ANALYSIS I. How do we get hold of a real number? Answer, we look at successive approximations. E.g.: 1000, , 1414

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1 ANALYSIS I 5 Real and complex sequences 5. Real numbers in practice How do we get hold of a real number? Answer, we look at successive approximations. E.g.: Our task is to make all this precise Sequences, 4 0, 4 00, ,... A sequence of real numbers is a function α : N R. A sequence of complex numbers is a function α : N C. E.g. σ(n) = ( ) n, e.g. ζ(n) = 0; e.g. ι(n) = n etc. etc. Note, we usually just give the values, and say the sequence,, 4,... if it is clear what the function must be. Or better we write the sequence (a n ) n= or the sequence (a n). Take care! 5.3 New sequences from old Suppose (a n ) and (b n ) are sequences of real numbers and c R. We define the sequences (a n + b n ), (ca n ), (a n b n ), (a n /b n ) in the obvious way. All are well defined except possibly the quotient. Example. a n = ( ) n, b n = n: 5.4 Tails (a n + b n ) = (0,, 0,,... ) ( a n ) = (( ) n+ ) (a n b n ) = (a n ) Let (a n ) be a sequence of real numbers and let k > 0. Define b n = a n+k then (b n ) is a sequence. Usually we write (a n+k ) n=. This we call the tail of the given sequence. 5.5 Definition of convergence Let (a n ) be a sequence of real numbers, and let l R. We say (a n ) converges to l, and write a n l as n, if for every positive real number ε > 0 there exists a natural number N N such that n N = a n l < ε If this happens, we say l is the limit of (a n ) and we say that (a n ) is convergent if for some l R, (a n ) converges to l. Note. Some people say tends to. We also write lim a n = l, lim n a n = l.

2 5.6 Examples (i) a n = n n. Then a n Proof. Look at a n = n n = n. Given ε > 0, how do we find N? Well notice that n < n [Prove n < n by induction!]. Use our Archimedean Property to find N such that N < ε. Then for n > N, n < N. So a n n n < N < ε (ii) The sequence is convergent. a n = n + n + 3n + 4 Proof. (a) How do we guess l? Well + n + n n 3 NOT PART OF OUR PROOF (b) Now suppose ε > 0 (c) Look at a n 3 = + n + n = n n 3 ( ) = 3n 3(3n n n + 4) 3n 3n 3(3n + 4) 3 3n n (d) By Archimedean Property there exists N N such that N > ε (e) n N = ε so done. n N (iii) Let a n = n + ( ) n n. + Proof. (a) Guess a = (b) Let ε > 0 (c) a n = n + ( ) n n n + n + n (d) By Archimedean Property there exists N such that N > ε (e) Done

3 5.7 Complex Sequences Let (z n ) be a sequence of complex numbers and let w C. We say that (z n ) converges to w and write z n w (or lim z n = w etc.) if for every positive real number ε > 0, there exists a natural number N such that n N = z n w < ε Theorem. Let z n = x n + iy n. (i) z n z = x n Rz, y n Iy (ii) x n x, y n y = z n x + iy Proof. (i) Put x = Rz. x n X = R(z n z) z n z. So given ε > 0 use the same N. (ii) z n z x n x + y n y by law Find N to ensure first term is less than ε/, and N to ensure second is less than ε/ then use N := min(n, N ). Note. A ε argument. 5.8 Example Let z n = ( +i) n. Then zn 0. Proof. ( ) z n 0 = n = + i + i n = + i n = = n n/ n < ε }{{} Arch. 5.9 Uniqueness of Limits Theorem. suppose that a n e and a n e. Then e = e. Proof. Suppose ε > 0. Then ε/ > 0 so there exists N such that Again ε/ > 0 so there exists N such that Suppose now n max(n, N ). Then n N = a n e < ε/ ( ) n N = a n e < ε/. ( ) e e e a n + e a n by the law < ε/ + ε/ = ε If e e then e e > 0. Chose ε = e e which contradicts to e e < ε by ( ) and ( ). 3

4 5.0 The secret of success Compare the sequence you re looking at with ones you already know about. 5. Inequalities preserved Theorem. Let (a n ) and (b n ) be real sequences, which a n a and b n b. Suppose that a n b n for all n. Then a b. Proof. Suppose, as the contrary, that b < a. Then a b > 0. Chose ε = a b > 0. Then there exists So So for n max(n, N ) N such that n N = a n a < ε N such that n N = b n b < ε n N = a ε < a n < a + ε n N = b ε < b n < b + ε a n > a ε > a a b b n < b + ε < b + b a Therefore a n > b n which is contradiction. = a + b = 3b a (as a > b) 5. A corollary Corollary. Suppose a n a, b n b and a n b n for all n K. Then a b. Proof. For you. 5.3 Sandwich rule Suppose x n a and y n a. If x n a n y n then a n a. 5.4 The tail wags the dog Theorem. Let (a n ) be a sequence, and let k N. Put b n = a n+k. Then the followings are equivalent: (i) a n l (ii) b n l Proof. 4

