The Limit of a Sequence of Numbers: Infinite Series


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1 Connexions module: m The Limit of a Sequence of Numbers: Infinite Series Lawrence Baggett This work is produced by The Connexions Project and licensed under the Creative Commons Attribution License Abstract Probably the most interesting and important examples of sequences are those that arise as the partial sums of an innite series. In fact, it will be innite series that allow us to explain such things as trigonometric and exponential functions. Probably the most interesting and important examples of sequences are those that arise as the partial sums of an innite series. In fact, it will be innite series that allow us to explain such things as trigonometric and exponential functions. Denition 1: Let {a n } 0 be a sequence of real or complex numbers. By the innite series a n we mean the sequence {S N } dened by S N = N a n. (1) n=0 The sequence {S N } is called the sequence of partial sums of the innite series a n, and the innite series is said to be summable to a number S, or to be convergent, if the sequence {S N } of partial sums converges to S.The sum of an innite series is the limit of its partial sums. An innite series a n is called absolutely summable or absolutely convergent if the innite series a n is convergent. If a n is not convergent, it is called divergent. If it is convergent but not absolutely convergent, it is called conditionally convergent. A few simple formulas relating the a n 's and the S N 's are useful: S N = a 0 + a 1 + a a N, (2) and S N+1 = S N + a N+1, (3) S M S K = Version 1.2: Dec 9, :05 pm US/Central M n=k+1 a n = a K+1 + a K A M, (4)
2 Connexions module: m for M > K. 1: REMARK Determining whether or not a given innite series converges is one of the most important and subtle parts of analysis. Even the rst few elementary theorems depend in deep ways on our previous development, particularly the Cauchy criterion. Theorem 1: Let {a n } be a sequence of nonnegative real numbers. Then the innite series a n is summable if and only if the sequence {S N } of partial sums is bounded. If a n is summable, then {S N } is convergent, whence bounded according to here 1. Conversely, we see from the hypothesis that each a n 0 that {S N } is nondecreasing (S N+1 = S N + a N+1 S N ). So, if {S N } is bounded, then it automatically converges by here 2, and hence the innite series a n is summable. The next theorem is the rst one most calculus students learn about innite series. Unfortunately, it is often misinterpreted, so be careful! Both of the proofs to the next two theorems use here 3, which again is a serious and fundamental result about the real numbers. Therefore, these two theorems must be deep results themselves. Theorem 2: Let a n be a convergent innite series. Then the sequence {a n } is convergent, and lima n = 0. Because a n is summable, the sequence {S N } is convergent and so is a Cauchy sequence. Therefore, given an ε > 0, there exists an N 0 so that S n S m < ε whenever both n and m N 0. If n > N 0, let m = n 1. We have then that a n = S n S m < ε, which completes the proof. 2: REMARK Note that this theorem is not an if and only if theorem. The harmonic series (part (b) of Exercise below) is the standard counterexample. The theorem above is mainly used to show that an innite series is not summable. If we can prove that the sequence {a n } does not converge to 0, then the innite series a n does not converge. The misinterpretation of this result referred to above is exactly in trying to apply the (false) converse of this theorem. Theorem 3: If a n is an absolutely convergent innite series of complex numbers, then it is a convergent innite series. (Absolute convergence implies convergence.) If {S N } denotes the sequence of partial sums for a n, and if {T N } denotes the sequence of partial sums for a n, then S M S N = M n=n+1 a n M n=n+1 a n = T M T N (5) 1 "The Limit of a Sequence of Numbers: Properties of Convergent Sequences", Theorem 1 < 2 "The Limit of a Sequence of Numbers: Existence of Certain Fundamental Limits", Theorem 1 < 3 "The Limit of a Sequence of Numbers: Subsequences and Cluster Points", Theorem 3: Cauchy Criterion <
