Section 10.2: Series Definition: An infinite series is a sum of the terms of an infinite sequence {an }∞ n=1 . ∞ X an = a1 + a2 + a3 + · · · + an + · · · . n=1 The nth partial sum of the series is given by s n = a1 + a2 + a3 + · · · + an = n X aj . j=1 That is, s 1 = a1 s 2 = a1 + a2 s 3 = a1 + a2 + a3 .. . These partial sums form a new sequence {sn }∞ n=1 , called the sequence of partial sums. Definition: The series ∞ X an is said to converge with sum S if the sequence of partial sums n=1 {sn }∞ n=1 converges to S. Otherwise, the series diverges. Example: List the first four partial sums of ∞ X 1n . Does the series converge or diverge? n=1 The partial sums of this series are s1 s2 s3 s4 = = = = .. . 1 1+1=2 1+1+1=3 1+1+1+1=4 sn = n. The sequence of partial sums diverges: lim sn = lim n = ∞. n→∞ Therefore, the series diverges. n→∞ ∞ X Example: List the first four partial sums of (−1)n . Does the series converge or diverge? n=1 The partial sums for this series are s1 s2 s3 s4 −1 −1 + 1 = 0 −1 + 1 − 1 = −1 0. = = = = In general, sn = −1, n odd 0, n even Since the limit lim sn does not exist, the series diverges. n→∞ ∞ X 1 Example: Show that the series converges. 2n n=1 The partial sums for this series are s1 s2 s3 sn 1 s1 = 1 − 2 1 s2 = 1 − 2 2 1 s3 = 1 − 3 2 .. . 1 sn = 1 − n . 2 1 = 2 1 1 3 = + = 2 4 4 1 1 1 7 = + + = 2 4 8 8 .. . 1 1 1 = + + ··· + n 2 4 2 The sequence of partial sums is convergent lim sn = lim n→∞ n→∞ 1 1− n 2 = 1. Thus, the series converges with sum 1. Example: If the nth partial sum of the series ∞ X n=1 an is sn = n+1 , find the sum of the series 2n + 4 and a general formula for the nth term an . The sum of the series is n+1 1 = . n→∞ 2n + 4 2 lim sn = lim n→∞ The general term an is given by an = sn − sn−1 n+1 n = − 2n + 4 2n + 2 n+1 n = − 2(n + 2) 2(n + 1) (n + 1)(n + 1) − n(n + 2) = 2(n + 1)(n + 2) 1 = . 2(n + 1)(n + 2) Note: There are certain types of series whose sum can be computed easily, provided that the series is convergent. Definition: A series is called a telescoping series if there is an internal cancellation in the partial sums. Example: Determine whether the given series converge. If so, find the sum of the series. ∞ X 1 1 (a) cos − cos n n+1 n=1 The nth partial sum of this series is 1 1 1 1 1 1 1 + cos − cos + cos − cos + · · · + cos − cos sn = cos 1 − cos 2 2 3 3 4 n n+1 1 = cos 1 − cos . n+1 Since lim sn = lim cos 1 − cos n→∞ n→∞ 1 n+1 = cos 1 − 1, the series converges with sum cos 1 − 1. (b) ∞ X n=1 ln n+1 n+2 Using properties of logarithms, X ∞ ∞ X n+1 = [ln(n + 1) − ln(n + 2)]. ln n + 2 n=1 n=1 The nth partial sum of this series is sn = (ln 2 − ln 3) + (ln 3 − ln 4) + (ln 4 − ln 5) + · · · + [ln(n + 1) − ln(n + 2)] = ln 2 − ln(n + 2) 2 = ln . n+2 Since lim sn = −∞, the series diverges. n→∞ (c) ∞ X n=1 1 n(n + 1) Using partial fraction decomposition, ∞ X n=1 ∞ X 1 = n(n + 1) n=1 1 1 − n n+1 . The nth partial sum of this series is 1 1 1 1 1 1 1 1 1 1 sn = 1 − − − − − . + + + +···+ = 1− 2 2 3 3 4 4 5 n n+1 n+1 Since lim sn = 1, the series converges with sum 1. n→∞ Definition: A geometric series is a series of the form ∞ X arn = a + ar + ar2 + ar3 + · · · , n=0 where a is a constant and r is called the ratio of the series. Theorem: (Geometric Series Test) P n If |r| < 1, the geometric series ∞ n=0 ar , where a 6= 0, converges with sum a . 