Math 117: Infinite Sequences

Transcription

1 Math 7: Infinite Sequences John Douglas Moore November, 008 The three main theorems in the theory of infinite sequences are the Monotone Convergence Theorem, the Cauchy Sequence Theorem and the Subsequence Theorem, all described below Convergence of infinite sequences A sequence of real numbers is simply a function s : N R We let s n denote the value of the function s at n N and sometimes write s n ) for the sequence The key definition in the subject is: Definition A sequence s n ) of real numbers is said to converge to a real number s if for every ɛ > 0, there is an N N such that n > N s n s < ɛ In this case, we write s = lim s n A sequence s n ) of real numbers which does not converge to a real number is said to diverge For example, the sequence s n ) defined by s n = /n converges to 0, because given ɛ > 0, there exists N N such that /N < ɛ by the Archimedean property of the real numbers, and n > N 0 < n < N s n 0 = n 0 < ɛ On the other hand, the sequence s n ) defined by s n = + ) n diverges Indeed, suppose that this sequence s n converges to s Then we can take ɛ =, and there exists N N such that n > N s n s < If n > N and n is even, then s n = and s n+ = 0 Hence, a contradiction = s n s n+ s n s + s s n+ < + =,

2 Proposition Suppose that s n ) is a sequence of real numbers which converges to s R Then s n ) is bounded Proof: Choose N N such that and let n > N s n s <, M = sup{ s, s,, s N, s } + If n N, then s n M, while if n N +, s n s n s + s + s M Thus s n < M for all n N, and s n ) is bounded Proposition Suppose that s n ) and t n ) are convergent sequences of real numbers with lim s n = s and lim t n = t Then lims n + t n ) = s + t, lims n t n ) = st, 3 lims n /t n ) = s/t, provided t n 0 for all n and t 0, We give a proof of the first of these assertions; proofs of the others can be found in the text []; see Theorem 7 Let ɛ > 0 be given Since s n ) converges to s, there exists an N N such that n N and n > N s n s < ɛ Since t n ) converges to s, there exists an N N such that n N and n > N t n t < ɛ Let N = maxn, N ) Then using the triangle inequality, we conclude that n N and n > N s n +t n ) s+t) s n s + t n t < ɛ + ɛ = ɛ, which is exactly what we needed to prove Application We can use this theorem, for example, to find the limit lim n + n + = lim + /n) + /n = lim + /n) lim + /n) = + lim/n) + lim/n) = It is useful to have criteria which guarantee convergence of sequences One of these concerns increasing and decreasing sequences

3 A sequence s n ) of real numbers is said to be increasing if s n s n+ for all n It is decreasing if s n s n+ for all n A sequence s n ) of real numbers is monotone if it is either increasing or decreasing Monotone Convergence Theorem A bounded monotone sequence s n ) of real numbers must converge Proof: We first consider the case in which s n ) is increasing By hypothesis, S = {s n : n N} is bounded By the completeness axiom, it must therefore have a supremum s Let ɛ > 0 Since s = sup S, there exists N N such that s N > s ɛ Then n > N s n s N > s ɛ and s n s s n s < ɛ Hence s = lim s n The case where s n ) is decreasing is proven in a similar fashion Definition A sequence s n ) of real numbers is said to diverge to and we write lim s n = ) if for every M R, there is an N N such that n > N s n > M A sequence s n ) of real numbers is said to diverge to and we write lim s n = ) if for every M R, there is an N N such that n > N s n < M It is not difficult to show that if an increasing sequence s n ) is not bounded, the sequence diverges to Similarly, if a decreasing sequence s n ) is not bounded, the sequence diverges to Cauchy sequences It is often useful to consider sequences from more general spaces than just R A sequence of elements from a metric space X, d) is just a function from N to X; we will denote such a sequence by x n ) Definition A sequence x n ) of elements in a metric space X, d) is said to converge to an element x X if for every ɛ > 0, there is an N N such that n > N dx n, x) < ɛ A sequence x n ) of elements in X, d) is called a Cauchy sequence if for every ɛ > 0 there exists an N N such that n, m N and n, m > N dx n, x m ) < ɛ For us, the most important cases are the cases in which X = R or X = R n In the case of R n, dx, y) = x y 3