3 Connexions module: m for all N and M. We are given that {T N } is convergent and hence it is a Cauchy sequence. So, by the inequality above, {S N } must also be a Cauchy sequence. (If T N T M < ε, then S N S M < ε as well.) This implies that a n is convergent. Exercise 1: The Innite Geometric Series Let z be a complex number, and dene a sequence {a n } by a n = z n. Consider the innite series an. Show that n=0 a n converges to a number S if and only if z < 1. Show in fact that S = 1/ (1 z), when z < 1. HINT: Evaluate explicitly the partial sums S N, and then take their limit. Show that S N = 1 z N+1 1 z. Exercise 2 a. Show that 1 n=1 n(n+1) converges to 1, by computing explicit formulas for the partial sums. HINT: Use a partial fraction decomposition for the a n 's. b. (The Harmonic Series.) Show that n=1 1/n diverges by verifying that S 2k > k/2. HINT: Group the terms in the sum as follows, ( ) + ( ) ( ) +..., (6) 16 and then estimate the sum of each group. Remember this example as an innite series that diverges, despite the fact that is terms tend to 0. The next theorem is the most important one we have concerning innite series of numbers. Theorem 4: Comparison Test Suppose {a n } and {b n } are two sequences of nonnegative real numbers for which there exists a positive integer M and a constant C such that b n Ca n for all n M. If the innite series a n converges, so must the innite series b n. We will show that the sequence {T N } of partial sums of the innite series b n is a bounded sequence. Then, by Theorem 1, p. 2, the innite series b n must be summable. Write S N for the Nth partial sum of the convergent innite series a n. Because this series is summable, its sequence of partial sums is a bounded sequence. Let B be a number such that S N B for all N. We have for all N > M that rclt N = N n=1 b n = n=1 b n + N n=m+1 b n n=1 b n + N n=m+1 Ca n = n=1 b n + C N n=m+1 a n n=1 b n + C N n=1 a n n=1 b n + CS N n=1 b n + CB, which completes the proof, since this nal quantity is a xed constant. (7) Exercise 3 a. Let {a n } and {b n } be as in the preceding theorem. Show that if b n diverges, then a n also must diverge.
4 Connexions module: m b. Show by example that the hypothesis that the a n 's and b n 's of the Comparison Test are nonnegative can not be dropped. Exercise 4: The Ratio Test Let {a n } be a sequence of positive numbers. a. If lim supa n+1 /a n < 1, show that a n converges. HINT: If lim supa n+1 /a n = α < 1, let β be a number for which α < β < 1. Using part (a) of here 4, show that there exists an N such that for all n > N we must have a n+1 /a n < β, or equivalently a n+1 < βa n, and therefore a N+k < β k a N. Now use the comparison test with the geometric series β k. b. If lim infa n+1 /a n > 1, show that a n diverges. c. As special cases of parts (a) and (b), show that {a n } converges if lim n a n+1 /a n < 1, and diverges if lim n a n+1 /a n > 1. d. Find two examples of innite series' a n of positive numbers, such that lima n+1 /a n = 1 for both examples, and such that one innite series converges and the other diverges. Exercise 5 a. Derive the Root Test: If {a n } is a sequence of positive numbers for which lim supa 1/n n < 1, then a n converges. And, if lim infa 1/n n > 1, then a n diverges. b. Let r be a positive integer. Show that 1/n r converges if and only if r 2. HINT: Use Exercise and the Comparison Test for r = 2. c. Show that the following innite series are summable. ( 1/ n ), n/2 n, a n /n!, (8) for a any complex number. Exercise 6 Let {a n } and {b n } be sequences of complex numbers, and let {S N } denote the sequence of partial sums of the innite series a n. Derive the Abel Summation Formula: N n=1 N 1 a n b n = S N b N + S n (b n b n+1 ). (9) The Comparison Test is the most powerful theorem we have about innite series of positive terms. Of course, most series do not consist entirely of positive terms, so that the Comparison Test is not enough. The next theorem is therefore of much importance. Theorem 5: Alternating Series Test Suppose {a 1, a 2, a 3,...