1−r If |r| ≥ 1, the series diverges. Proof: The nth partial sum of the series is sn = a + ar + ar2 + · · · + arn−1 . Then rsn = ar + ar2 + ar3 + · · · + arn and rsn − sn = ar + ar2 + ar3 + · · · + arn − (a + ar + ar2 + · · · + arn−1 ) sn (r − 1) = a(rn − 1) a(rn − 1) sn = . r−1 If |r| < 1, then lim sn = n→∞ a . 1−r If |r| > 1, then limn→∞ sn = ∞. If r = 1, then sn = an and {sn } diverges. If r = −1, then {sn } = {a, 0, a, 0, . . .} and limn→∞ sn does not exist. Example: Determine whether the given series converge. If so, find the sum. (a) n ∞ X 1 1 n=0 2 3 Since |r| = 1 < 1, the series converges with sum 3 1 1 1 3 3 = = . 1 2 2 2 4 1− 3 n ∞ X 2 (b) 5 7 n=1 The series can be rewritten as n X n+1 X n ∞ ∞ ∞ X 2 2 2 10 2 5 5 = = . 7 7 7 7 7 n=0 n=0 n=0 Since |r| = 2 < 1, the series converges with sum 7 10 1 10 7 = = 2. 2 7 7 5 1− 7 ∞ X 23n (c) 5n+1 n=0 The series can be rewritten as ∞ X (23 )n n=0 Since |r| = (d) 5 · 5n = n ∞ X 1 8 n=0 5 5 . 8 > 1, the series diverges. 5 1 1 1 1 − + − + −··· 2 4 8 16 The series can be rewritten as X n ∞ 1 1 1 1 1 1 . 1 − + − + −··· = − 2 2 4 8 2 2 n=0 Since |r| = 1 < 1, the series converges with sum 2 1 1 1 2 1 = . = 1 2 2 3 3 1+ 2 Example: Find all values of x for which the series ∞ X (x − 3)n converges. Find the sum of n=1 the series for these values of x. By the Geometric Series Test, the series converges if |x − 3| < 1. That is, −1 < x − 3 < 1 2 < x < 4. For these values of x the sum is x−3 x−3 = . 1 − (x − 3) 4−x Note: In the following sections, we will discuss more general series and convergence tests convergence. The following theorem provides a quick way to determine if a series diverges. Theorem: (The P Divergence Test) The series an diverges if lim an 6= 0. n→∞ Proof: Suppose the series P an converges with sum S. Then lim sn = S, n→∞ lim sn−1 = S. n→∞ It follows that lim an = lim (sn − sn−1 ) = S − S = 0. n→∞ We have P shown that if series an diverges. P n→∞ an converges, then limn→∞ an = 0. Thus, if limn→∞ an 6= 0, the Note: If lim an = 0, the Divergence Test does not provide any information. n→∞ Example: Determine whether the following series converge or diverge. (a) ∞ X n=0 ln 2n + 3 5n − 7 The series diverges by the Divergence Test 2n + 3 2 lim ln = ln 6= 0. n→∞ 5n − 7 5 (b) ∞ X tan−1 (n) n=1 The series diverges by the Divergence Test lim tan−1 (n) = n→∞ π 6= 0. 2 ∞ X n (c) ln n n=1 The series diverges by the Divergence Test n 1 = lim n = ∞. = lim n→∞ ln n n→∞ n→∞ 1 n lim n ∞ X 1 (d) 1+ n n=1 The series diverges by the Divergence Test n 1 = e 6= 0. lim 1 + n→∞ n

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