2.7 Subsequences. Definition Suppose that (s n ) n N is a sequence. Let (n 1, n 2, n 3,... ) be a sequence of natural numbers such that

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1 2.7 Subsequences Definition Suppose that (s n ) n N is a sequence. Let (n 1, n 2, n 3,... ) be a sequence of natural numbers such that n 1 < n 2 < < n k < n k+1 <. Let t k = s nk, k N. Then the sequence (t k ) is called a subsequence of (s n ). We will write (t k ) (s n ). But one should keep in mind that this is different from the inclusion relation between sets. Examples. 1. If n k = k for k N, then t k = s k. So (s n ) is a subsequence of itself. 2. Suppose s n = ( 1) n, n N. Then (s n ) has two convergent subsequence: (s 2k ) = (1, 1, 1,... ) and (s 2k 1 ) = ( 1, 1, 1,... ), which converge to different limits. So lim s n does not exist. Remark. 1. Since n 1 1 and n k+1 n k + 1, we have n k k for all k N. 2. If (t k ) is a subsequence of (s n ), and (r m ) is a subsequence of (t k ), then (r m ) is a subsequence of (s n ). Theorem If lim s n exists, and (t k ) is a subsequence of (s n ), then lim t k also exists and equals lim s n. Proof. Suppose t k = (s nk ). First suppose that lim s n = s R. Let ε > 0. There is N N such that if n > N, then s n s < ε. If k > N, then n k k > N, so s nk s < ε. Thus, t k s. The cases that lim s n = ± are similar. Definition Let (s n ) be a sequence. A subsequential limit of (s n ) is the limit of some subsequence of (s n ), whose limit exists. Example. 1. If lim s n exists, the only subsequential limit of (s n ) is lim s n. Thus, if (s n ) has more than one subsequential limit, then lim s n does not exist. 2. Suppose lim a n and lim b n both exist, which could be a real number,, or. Construct a new sequence (s n ) by (s n ) = (a 1, b 1, a 2, b 2, a 3, b 3, a 4, b 4, a 5, b 5, a 6, b 6,... ). Then both (a n ) and (b n ) are subsequences of (s n ). So lim a n and lim b n are subsequential limits of (s n ). If lim a n lim b n, then (s n ) has two different subsequential limits. So lim s n 38

2 does not exist. Now we show that (s n ) has no other subsequential limits. Suppose (s nk ) is a subsequence of (s n ) with lim s nk exists. Then the index sequence (n k ) contains either infinitely many odd numbers or infinitely many even numbers. In the first case, we find a subsequence of (s nk ), say (t k ), which is also a subsequence of (a n ). As a subsequences, we have lim t k = lim s nk and lim t k = lim a n. So lim s nk = lim a n. In the second case, a similar argument gives lim s nk = lim b n. Similarly, we may combine more than 2 sequences with limits into one sequence such as (s n ) = (a 1, b 1, c 1, a 2, b 2, c 2, a 3, b 3, c 3, a 4, b 4, c 4,... ), where lim a n, lim b n, lim c n exist. Then the set of subsequential limits of (s n ) is {lim a n, lim b n, lim c n }. 3. The sequence ( 1) n breaks into a constant 1 subsequence and a constant 1 subsequence. From the above, the only subsequential limits of ( 1) n are 1 and 1. Now let s consider the sequence (x n ) for x < 1. We may decompose the sequence into an odd sequence and an even sequence. (x 2k 1 ) = (x 1, x 3, x 5, x 7, x 9, ), (x 2k ) = (x 2, x 4, x 8, x 1 0, ). Since x > 1, x 2k 1 = x 2k 1 and x 2k = x 2k as n. Thus, the only subsequential limits of (x n ) are and, and lim x n does not exist. In summary, lim x n does not exist if x 1. Theorem Let (s n ) be a sequence. Both lim sup s n and lim inf s n are subsequential limits of (s n ). Proof. We will show that lim sup s n is a subsequential limit of (s n ). The proof for lim inf s n is similar. Recall that lim sup s n = lim m sup{s n : n m} = inf{sup{s n : n m} : m N}. Case 1. lim sup s n =. From lim inf s n lim sup s n, we have lim inf s n = lim sup s n =. Thus, lim s n =. So lim sup s n = is the limit of (s n ). Case 2. lim sup s n = +, then sup{s n : n m} = + for all m N. This means that {s n : n m} is not bounded above for any m N. We now construct a subsequence which diverges to +. Since {s n : n 1} is not bounded, there is n 1 1 such that s n1 > 1. Suppose we have integers n k > n k 1 > > n 1 1 such that s nj > j for 1 j k. Since {s n : n n j +1} is not bounded, there is n j+1 n j +1 such that s nj+1 > j +1. From induction, we get a sequence n 1 < n 2 < n 3 < such that s nk > k for all k N. Since lim k = +, lim s nk = +. Thus, + is a subsequential limit of (s n ). Case 3. lim sup s n = s R. We will construct a subsequence (s nk ) such that s nk > s 1/k for every k. Recall that lim sup s n = inf{sup{s n : n m} : m N}. So sup{s n : n m} lim sup s n = s for every m. For any m, if x < s, then x < sup{s n : n m}, which implies that x is not an upper bound of {s n : n m}, and so there exists n m such that 39