} is an alternating sequence of real numbers; i.e., their signs alternate. Assume further that the sequence { a n } is nonincreasing with 0 = lim a n. Then the innite series an converges. Assume, without loss of generality, that the odd terms a 2n+1 of the sequence {a n } are positive and the even terms a 2n are negative. We collect some facts about the partial sums S N = a 1 +a a N of the innite series a n. n=1 1. Every even partial sum S 2N is less than the following odd partial sum S 2N+1 = S 2N + a 2N+1, And every odd partial sum S 2N+1 is greater than the following even partial sum S 2N+2 = S 2N+1 + a 2N+2. 4 "The Limit of a Sequence of Numbers: Subsequences and Cluster Points", Exercise 5 <
5 Connexions module: m Now 2. Every even partial sum S 2N is less than or equal to the next even partial sum S 2N+2 = S 2N +a 2N+1 +a 2N+2, implying that the sequence of even partial sums {S 2N } is nondecreasing. 3. Every odd partial sum S 2N+1 is greater than or equal to the next odd partial sum S 2N+3 = S 2N+1 + a 2N+2 + a 2N+3, implying that the sequence of odd partial sums {S 2N+1 } is nonincreasing. 4. Every odd partial sum S 2N+1 is bounded below by S 2. For, S 2N+1 > S 2N S 2. And, every even partial sum S 2N is bounded above by S 1. For, S 2N < S 2N+1 S Therefore, the sequence {S 2N } of even partial sums is nondecreasing and bounded above. That sequence must then have a limit, which we denote by S e. Similarly, the sequence {S 2N+1 } of odd partial sums is nonincreasing and bounded below. This sequence of partial sums also must have a limit, which we denote by S o. S o S e = lims 2N+1 lims 2N = lim (S 2N+1 S 2N ) = lima 2N+1 = 0, (10) showing that S e = S o, and we denote this common limit by S. Finally, given an ε > 0, there exists an N 1 so that S 2N S < ε if 2N N 1, and there exists an N 2 so that S 2N+1 S < ε if 2N + 1 N 2. Therefore, if N max (N 1, N 2 ), then S N S < ε, and this proves that the innite series converges. Exercise 7: The Alternating Harmonic Series a. Show that n=1 ( 1)n /n converges, but that it is not absolutely convergent. b. Let {a n } be an alternating series, as in the preceding theorem. Show that the sum S = a n is trapped between S N and S N+1, and that S S N a N. c. State and prove a theorem about eventually alternating innite series. d. Show that z n /n converges if and only if z 1, and z 1. HINT: Use the Abel Summation Formula to evaluate the partial sums. Exercise 8 Let s = p/q be a positive rational number. a. For each x > 0, show that there exists a unique y > 0 such that y s = x; i.e., y p = x q. b. Prove that 1/n s converges if s > 1 and diverges if s 1. HINT: Group the terms as in part (b) of Exercise. Theorem 6: Test for Irrationality Let x be a real number, and suppose that {p N /q N } is a sequence of rational numbers for which x = limp N /q N and x p N /q N for any N. If limq N x p N /q N = 0, then x is irrational. We prove the contrapositive statement; i.e., if x = p/q is a rational number, then limq N x p N /q N 0. We have x p N /q N = p/q p N /q N = pq N qp N qq N. (11) Now the numerator pq N qp N is not 0 for any N. For, if it were, then x = p/q = p N /q N, which we have assumed not to be the case. Therefore, since pq N qp N is an integer, we have that x p N /q N = pq N qp N qq N 1 qq N. (12)
6 Connexions module: m So, and this clearly does not converge to 0. Exercise 9 q N x p N /q N 1 q, (13) a. Let x = n=0 ( 1)n /2 n. Prove that x is a rational number. b. Let y = n=0 ( 1)n /2 n2. Prove that y is an irrational number. HINT: The partial sums of this series are rational numbers. Now use the preceding theorem and part (b) of Exercise (The Alternating Harmonic Series).
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