MTH 1035 Solution to Handout – Cauchy sequences A sequence {xn} is Cauchy if for every ǫ > 0, there is an integer N such that |xm − xn|< ǫ for all m > n > N. Fundamental Property of Real Numbers Every sequence of real numbers is convergent if and only if it is a Cauchy sequence. = 2n+1 Example. The sequence {xn} 3n is Cauchy. Before proving this, first note that 2m + 1 2©n + 1 ª 6mn + 3n − 6mn − 3m n − m − = = . 3m 3n 9mn 3mn 1 Given any ǫ > 0, we may choose an integer N such that N > ǫ . If m > n > N, then |n − m| m 1 1 |xm − xn| = ≤ ≤ ≤ < ǫ. |3mn| mn n N Since {xn} is Cauchy, it is convergent. Note that we proved this without the actual limit of the sequence ever appearing in our proof. Exercises. (1) Show that each of the following sequences is Cauchy. 1 n2 + 1 sin(1) sin(2) sin(n) + +···+ ½n ¾ ½ n2 ¾ ½ 2 4 2n ¾ Answer. Note that there are many correct ways to choose an N and your answer may differ from those here. 2 (a) Given ǫ > 0, choose N such that N > ǫ . If m > n > N, then 1 1 1 1 1 1 − ≤ + < + < ǫ. ¯m n ¯ m n N N ¯ ¯ 1 2 1 (b) Given ǫ > 0, choose¯N such¯ that N > 1 and N > ǫ . Then N > N > ǫ as well. If m > n > N, then m2 + 1 n2 + 1 m2 − n2 m2 1 1 − = < = < < ǫ. ¯ m2 n2 ¯ n2m2 n2m2 n2 N 2 ¯ ¯ (c) Let sn denote¯ the sum ¯ sin(1) sin(2) sin(n) sn = + +···+ . 2 4 2n If m > n, then sin(n + 1) sin(m) sm − sn = +···+ 2n+1 2m 1 2 and sin(n + 1) sin(m) |s − s |= +···+ m n ¯ n+1 m ¯ ¯ 2 2 ¯ ¯sin(n + 1) sin(m¯) ≤ ¯ +···+ ¯ ¯ n+1 ¯ ¯ m ¯ ¯ 2 ¯ ¯ 2 ¯ ¯ 1 ¯ 1 ¯ ¯ ≤ ¯ +···+¯ ¯ ¯ 2n+1 2m 1 1 = − 2n 2m 1 < . 2n Given ǫ > 0, choose N such that 2−N < ǫ. Then 1 1 |sm − sn|< < < ǫ 2n 2N for all m > n > N. (2) Show that the sum of two Cauchy sequences is again a Cauchy sequence. Answer. Let xn and yn be Cauchy sequences and consider the sum xn + yn . © ª © ª ǫ ǫ © ª ǫ Suppose we are given a positive number . Then 2 is also a positive number. Using 2 in ǫ | − |< ǫ > > place of in the definition, there is an integer N1 such that xm xn 2 for all m n N1. | − |< ǫ > > Similarly, there is an integer N2 such that ym yn 2 for all m n N2. Let N be the maximum of N1 and N2. Now if n > N, then both m > n > N1 and m > m > n > N2 hold and |(xm + ym) − (xn + yn)|=|(xm − xn) + (ym − yn)| ≤|xm − xn|+|ym − yn| ǫ ǫ < + < ǫ. 2 2 ∞ ǫ A series n=1 an satisfies the Cauchy criterion if for every > 0, there is an integer N such that P |an+1 +···+ am|< ǫ for all m > n > N. Note that the series satisfies the Cauchy criterion if and only if its sequence of partial sums n k=1 ak is a Cauchy sequence. Therefore, the series converges if and only if it satisfies the Cauchy©P criterion.ª Exercises. (3) Using the Cauchy criterion, prove the Comparison Test: ∞ ∞ if n=1 bn converges and |an|≤ bn for all n, then n=1 an converges. P P ∞ ǫ Answer. As n=1 bn converges, it satisfies the Cauchy criterion. Let > 0. Then there is N such that P |bn+1 +···+ bm|< ǫ 3 for all m > n > N. Further, |an+1 +···+ am|≤|an+1|+···+|am| ≤ bn+1 +···+ bm =|bn+1 +···+ bm| < ǫ ∞ ∞ and so n=1 an also satisfies the Cauchy criterion. Hence, n=1 an converges. (4) Show, usingP the definition of a limit, that if {xn} is a sequenceP of positive numbers and lim xn < 1 n→∞ then there is 0 < λ < 1 and an integer N such that 0 < xn < λ for all n > N. Answer. Let L = limn→∞ xn. ≥ < ǫ = −L First note that L 0. Indeed, suppose L 0 and set 0 2 , then there is N0 such that |xn − L|< ǫ0 for n > N0. Then L L xn < L + ǫ0 = L − = < 0 2 2 for n > N0. Since xn is positive, this is a contradiction. ≤ < ǫ = 1 − > λ = + ǫ < λ < Therefore, 0 L 1. Define 2 (1 L) 0 and L . Note that 0 1. Now, there is N such that |xn − L|< ǫ ⇒ xn < L + ǫ = λ for all n > N. (5) Suppose {an} is a sequence such that |an+1| lim < 1. →∞ n |an| n Show that there are constants 0 < λ < 1 and C > 1 such that |an|< Cλ for all n. |an+1| Answer. Set xn = . By the last exercise, there is N and 0 < λ < 1 such that |an | |an+1|< λ|an| for all n > N. Define C > 1 such that |a1| |a2| |aN | C > max , , ..., . ½ λ λ2 λN ¾ n Then, |an|< Cλ for all 1 < n ≤ N. One may show that this holds for all integers n by induction. Indeed, the inductive step is n n+1 |an+1|< λ|an|< λCλ = Cλ . (6) Prove the Ratio Test: ∞ |an+1| if lim < 1 then an converges. n→∞ | | an nX=1 n Answer. By the previous exercise, there are C > 1 and 0 < λ < 1 such that |an|< Cλ for all n. As the geometric series ∞ Cλn nX=1 ∞ converges, the series n=1 an converges by the Comparison Test. P 4 Bonus Exercises. (7) Prove that the product of Cauchy sequences is again a Cauchy sequence. (8) Prove the Root Test: ∞ n if lim |an|< 1 then an converges. n→∞ p nX=1 (9) Construct a sequence for which the Ratio Test fails to show convergence, but the Root Test succeeds. (10) If f ,g : [0,1] → R are continuous functions then, by the Maximum Value Theorem, the expression |f (x) − g(x)| has a maximum value on [0,1], define |f − g|= max |f (x) − g(x)|. x∈[0,1] Using this in place of the absolute value on R, show that the sequence of functions {fn} is a Cauchy sequence where x2 x3 xn fn(x) = 1 + x + + +···+ . 2 3! n! (11) Consider a sequence of functions fn where each fn : [0,1] → R is continuous. We say fn converges if there is a function© g ªsuch that © ª lim |fn − g|= 0. n→∞ In this setting, does every Cauchy sequence converge?.