4 Cauchy Theorem A Cauchy sequence in R n must converge Note It is not true that a Cauchy sequence in an arbitrary metric space is convergent It is true in R or in R n with the standard metric, because the Heine-Borel theorem and the Bolzano-Weierstrass Theorem are available Sketch of Proof: Suppose that x n ) is a Cauchy sequence in R n We divide into two cases Case I: S = {x n : n N} is finite In this case, we can set ɛ = min { x n x m ) : x n x m, n, m N}, a positive number, since there are only finitely many distances between distinct elements of S By definition of Cauchy sequence, there exists an element N N such that n, m N and n, m > N x n x m < ɛ But this implies that x n = x m for n, m > N, and hence the Cauchy sequence converges Case II: S = {x n : n N} is infinite In this case, we first show that S is bounded To do this, we first use the fact that x n ) is a Cauchy sequence to choose N N such that We then let n, m N and n, m > N x n x m < M = sup{ x, x,, x N, x N+ } + If n N, then x n M, while if n N +, Thus x n < M for all n N x n x n x N+ + x N+ + x N+ M Continuing with case II, we note that since x n ) is a Cauchy sequence, there exists N N such that n, m > N x n x m < ɛ/ It follows from the Bolzano-Weierstrass Theorem that S has an accumulation point x Since x is an accumulation point, infinitely many points x n ) must lie in Nx; ɛ/) Thus there exists an n > N such that x n Nx; ɛ/) Then m N and m > N x m x < x m x n + x n x < ɛ + ɛ = ɛ Thus x n ) converges to x, and hence x n ) is bounded We say that a metric space X, d) is complete if every Cauchy sequence in X converges Thus the above theorem states that R n, d) is complete when d is the standard metric on R n 4

5 3 Subsequences Suppose that s n ) is a sequence of real numbers If n < n < < n k < is an increasing sequence of natural numbers, then s nk ) is called a subsequence of s n ) Even though a bounded sequence of real numbers need not converge, we have: Subsequence Theorem A bounded sequence in R n has a convergent subsequence As in the Cauchy theorem, the proof rests on the Bolzano-Weierstrass Theorem Suppose that x n ) is a bounded sequence in R n We divide into two cases Case I: S = {x n : n N} is finite In this case, x n is a fixed element x 0 of R n for infinitely many n N We can let n be the smallest element of N such that x n = x 0, n be the second smallest element of N such that x n = x 0, and so forth We thereby obtain a subsequence x nk ) of x n ) which converges to x 0 Case II: S = {x n : n N} is infinite Since S is bounded, it follows from the Bolzano-Weierstrass Theorem that S has an accumulation point x We choose n so that x n N x; ) and let ɛ = min x n x, ) We next choose n so that x n N x; ɛ ) and let ) ) ɛ = min x n x, Continuing in this fashion, we obtain a subsequence x nk ) of x n ) such that x nk x < The subsequence x nk ) converges to x ) k Definition Let s n ) be a sequence of real numbers, and let M n = sup{s n, s n+, s n+, }, m n = inf{s n, s n+, s n+, } Then M n ) is a monotone decreasing sequence, while m n ) is a monotone increasing sequence We let lim sups n ) = lim M n, lim infs n ) = lim m n Note that lim infs n ) lim sups n ), with equality holding if and only if s n ) converges 5

6 Proposition Suppose that s n ) is a sequence of real numbers and let S = { all limits of subsequences of s n ) } Then sup S = lim sups n ) and inf S = lim infs n ) The proof is beyond the scope of the course; we refer to [], 9 for many examples 4 Infinite series Definition An infinite series is a sum of the form a n, where the a n s are real numbers The infinite series is said to converge to a real number s if the partial sum s n = converges to s In this case, we write s = a m a n An infinite series which does not converge to a real number is said to diverge Example One of the most important infinite series is the geometric series x n, where x < In this case, we have the partial sum Since we find that s n = x m = + x + + x n xs n = x + x + + x n+, s n xs n = x n+, or s n = xn+ x Under the assumption that 0 < x <, lim x n+ = 0, from which one can conclude that lim s n = x or x m = x 6

