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4-3: Alternating Series, and the Alternating Series Theorem

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4-3: Alternating Series, and the Alternating Series Theorem 4-3: Alternating Series, and the Alternating Series Theorem Prakash Balachandran Department of Mathematics Duke University February 5, 2010 • Due to some short term-amnesia over break, I’ve decided that you’ll be allowed a one page cheat sheet for 1 this exam only (on an 8 2 × 11 piece of paper). Use this opportunity to review the essential material from 31L. • Let’s begin with a prototype example: 1 X (−1)k+1 : k k=1 Does it converge? • Clearly none of the tests we’ve developed so far can help us (I dare you to try them). So how can we determine if this series converges? Pn (−1)k+1 • Idea: let’s look at Sn = k=1 k for n = 1; 2; 3; 4; 5; 6; 7; 8; 9; 10. • Obviously, if we continue, we find the “hopping” behavior continues infinitely often. But, as we go higher 1 in n, the magnitude of each additional “hop” is getting smaller and smaller by n , so that we eventually settle down to a value as n ! 1 (if you’re curious, the above series converges to ln 2 which we’ll prove later in the semester). • Definition: A series is called an alternating series if the terms alternate in sign. That is, an alternating P k+1 series is a series of the form (−1) ak where ak > 0 for all k. • The series above is thus an example of an alternating series, and is called the alternating harmonic series. • The idea of hopping back and forth to a limit is basically the proof of: Theorem 1 (The Alternating Series Theorem) The alternating series 1 X n−1 n−1 (−1) an = a1 − a2 + a3 − a4 + ··· + (−1) an + ··· n=1 converges if 0 < an+1 < an for all n and limn!0 an = 0. • As usual, once we have a test for convergence, we ask: what’s the error in approximating the convergent sum by an nth partial sum? The answer is in the Pn i−1 th Theorem 2 (Error Estimate for Alternating Series) Let Sn = i=1(−1) ai be the n partial sum of an alternating series and let S = limn!1 Sn. Suppose that 0 < an+1 < an for all n and limn!1 an = 0. Then, jS − Snj < an+1. Proof of Theorem 2: 1 P1 i−1 – S − Sn = i=n+1(−1) ai. P1 i−1 P1 i−1 – When n is odd: 0 > S − Sn = i=n+1(−1) ai = −an+1 + i=n+2(−1) ai > −an+1 P1 i−1 P1 i−1 – When n is even: 0 < S − Sn = i=n+1(−1) ai = an+1 + i=n+2(−1) ai < an+1 – Both of these say that jS − Snj < an+1. • Examples: P1 (−1)n−1 – The series n=1 np converges for any p > 0. Contrast this with the p-series. P1 (−1)k−1 – What’s the error in approximating ln 2 = k=1 k by S9? What does this tell us about the interval in which S lies? 1 ∗ By the error estimate for alternating series: jS − S9j < a10 = 10 . ∗ From my calculator: S9 = 0:7456 so that 1 1 1 1 − < S − S < , S − < S < S + , 0:6456 < S < 0:8456 10 9 10 9 10 9 10 and so S lies somewhere in the interval (0:6456; 0:8456) (ln 2 ≈ 0:6931). P1 n n2 – Does n=1(−1) 1+n2 converge? ∗ This diverges by the nth term test. P1 – Does n=1 sin(ln n) converge? ∗ Ditto. P1 (−1)n – Does n=1 1 converge? ln(1+ n ) ∗ Ditto. P P P • Definition: A series ak is called conditionally convergent if jakj diverges, but ak converges. • (∗) A remarkable theorem of Cauchy shows that if you have a conditionally convergent series, then I can add all the terms up in a different order to obtain any number I want in the limit. Obviously this contrasts to what happens in the case of summing a finite number of numbers. – The general proof is too hard for this class, but let’s get a flavor for it by using the alternating harmonic P1 (−1)n−1 series: n=1 n . – Let’s isolate the positive and negative terms: 1 1 1 1; ; ; ;::: = fa ; a ; a ; a ;:::g 3 5 7 1 3 5 7 1 1 1 1 − ; − ; − ; − ;::: = fa ; a ; a ; a ;:::g : 2 4 6 8 2 4 6 8 – Now, here’s the idea: I give you a number x. For the sake of simplicity, suppose x > 0. Start adding the positive numbers a1; a3;::: until I just get above x and stop. Say I add M1 terms, and let b1 = a1; b2 = a3; : : : ; bM1 = a2M1+1: P1 P1 1 Note: I can always do this since n=1 a2n−1 = n=1 2n−1 diverges. 2 – Now start adding the negative terms a2; a4;::: until I just pass below x and stop. Suppose I add M2 terms, and let bM1+1 = a2; bM1+2 = a4; : : : bM1+M2 = a2M2 : P1 P1 1 Note: Again, I can always do this since n=1 a2n = − n=1 2n diverges. – Now start adding the positive terms a2M1+3; a2M1+5;::: until I just get above x and stop. Rinse and repeat. – The idea is to continue getting just above x and just below x. The result is a new sequence of terms th bk from the original sequence ak that are in a different order. By the construction of bk, its n partial sums will continue to hop just above x and just below x, so that since the terms I’m adding are getting closer and closer to 0, I hit x in the limit. Awesome!!!!!!!!!!! 3.
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