3 s n > x. Taking m = 1 and x = s 1, we get n 1 1 such that s n1 > x 1. Taking m = n and x = s 1/2, we get n 2 > n 1 such that s n2 > s 1/2. After finding 1 n 1 < n 2 < < n m such that s nk > x 1/k for 1 k m, we may then find n m+1 n m + 1 such that s nm+1 > s 1/(m + 1). By induction, we get the desired subsequence (s nk ). We now have s 1/k < s nk sup{s n : n n k } for all k. Since both s 1/k and sup{s n : n n k } tend to s as k, we must have s nk s. So s is a subsequential limit. Theorem [Bolzano-Weierstrass Theorem] Every bounded sequence has a convergent subsequence. Proof. Suppose that the sequence (s n ) is bounded above by M, and bounded below by m. By comparison theorem for upper limits, we have m lim sup s n M. Thus, lim sup s n R. From the above theorem, there is a subsequence (t k ) of (s n ) such that lim t k = lim sup s n. Since lim sup s n R, (t k ) converges. Theorem Let (s n ) be a sequence. Let S be the set of all subsequential limits of (s n ). Note that S R {+, }. Then max S = lim sup s n, and min S = lim inf s n. Proof. We have already proved that lim sup s n, lim inf s n S. Let s S. Then there is a subsequence (s nk ) of (s n ) such that s = lim s nk. Since n k k > k 1, we have s nk {s n : n > k 1}, so s nk sup{s n : n > k 1}. Taking a limit, we get s lim k sup{s n : n > k 1} = lim sup s n. Similarly, s lim inf s n. Since this is true for every s S, we have lim sup s n = max S and lim inf s n = min S. If (s n ) has only one subsequential limit, then lim sup s n = lim inf s n. So lim s n exists, which is the only subsequential limit. Corollary lim s n exists if and only if (s n ) has only one subsequential limit. Proof. If lim s n = L exists, then every subsequence of (s n ) tends to L, and so (s n ) has only one subsequential limit: L. On the other hand, if (s n ) has only one subsequential limit: L, then lim sup s n = lim inf s n = L, which implies that lim s n = L. Example. Now we consider again the sequence (s n ) = (a 1, b 1, a 2, b 2, a 3, b 3, a 4, b 4, a 5, b 5, a 6, b 6,... ) where lim a n and lim b n both exist. We have seen that lim a n and lim b n are all subsequential limits of (s n ). From the theorem above we get lim sup s n = max{lim a n, lim b n } and lim inf s n = min{lim a n, lim b n }. If lim a n = lim b n, then lim s n exists. Similarly, if (s n ) = (a 1, b 1, c 1, a 2, b 2, c 2, a 3, b 3, c 3, a 4, b 4, c 4,... ), where lim a n, lim b n, lim c n exists, then since the set of subsequential limits of (s n ) is {lim a n, lim b n, lim c n }, we have lim sup s n = max{lim a n, lim b n, lim c n } and lim inf s n = min{lim a n, lim b n, lim c n }. Moreover, lim s n exists if and only if lim a n = lim b n = lim c n. Now we summarize the relation between lim, lim sup, and lim inf. 40

