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series

Course: MATH 31B, Fall 2007
School: UCLA
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Word Count: 755

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Series (12.2) A series is the sum of a sequence. If {an } is a sequence, we may add it up as a1 + a2 + a3 + + an + = n=1 an . What does it mean to add up infinitely many numbers? We define the n-th partial sum sn of the series by n sn = a1 + a2 + a3 + + an = i=1 ai . We say that the series converges if the sequence {sn } of partial sums converges; otherwise, n we say the series diverges. That is,...

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Series (12.2) A series is the sum of a sequence. If {an } is a sequence, we may add it up as a1 + a2 + a3 + + an + = n=1 an . What does it mean to add up infinitely many numbers? We define the n-th partial sum sn of the series by n sn = a1 + a2 + a3 + + an = i=1 ai . We say that the series converges if the sequence {sn } of partial sums converges; otherwise, n we say the series diverges. That is, the sum of the series is lim sn = lim n n ai , if that i=1 limit exists. Example. Consider 1 1 1 1 1 = + + + + n + . The first four partial sums are 2n 2 4 8 2 n=1 1 3 7 15 1 . It turns out sn = 1 - n . So the sum of the series s1 = , s2 = , s3 = and s4 = 2 4 8 16 2 1 1 is lim sn = lim 1 - n = 1 - 0 = 1. Why is sn = 1 - n ? Here's a simple way of n n 2 2 verifying this: Write 1 1 1 1 + + + + n 2 4 8 2 1 1 1 so 2sn = 1 + + + + n-1 2 4 2 1 2sn - sn = 1 - n (things cancel) 2 1 so sn = 1 - n . 2 sn = so subtract to get Example. A geometric series is any series of the form arn = a + ar + ar2 + ar3 + , n=0 where a and r are constants. We call a the initial term and r the ratio of the series. The partial sums of this series turn out to be n sn = i=1 ari = 1 a(1 - rn+1 ) . 1-r This is verified using the same trick as the example above. (Subtract sn from rsn and things cancel.) The limit lim sn = lim a(1 - rn+1 ) converges exactly when lim rn converges; this n n n 1-r a a(1 - rn+1 ) = . Therefore, the occurs when |r| < 1. In that case, the limit is lim n 1-r 1-r a arn converges to geometric series if |r| < 1 but diverges otherwise. 1-r n=0 n 1 3 1 1 3 = . Since Example. Try a = 3 and r = . This is the series < 1, n 5 5 5 5 n=0 n=0 3 15 a = . Try a few partial sums on a calculator and the series converges to 1 = 1-r 4 1- 5 see how rapidly they converge to the exact sum 3.75. For example, s5 = 3 + 3 3 3 3 3 + + + + = 3.74976. 5 25 125 625 3125 We can also find s5 using the formula for the n-th partial sum of a geometric series: 3 1 - ( 1 )6 a(1 - r6 ) 11718 5 = . = s5 = 1 1-r 3125 1- 5 Example. Repeating decimals can be written as geometric series. For example, 0.454545 = 45 45 45 + + + 100 10000 1000000 45 45 45 + = + + . 2 100 100 1003 This is a geometric series with a = 45/100 and r = 1/100. So it converges to 45 45 45 5 45 a 45 + = 100 1 = = . + + = 2 3 100 100 100 1-r 99 11 1 - 100 1 1 1 1 1 Example. A "telescoping" series: + + + + + = . 12 23 34 n(n + 1) n(n + 1) n=1 We can find an exact formula for the n-th partial sum. We use partial fractions to rewrite the n-th term: A B 1 = + n(n + 1) n n+1 which leads to 1 = A(n + 1) + Bn or 1 = (A + B)n + A. This forces A + B = 0 and A = 1, so B = -1. Therefore, 1 1 1 = - n(n + 1) n n+1 2 Now the n-th partial sum becomes sn = 1 1 1 1 + + + + 12 23 34 n(n + 1) 1 1 1 1 1 1 = - + - + - + 1 2 2 3 3 4 1 =1- n+1 1 1 - n-1 n + 1 1 - n n+1 For example, s4 = 1 1 1 1 + + + 12 23 34 45 1 1 1 1 1 1 - + - + - = 1 2 2 3 3 4 1 4 =1- = . 5 5 + 1 1 - 4 5 Now to find the sum of the series, we compute 1 1 = 1 - 0 = 1. So it adds up to 1. = lim sn = lim 1 - n n(n + 1) n n+1 n=1 Example. A simple divergent series: (-1)n = -1 + 1 - 1 + 1 - 1 + 1 - 1 + . The n=1 n n-th partial sum is sn = 0 if n is even, but it is -1 if n is odd. So lim sn does not exist. Example. Another divergent series: n=0 2n . The n-th partial sum is sn = 2n+1 - 1 (using the formula for the partial sum of a geometric series, with a = 1 and r = 2). Now lim sn = . The famous mathematician Leonard Euler speculated (back in the 1700s) n a that there was a connection between negative numbers and infinity since the formula 1-r 1 is = -1 so -1 = . This is of course nonsense we aren't allowed to use the formula 1-2 unless |r| < 1. Theorem. If n=1 an converges, then lim an = 0. n Proof Suppose s = n=1 n an , so lim sn = s. Then lim sn-1 = s too, so lim sn = n n n n n lim (sn-1 + an ), which forces s = s + lim an , so lim an = 0. So we're done. 3 This theorem gives a quick way of verifying some series diverge. For example, diverges since lim n 2n + 1 n=1 1 n = = 0. This is called the divergent series test. But n 2n + 1 2 WARNING, you cannot use this test to show series converge. For example, you might 1 1 converges since 0, but in fact, it diverges. suppose that the "harmonic series" n n n=1 Theorem. Suppose n=1 an = a and n=1 bn = b. Then n=1 can = ca and also (an +bn ) = n=1 a + b. We can use this theorem to sum some series up. For example, n=1 5 -4 n(n + 1) 3 5n = 5(1) - 4 15 4 = 5 - 15 = -10. 4
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