. However, if we consider the related function f (x) = x 2 /(x 3 + 1), we find that: Example 3
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ALTERNATING SERIES Since we are considering only positive x, we see that f ’(x). Thus, f is decreasing on the interval (, ∞). Example 3
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ALTERNATING SERIES This means that f (n + 1) < f (n) and, therefore, b n+1 < b n when n 2. The inequality b 2 < b 1 can be verified directly. However, all that really matters is that the sequence {b n } is eventually decreasing. Example 3
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ALTERNATING SERIES Condition ii is readily verified: Thus, the given series is convergent by the Alternating Series Test. Example 3
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ESTIMATING SUMS A partial sum s n of any convergent series can be used as an approximation to the total sum s. However, this is not of much use unless we can estimate the accuracy of the approximation. The error involved in using s s n is the remainder R n = s – s n.
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ESTIMATING SUMS The next theorem says that, for series that satisfy the conditions of the Alternating Series Test, the size of the error is smaller than b n+1. This is the absolute value of the first neglected term.
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ALTERNATING SERIES ESTIMATION THEOREM If s = (–1) n-1 b n is the sum of an alternating series that satisfies i. 0 b n+1 b n ii. then |R n | = |s – s n | b n+1
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ALTERNATING SERIES ESTIMATION THM.—PROOF From the proof of the Alternating Series Test, we know that s lies between any two consecutive partial sums s n and s n+1. It follows that: |s – s n | |s n+1 – s n | = b n+1
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ALTERNATING SERIES ESTIMATION THEOREM You can see geometrically why the theorem is true by looking at this figure. Notice that s – s 4 < b 5, |s – s 5 | < b 6, and so on. Notice also that s lies between any two consecutive partial sums.
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ESTIMATING SUMS Find the sum of the series correct to three decimal places. By definition, 0! = 1. Example 4
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ESTIMATING SUMS First, we observe that the series is convergent by the Alternating Series Test because: i. ii. Example 4
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ESTIMATING SUMS To get a feel for how many terms we need to use in our approximation, let us write out the first few terms of the series: Example 4
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ESTIMATING SUMS Notice that and Example 4
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ESTIMATING SUMS By the Alternating Series Estimation Theorem, we know that: | s – s 6 | ≤ b 7 < 0.0002 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. Example 4
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NOTE The rule that the error (in using s n 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.
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