Alternating Series. L. Marizza A. Bailey. February 28, L. M. A. Bailey Alternating Series February 28, / 22

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1 Alternating Series L. Marizza A. Bailey February 28, 2018 L. M. A. Bailey Alternating Series February 28, / 22

2 Warm-Up: AP 2013 # 8 L. M. A. Bailey Alternating Series February 28, / 22

3 Answer: D The arclength of a parametric curve on the interval from t = 0 to t = 1 is given by 1 0 (x (t)) 2 + (y (t) 2 dt. So, the arclength is given by 1 0 x (t) = 3 y (t) = 2t 4t 2 + 9dt L. M. A. Bailey Alternating Series February 28, / 22

4 Answers to Integration Hwk 1 9 4t C sin(θ) + C 3 arctan(t + 2) + C or arccot(t + 2) + C 4 2 ( 1 x) ( 1 x) C 5 x + cot(x) csc(x) + C 6 x 3 3 ln(x) x C L. M. A. Bailey Alternating Series February 28, / 22

5 1 Converge: limit compare to 1 n 6 2 Diverge: limit compare to 1 n Diverge: limit compare to harmonic, 1 n 4 Diverge: Divergence Test 5 Diverge: Divergence Test 6 Converge: Ratio test or limit comparison 1 4 n L. M. A. Bailey Alternating Series February 28, / 22

6 Alternating Series A series in which terms are alternately positive and negative is called an alternating series The Alternating Harmonic Series is much more well behaved than the regular Harmonic Series. L. M. A. Bailey Alternating Series February 28, / 22

7 Bounded Monotonic Sequence First Recall: Every bounded monotonic sequence converges Monotonic means stricly increasing or strictly decreasing. L. M. A. Bailey Alternating Series February 28, / 22

8 The Alternating Harmonic Series First, we will look at even partial sums: Look at the second partial sum: and the fourth partial sum: S 2 = = 1 2 and the sixth partial sum: S 4 = 1 ( ) (1 4 ) = 7 12 S 6 = 1 ( ) ( ) 1 6 The sequence of even partial sums is always less than 1. L. M. A. Bailey Alternating Series February 28, / 22

9 Blue Dress with Black Stripes or White Dress with Gold Stripes If we look at the even partial sums a different way: Look at the second partial sum: and the fourth partial sum: = 1 2 (1 1 2 ) + ( ) > 1 2 is 1 2 plus something positive added to it, so it s greater than 1 2. The sixth partial sum: (1 1 2 ) + ( ) ( ) > (1 1 2 ) + ( ) because something positive was added to the 4 th partial sum. Each subsequent even partial sums adds a positive number to the previous partial sum, thereby, increasing the sequence. This means the sequence is strictly increasing and bounded above by 1, L. M. A. Bailey Alternating Series February 28, / 22

10 The odd partial sums Now let us see what happens to the odd partial sums: The first partial sum S 1 = 1 The third partial sum is The fifth partial sum is S 3 = 1 ( ) S 5 = 1 ( ) ( ) Since the sequence is decreasing because every odd partial sum takes away a positive number from the previous partial sum. L. M. A. Bailey Alternating Series February 28, / 22

11 It s the dress again Or we can look at it this way: The first partial sum is positive: S 1 = 1 > 0 The third partial sum is the sum of two positive numbers, and is therefore, positive: S 3 = (1 1 2 ) > 0 The fifth partial sum is the sum of three positive numbers, and is therefore positive. S 5 = (1 1 2 ) + ( ) > 0 Each partial sum is still the sum of positive numbers, and is therefore positive. Therefore, the odd partial sums are a monotonic decreasing sequence which is bounded below by zero, and therefore converges. L. M. A. Bailey Alternating Series February 28, / 22

12 But how do we know they have the same limit? Let s look at what happens to the difference between an odd partial sum and the subsequent even partial sum: lim s 2n+1 s 2n = lim (1 1 n n n 1 2n + 1 ) ( = 0 because the difference between consecutive partial sums is just the last term of the sequence. Therefore, the odd and even partial sums converge to the same limit. L. M. A. Bailey Alternating Series February 28, / 22

13 Note that the only thing needed to make the alternating harmonic series converge was that: It was really alternating The absolute value of the terms were decreasing The sequence converged to zero L. M. A. Bailey Alternating Series February 28, / 22

14 Alternating Series Test: Leibniz s Test L. M. A. Bailey Alternating Series February 28, / 22

15 Sequence of Partial Sums of Alternating Harmonic Series L. M. A. Bailey Alternating Series February 28, / 22

16 Practice Does the series below converge? k=0 ( 1) k 2k + 1 L. M. A. Bailey Alternating Series February 28, / 22

17 Absolute or Conditional Convergence If a k is an alternating series and a k converges, then we say a k converges absolutely. If a k diverges but a k converges, then we say a k converges conditionally. It is clear that if a k converges, then a k converges. L. M. A. Bailey Alternating Series February 28, / 22

18 Practice Do the series below converge conditionally or converge absolutely? k=1 k=1 ( 1) k+1 k sin(k) k 2 L. M. A. Bailey Alternating Series February 28, / 22

19 Alternating Series Error Bound What is the difference between the n th partial sum and the actual sum? n ( 1) k a k ( 1) k a k k=0 k=0 will only be the sum of the terms after n. We call this the remainder. ±(a n+1 a n+2 + a n+3 a n+4 + a n+5 a n+6 ) if we look group the terms differently, we can see that ±(a n+1 (a n+2 a n+3 ) (a n+4 a n+5 ) ) the absolute value of the remainder is always less than a n+1. L. M. A. Bailey Alternating Series February 28, / 22

20 Alternating Series Error Bound L. M. A. Bailey Alternating Series February 28, / 22

21 Practice Estimate the value of the following convergent series with an absolute error of less than.001 ( 1) k k 5 k=1 L. M. A. Bailey Alternating Series February 28, / 22

22 Alternating Series Rearrangement Theorem L. M. A. Bailey Alternating Series February 28, / 22