Lecture 10: The Completeness of the Real Numbers and Basic Properties of Infinite Series

18.100A: Complete Lecture Notes Lecture 10: The Completeness of the Real Numbers and Basic Properties of Infinite Series Cauchy Sequences Definition 1 Cauchy A sequence {xn } is Cauchy if ∀ > 0 ∃M ∈ N such that for all n, k ≥ M , |xn − xk | < . Example 2 Show the sequence xn = 1 n is Cauchy. Proof : Let  > 0 and choose M ∈ N such that 1 M < 2 . Then, if n, k ≥ M , then 1 1 1 2 1 − ≤ + ≤ < . n k n k M  Negation 3 (Not Cauchy) By the negation of the definition, a sequence {xn } is not Cauchy if ∃0 > 0 such that for all M ∈ N, ∃n, k ≥ M such that |xn − xk | ≥ 0 . Example 4 Show the sequence xn = (−1)n is not Cauchy. Proof : Choose  = 1 and let M ∈ N. Choose n = M and k = M + 1. Then, |(−1)n − (−1)k | = 2 ≥ 1.  Theorem 5 If {xn } is Cauchy, then {xn } is bounded. Proof : If {xn } is Cauchy then ∃M ∈ N such that for all n, k ≥ M , |xn − xk | < 1. 1 Then, for all n ≥ M , |xn − xM | < 1. Hence, |xn | ≤ |xn − xM | + |xM | < |xM | + 1. Let B = |x1 | + · · · + |xM | + 1. Then, for all n ∈ N, |xn | ≤ B. Theorem 6 If {xn } is Cauchy and a subsequence {xnk } converges, then {xn } converges. Proof : Suppose that {xnk } is a subsequence of {xn } such that limk→∞ xnk = x. We claim that xn → x. Let  > 0. Since xnk → x, there exists M0 ∈ N such that ∀k ≥ M0 , |xnk − x| <  . 2 Since {xn } is Cauchy, there exists an M1 ∈ N such that for all n ≥ M1 and m ≥ M1 , |xn − xm | <  . 2 Choose M = M0 + M1 . If n ≥ M , then nM ≥ M ≥ M0 and n≥ M1 . Therefore, |xn − x| ≤ |xn − xnM | + |xnM − x| <   + = . 2 2 Theorem 7 A sequence of real numbers {xn } is Cauchy if and only if {xn } is convergent. Proof : ( =⇒ ) If {xn } is Cauchy, then {xn } is bounded. Therefore, {xn } has a convergent subsequence by Bolzano-Weierstrass. By the previous theorem, we thus have that {xn } is convergent. ( ⇐= ) Suppose that {xn } is convergent and x = limn→∞ xn . Let  > 0. Since xn → x, ∃M0 ∈ N such that ∀n ≥ M0 ,  |xn − x| < . 2 Choose M = M0 . Then, if n, k ≥ M , |xn − xk | ≤ |xn − x| + |xk − x| <   + = . 2 2 Therefore, {xn } is Cauchy. Series Remark 8. Series were the original motivation for analysis. Definition 9 Given {xn }, the symbol sequence P∞ n=1 xn or P xn is the series associated to {xn }. We say ( sm = m X n=1 P )∞ xn m=1 converges. We call the terms of {sm } the partial sums. If limm→∞ sm = s, we write s = as a number. 2 xn converges if the P xn and treat P xn Remark 10. A series need not start at n = 1. Example 11 P∞ 1 n=1 n(n+1) converges. Proof : We may do show this directly by consider the partial sums: sm Thus, sm = 1 − 1 m+1 m X m X 1 1 1 = = − n(n + 1) n n + 1 n=1 n=1     1 1 1 1 1 = 1 + + ··· + − + ... + 2 m 2 m m+1 1 . =1− m+1 → 1. Hence, the partial sums converge and thus the series converges. Theorem 12 If |r| < 1 then P∞ n=0 rn converges and ∞ X rn = n=0 Proof : We have ∀m ∈ N, sm = m X 1 . 1−r rn = n=0 m+1 by induction. Since |r| < 1, limm→∞ |r| = 0. Therefore, lim sm = m→∞ Remark 13. Series of the form P∞ n=0 1 − rm+1 1−r 1 1−0 = . 1−r 1−r α(r)n for α ∈ R and r ∈ R are called geometric series. Theorem 14 Let {xn } be a sequence and let M ∈ N. Then, P∞ n=1 xn converges if and only if Proof : The partial sums satisfy, for all m ∈ N, m X n=1 xn = m X xn + n=M Definition 15 P xn is Cauchy if the sequence of partial sums is Cauchy. Theorem 16 P P xn is Cauchy ⇐⇒ xn is convergent. 3 M X n=1 xn . P∞ n=M xn converges.  Proof : This follows by the analogous theorem for regular sequences of real numbers proven earlier. Theorem 17 P xn is Cauchy if and only if ∀ > 0, ∃M ∈ N such that for all m ≥ M and ` > m, ` X xn < . n=m+1 Proof : ( =⇒ ) Suppose P xn is Cauchy. Let  > 0. Then, ∃M0 ∈ N such that ∀m, ` ≥ M0 , |sm − s` | < . Choose M = M0 . Then, if m ≥ M and ` > m, then ` X xn = |s` − sm | < . n=m+1 The other direction is left as an exercise. Theorem 18 P If xn converges then limn→∞ xn = 0. Proof : Suppose all ` > m ≥ M0 , P xn converges. Then, P xn is Cauchy. Let  > 0. Since ` X P xn is Cauchy, ∃M0 ∈ N such taht for xn < . n=m+1 Choose M = M0 + 1. Then, if m ≥ M =⇒ m − 1 ≥ M0 . Therefore, |xm | = m X xn <  n=m by taking ` = m. Theorem 19 If |r| ≥ 1, then P∞ n=0 rn diverges. P∞ Proof : If |r| ≥ 1, then limm→∞ rm 6= 0. Therefore, n=0 rn diverges, as if this wasn’t the case then limm→∞ rm = 0 by the previous theorem which is a contradiction. Corollary 20 P∞ The series n=0 α(r)n converges if and only if |r| < 1. 4