4 1. lim sup s n and lim inf s n always exist, and lim sup s n lim inf s n. 2. lim s n may not exist; lim s n exists if and only if lim sup s n = lim inf s n, in which case the three limits are equal. 3. lim sup s n (resp. lim inf s n ) is the greatest (resp. least) subsequential limit of (s n ). Homework (b,c,d), 11.4 (b,c,d). No proof needed for these two problems. Additional: 1. Construct a sequence (s n ) that is bounded below and not bounded above, such that lim s n does not exist. 2. Find all subsequential limits of (sin( nπ 3 lim sup(sin( nπ 3 ) and lim inf(sin( nπ 3 ). )), and briefly explain it. Then use this to derive 2.8 Series Suppose (a n ) is a sequence. For n m, we use the symbol n a k to denote the sum a m + a m a n. To assign meaning to a k, we consider the sequences (s n ) of partial sums: s n = n a k. For example, s m = a m s m+1 = a m + a m+1 s m+2 = a m + a m+1 + a m+2 s m+3 = a m + a m+1 + a m+2 + a m+3 If lim s n = S, where S could be a real number or + or, then we say that a k = S. If S R, we say that a k converges to S. If S {+, }, we say that a k diverges to S. If lim s n does not exist, then a k has no meaning. If lim s n is infinite or does not exist, we say that a k diverges. In conclusion, when talking about the convergence/divergence/value of a series a k, we are talking about the convergence/divergence/limit of the sequence of partial sums s n = a k, k m. Example. Let r R and r 1. We have shown in homework that k=0 r k = 1 rn+1 1 r. 41

5 Suppose that r < 1. Then r n+1 0. So k=0 r k 1 r n+1 = lim n 1 r = 1 1 r. If r 1, then n k=0 rk n + 1, so k=0 rk = +. We will see later, if r 1, also diverges. k=0 rk Theorem Suppose the series a n and b n both converge. Let C R. Then the series (a n + b n ) and Ca n both converge, and (a n + b n ) = a n + Ca n = C a n. b n ; Proof. Let s n = n a k and t n = n b k be the partial sums for the the series a n and b n, respectively. By assumption, (s n ) and (t n ) converge to a n and b n, respectively. So (s n + t n ) and (Cs n ) converge to a n + b n and C a n, respectively. Since s n + t n = n (a k + b k ) and Cs n = n Ca k are the partial sums for the the series (a n + b n ) and Ca n, respectively, we then get the conclusion. Remark. 1. Except for very few cases such as n=0 rn, r < 1, we will not try to find the exact value of a series. We will only try to determine whether a series converges or diverges. Sometimes, we want to know whether the series diverges to or. Let (a n ) is a sequence. Let m < m 1 < n. Then we have a k = m 1 1 a k + 1 a k. Suppose that a k exists. Then from the above equality, we have a k = m 1 1 a k + 1 a k. This means that a k and 1 a k either both converge, or both diverge to +, or both diverge to, or both have no meaning. Thus, for our main purpose, it does not matter whether the summation starts from m or m 1. So we often write a n simply as a n. 42

6 2. We will learn several criterions to determine whether a series converges. For example, the above theorem tells us that if a n and b n both converge, then so do (a n + b n ) and Can, C R. 3. Consider the series a n with a n 0 for all n. It is clear that the partial sums s n 0 for all n and (s n ). So a n is always meaningful, which could either converges to a nonegative number, or diverges to. The series converges if and only if the partial sum sequence is bounded above. Homework. Suppose that a n diverges and b n converges. Let C R \ {0}. Show that (an + b n ) and Ca n both diverge. Theorem [Comparison Test] Let a n and b n be two series. Suppose that b n a n 0 for all n. If b n converges, then an also converges. If a n =, then b n =. Proof. Let (s n ) and (t n ) be the partial sums for the the series a n and b n, respectively. Since b n converges, (t n ) is bounded above. Since a n b n for all n, we have s n t n for all n. So (s n ) is also bounded above, which implies that a n converges. Since a n and b n either converge or diverge to, the rest statement is the contrapositive statement of the one we just proved. 43