5 (i) (ii) Let ε > 0. Then there exists N such that n N = a n l < ε. With N = N n N = n + k N = a n+k l < ε = b n l < ε So b n l. (ii) (i) Let ε > 0. Then there exists N such that n N = b n l < ε. With N = N + k n N = (n k) N = b n k l < ε = a n l < ε. 5.5 Some notation Let a n, b n be sequences. We write a n = O(b n ) if there exist c such that for some N We write a n = o(b n ) if an b n Example. (i) n = O(n ) (ii) n = o(n ) (iii) sin nθ = O() (iv) sinnθ = o(n) is defined and n N = a n < cb n a n b n 0 5

6 6 What it s about 6. The Algebra of limits Most sequences can be built up from simpler ones using the addition, multiplication, and so on. The algebra of limits tells us how the limits behave. 6. AOL: Constants If a n = a for all n then a n a. Proof. Take N = ; n N = a n a = 0 < ε 6.3 AOL: Sums If a n a and b n b as n then a n + b n a + b. Proof. Let ε > 0 Then ε/ > 0. So Put N 3 = max(n, N ). Then [A ε-proof] 6.4 AOL: Differences N : n N = a n a < ε/ N : n N = b n b < ε/ n N 3 = (a n + b n ) (a + b) a n a + b n n by the law < ε/ + ε/ = ε If a n a and b n b as n then a n b n a b. Proof. You do it! 6.5 AOL: Translation If n a as n then a n c a c. Proof. (a n c) (a c) = a n a so proof is done. 6.6 AOL: Scalar product If a n a as n and λ R then λa n λa. Proof. Let ε > 0. Then there exists N : a n a < ε for all n N and for all n N. So case λa n λa = λ a n a λ ε 6

7 (i) if λ = 0 then λa n = 0, λa = 0. (ii) if λ 0 then ε/ λ > 0 so there exists N such that n N = a n a < ε/ λ. Then λa n λa λ ε/ λ = ε. 6.7 AOL: Products If a n a and b n b as n then a n b n ab. Proof. a n b n ab = (a n a)(b n b) + b(a n a) + a(b n b) Lemma. If x n 0 and y n 0 then x n y n 0. Proof of Lemma. Given ε > 0, let ε = min(, ε) then N, n N = x n < ε Then N, n N = y n < ε n max(n, N ) = x n y n x n y n ε < ε ε Now and a n a 0 by (6.5) b n b 0 by (6.5) Then the proof is completed by (6.3). 6.8 AOL: Reciprocals } = (a n a)(b n b) 0 by Lemma b(a n a) 0 by (6.6) a(b n b) 0 by (6.6) If a n a n 0 and a n 0 for all n. Then a n a. Needs care. We prove it first for the case a > 0. (i) As a/ > 0 there exists N such that n N = a n a < a/ = a n > a/ = < a n a (ii) Now let ε > 0; so a ε > 0. So there exist N such that n N = a n a < a ε (iii) Put N 3 = max(n, N ). Then n N 3 = a n a a n a a n a ( a ε ) a a = ε Now if a < 0 we can deduce the result from the above and scalar multiplication by λ =. 7

8 6.9 AOL: Quotients If a n a, b n b 0 and b n 0 for all n. Then a n /b n a/b Proof. (6.8) and (6.7). 6.0 AOL: Modulus If a n a then a n a. Proof. an a an a by Exercise sheet. 6. Examples (i) n + n + 3n Proof. n 0 [Proof: By Arch.];, 3 3 n 0 by (6.7) + n + n by (6.3) n 3 by (6.3) 3+ 4 n by (6.8) (ii) Suppose a =, a = and a n+ = a n+ + a n, n [Fibonacci numbers.] Then (a n ) is convergent. Proof. By induction, a n for all n.so for n ( an+ a n+ ) ( an + = + a n ) Write x n = a n+ a n for n. x n > for all n. Then x = and x n+ = + x n. Suppose x n x. By Tails: x n+ x and + x n + x by AOL. So x = + x by Uniqueness of Limits. So x x = 0. So x = ± 5. But x n > 0 for all n, and so x > 0 by 5.. So x = + 5 >. Now as τ = τ +. So x n+ + 5 = + τ = + }{{ } x n x n τ = x n τ =:τ x n+ τ x n τ = x n τ = τx n τ 8

9 Then τ }{{} 0 (x n+ τ) τ n }{{} 0 so we get x n+ τ by Sandwich rule and x n τ by Tail. 9