7 It is customary to define the real number e by means of the infinite series, To see that this infinite series converges, one can compare its partial sums with those of a geometric series, s n = n! + + ) ) n We see that the increasing sequence s n ) is bounded by the sum of the geometric series It therefore follows from the Monotone Convergence Theorem that s n ) is a convergent series We are therefore justified in letting e denote the limit, Theorem lim + /n)) n = e e = n! Our proof follows Rudin [], 33 We let s n = Then by the binomial theorem t n = + n n + nn )! m! m! and t n = + n) n ) ) 3 nn )n ) + + n 3! n = + + ) + ) )! n 3! n n + + ) n! n n ) n It follows t n t n+, so that t N ) is an increasing sequence, and it also follows from the above expressions that t n s n Hence t n ) is a bounded increasing sequence and lim t n exists and lim t n e On the other hand, if n m, t n + +! ) + ) n 3! n If we fix m and let n, we obtain n + + m! ) lim t n + +! + + m! ) m ) n n 7

8 Letting m now yields lim t n e Thus t n converges and its limit is e More generally, if x R, we can define e x by means of the infinite series e x = x n n! ) To see that this infinite series converges, we find it useful to make use of the following: Proposition Suppose that a n b n, where b n > 0, for each n N If b n converges, then a n converges Proof: To prove this, suppose that b n converges, and let ɛ > 0 be given Then N N such that n > m > N b k < ɛ It follows that when n > m > N, a k Thus if s n = k=m a k k=m k=m b k < ɛ k=m a k, n > m > N s n s m < ɛ, k=0 which implies that s n ) is a Cauchy sequence By the Cauchy Theorem, s n ) converges and hence a n converges We can apply this to show that this series ) converges for x < M, where M N is given Indeed, we can write x n M n! = x n n! + xm a k, where a k = k=0 We can compare the last sum on the right with k=0 c k, where c k = M)! ) k x, M x k M + k)! a constant multiple of a convergent geometric series with positive terms The preceding proposition shows that the series ) converges By modifying the proof of the preceding theorem, one could now establish the following limit: lim + n) x n = e x, for x 0 8

9 5 The Contraction Mapping Theorem We now discuss a slightly more advanced topic A metric space X, d) is said to be complete if every Cauchy sequence in X, d) converges Thus, for example, it follows from the Cauchy Sequence Theorem that R and R n are complete A function f : X X is called a contraction if for all x, y X, dfx), fy)) < αdx, y), where 0 < α < ) A point x X is said to be a fixed point of the contraction if fx) = x Contraction Mapping Theorem If X, d) is a complete metric space and f : X X is a contraction, then f has a unique fixed point Sketch of proof: We start by picking a point x 0 X, and for n N, let x n = fx n ) This gives a sequence of points x n ) in X Suppose that dx 0, x ) = β Then it follows from ) that dx, x ) < αβ, dx, x 3 ) < α β,, dx n, x n+ ) < α n β, Hence if k > 0, dx n, x n+k ) dx n, x n+ ) + dx n+k, x n+k ) < α n β + α + + α k ) < α n β α, where we have used the formula for the sum of a geometric series Given ɛ > 0, we can choose N sufficiently large that when n > N, α n < ɛ α β and hence α n β α < ɛ Hence for n, m > N, dx n, x m ) < ɛ and x n ) is a Cauchy sequence Since X, d) is complete, the Cauchy sequence x n ) converges to an element x X One shows that this x is a fixed point of f If x and y are two fixed points of f, then it follows from ) that so the fixed point is unique dx, y) = dfx), fy)) < αdx, y) x = y, References [] Steven R Lay, Analysis: with an introduction to proof, Pearson Prentice Hall, Upper Saddle Riven, NJ, 005 [] W Rudin, Principles of mathematical analysis, Third edition, McGraw- Hill, New York, 976 9