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Math 231E, Lecture 27. Alternating Series

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Math 231E, Lecture 27. Alternating Series Math 231E, Lecture 27. Alternating Series 1 Definition Definition 1.1. An alternating sequence/series is a sequence/series where the signs of the term are alter- natively positive and negative. Examples of alternating series include 1 1 1 1 1 − 2 + 3 − 4 + 5 − 6 + 7 − :::; 1 − + − + − ::: 2 4 8 16 We continue the convention that an is the nth term in the sequence, and we will always write bn = janj. Then if an is an alternating sequence then either n n+1 an = (−1) bn; an = (−1) bn: 2 Alternating Series Test We have a nice theorem to test for convergence of an alternating series: Theorem 2.1 (Alternating Series Test). Let an be an alternating series, and bn = janj. If 1. bn+1 ≤ bn 2. limn!1 bn = 0, P then n an converges. Note! Remember how we said you cannot deduce that a series converges just because its terms go to zero? If the series is decreasing and alternating, then you can! Proof. Let us say that a1 ≥ 0 (the proof is similar if the first term is negative.) We have s2 = b1 − b2; s4 = b1 − b2 + b3 − b4 = s2 + (b3 − b4) ≥ s2: More generally, s2n = s2n−2 + b2n−1 − b2n ≥ s2n−2; so s2n is an increasing sequence. Also notice that s2n = b1 − (b2 − b3) − (b4 − b5) − ::: 1 and every term in a pair of parentheses is positive, so s2n ≤ b1. Therefore the sequence s2n is increasing and bounded above, and by the Monotone Convergence Theorem, has a limit. Let us write lim s2n = S: n!1 We then compute lim s2n+1 = lim (s2n + b2n+1) n!1 n!1 = lim s2n + lim b2n+1 n!1 n!1 = S + 0 = S; so lim s2n+1 = S: n!1 From this we see that sn converges to S, and we are done. Remark 2.2. There is one way in which we can weaken the assumptions of the AST slightly. If it turns out that there exists an N such that for all n > N, bn+1 ≤ bn, then this is good enough to obtain convergence. Example 2.3. Let us consider the Alternating Harmonic Series 1 1 1 1 1 X (−1)n+1 1 − + − + − · · · = : 2 3 4 5 n n=1 Here bn = 1=n which is clearly decreasing, and bn ! 0. So by the AST, this converges. In fact, it converges to ln(2). (If you are interested to see how to compute this, see here.) Example 2.4. Consider 1 X 6n + 7 (−1)n : 8n + 9 n=1 This is an alternating series, but notice that 6 3 lim bn = = 6= 0: n!1 8 4 Since the limit of the bn is not zero, then the series does not converge! Example 2.5. What about 1 X n2 (−1)n 2n3 + 3 n=1 Now, the limit of these terms is zero. But are the terms decreasing? Let us consider the function x2 f(x) = ; 2x3 + 3 which is continuous on the whole real line. Now we compute 2x(2x3 + 3) − x2(6x) 2x(3 − x3) f 0(x) = = : (2x3 + 3)2 (2x3 + 3)2 0 Now, notice that we do not have f (x) < 0 for all px, and therefore x is not a decreasing function. However, if x > 0 and 3 − x3 < 0, then f 0(x) < 0, so for x > 3 3, we have f 0(x) < 0. Therefore, for n ≥ 3, the sequence an is decreasing. Also we can see that an ! 0, so by Remark 2.2 this series converges. 2 3 Something to be careful about! As mentioned above, we need that the sequence bn is decreasing, or at least eventually decreasing. Is that really necessary? As it turns out, this is in fact necessary. Consider the (admittedly weird) example where 1 1 b = ; b = ; 2n n 2n+1 2n n and let an = (−1) bn. Then this is an alternating sequence, and definitely lim bn = 0; n!1 but just as clearly, this is not a decreasing function. For example, the first few terms of bn are 1 1 1 1 1 1 1 1 ; ; ; ; ; ; ; ;::: 2 2 4 3 8 5 16 7 and we can see from this pattern that it is not decreasing. So the sum we are considering is 1 1 1 1 1 1 1 1 − + − + − + − + + ::: 2 2 4 3 8 5 16 7 Let us compute some of the partial sums of this series: 1 1 1 1 1 1 3 19 s = 0; s = − = ; s = s + − = + = : 2 4 3 4 12 6 4 5 8 12 40 120 More generally, let’s just look at the even partial sums: we obtain 1 1 s = s + − : 2n 2(n−1) n 2n−1 It’s not too hard to check that for n ≥ 5, 1 1 1 − > ; n 2n−1 2n and therefore 1 s > s + 2n 2(n−1) 2n If we start this at k = 5, then this means that n X 1 s > s + : 2n 10 2k k=5 Now take the limit n ! 1. The right-hand side of this inequality goes to 1, since the harmonic series diverges, and so then s2n ! 1. Therefore the limit of the partial sums cannot converge, and therefore the sum cannot converge. tl;dr: The decreasing condition is actually quite important. 4 Estimating alternating sums Theorem 4.1. Let us consider the convergent alternating sum 1 X an; n=1 3 and call the sum S. Define the remainder Rn = S − sn as in last lecture, so 1 n 1 X X X Rn = ak − ak = ak: k=1 k=1 k=n+1 Then jRnj ≤ bn+1 = jan+1j : Example 4.2. Let us consider the alternating series 1 X (−1)n S = n! n=1 From the Alternating Series Test, this converges: notice that 1=n! is both decreasing and converges to zero. Let’s imagine that we want to compute this sum to within 2 decimal places, i.e. to have a remainder less than 1=100. Basically, we need to compute which term is less than 1=100 and we can stop there. But since 5! > 100 we can just go to five terms: 1 1 1 1 44 s = 1 − 1 + − + − = = 0:366666 ::: (1) 5 2 6 24 120 120 So how close is this? In fact, 1 S = : e (Why is this??) Then we can compute S = 0:367879 ··· and we got the first two digits correct! 4.
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