Lecture 11: Absolute Convergence and the Comparison Test for Series

18.100A: Complete Lecture Notes Lecture 11: Absolute Convergence and the Comparison Test for Series Recall 1 Last time we showed that if P xn converges then limn→∞ xn = 0. Question 2. Is the converse true? Does limn→∞ xn = 0 =⇒ Theorem 3 P∞ The series n=1 1 n P xn converges? does not converge. Proof : We will show that there exists a subsequence of sm = series diverges. Consider, for ` ∈ N, 2` X 1 . s2` = n n=1 Pm 1 n=1 n which is unbounded, which will imply the Then, s2` = 1 + =1+         1 1 1 1 1 1 1 + + ··· + + · · · + + + + ... 2 3 4 5 8 2`−1 + 1 2` ` X λ 2 X λ=1 n=2λ−1 +1 ≥1+ ` X λ 2 X λ=1 n=2λ−1 +1 =1+ 1 n 1 2λ ` X 1 λ (2 − (2λ−1 + 1) + 1) 2λ λ=1 =1+ ` X 2λ−1 2λ λ=1 ` =1+ . 2 Thus, {s2` }∞ `=1 is unbounded which implies {s2` } does not converge. P1 Remark 4. The series n is called the harmonic series. Theorem 5 Let α ∈ R and P xn and P yn be convergent series. Then the series P (αxn + yn ) converges and X X X (αxn + yn ) = α xn + yn . 1 Proof : The partial sums satisfy m X (αxn + yn ) = α n=1 m X xn + n=1 m X yn . n=1 By linear properties of limits, it follows that lim m→∞ m X (αxn + yn ) = α X xn + X yn . n=1 Series with non-negative terms are easier to work with than general series as then {sn } is a monotone sequence. Theorem 6 If ∀n ∈ N xn ≥ 0, then P xn converges if and only if {sm } is bounded. Proof : If xn ≥ 0 for all n ∈ N then sm+1 = m+1 X xn = n=1 m X xn + xm+1 = sm + xm+1 ≥ sm n=1 Thus, {sm } is a monotone increasing sequence. Therefore, {sm } converges if and only if {sm } is bounded. Definition 7 P P xn converges absolutely if |xn | converges. Theorem 8 P P If xn converges absolutely then xn converges. P P Proof : Suppose |xn | converges. We will then show that xn is Cauchy. Pm Pm Claim: ∀m ≥ 2, | n=1 xn | ≤ n=1 |xn |. We prove this claim by induction. For m = 2, this states that P` P` |x1 + x2 | ≤ |x1 | + |x2 |, which follows by the Triangle Inequality. Suppose for all n=1 xn ≤ n=1 |xn |. Then, `+1 X n=1 xn ≤ ` X xn + |x`+1 | ≤ n=1 ` X |xn | + |x`+1 | = n=1 `+1 X |xn |. n=1 P P P We now prove that xn is Cauchy. Let  > 0. Since |xn | converges, |xn | is Cauchy. Therefore, there exists an M0 ∈ N such that for all ` > m ≥ M0 , ` X |xn | < . n=m+1 Choose M = M0 . Then, for all ` > m ≥ M , ` X n=m+1 ` X xn ≤ |xn | < . n=m+1 P xn is Cauchy, and thus converges. n P∞ Remark 9. We will see that n=1 (−1) is convergent but not absolutely convergent. n Hence, Notice that it is immediately clear that this series is not absolutely convergent as harmonic series), which doesn’t converge. 2 P (−1)n n = P 1 n (the Convergence tests Theorem 10 (Comparison Test) Suppose for all n ∈ N 0 ≤ xn ≤ yn . Then, P P 1. if yn converges, then xn converges. P P 2. if xn diverges, then yn diverges. Proof : P Pm ∞ 1. If yn converges, then { n=1 yn }m=1 is bounded. In other words, there exists a B ≥ 0 such that for all m ∈ N, m X yn ≤ B. n=1 Pm Pm Thus, for all m ∈ N, n=1 xn ≤ n=1 yn ≤ B. Therefore, the partial sums of {xn } are bounded, which P implies xn converges. P Pm ∞ 2. If xn diverges, then { n=1 xn }m=1 is unbounded. We now prove that ( m X )∞ yn n=1 m=1 is also unbounded. Let B ≥ 0. Then, ∃m ∈ N such that m X xn ≥ B. n=1 Therefore, Pm n=1 yn ≥ Pm n=1 Pm P ∞ xn ≥ B. Thus, { n=1 yn }m=1 is unbounded, which implies yn diverges. Remark 11. We will see that geometric series and the Comparison Test imply everything! Theorem 12 For p ∈ R, the series P∞ 1 n=1 np converges if and only if p > 1. P∞ Proof : ( =⇒ ) We prove this direction through contradiction. Suppose n=1 n1p converges and p ≤ 1. Then, P1 P 1 P 1 1 1 np ≥ n , and n diverges. Therefore, by the Comparison Test, np also diverges. Hence, if np converges, then p > 1. ( ⇐= ) Suppose p > 1. We first prove that a subsequence of the partial series is bounded. 3 Claim 1: ∀k ∈ N, s2k ≤ 1 + 1 . 1−2−(p−1) Proof: s2k = 1 + k X ` 2 X `=1 n=2`−1 +1 ≤1+ k X ` 2 X `=1 n=2`−1 +1 ≤1+ k X 1 np 1 + 1)p (2`−1 2−p(`−1) (2` − (2`−1 + 1) + 1) `=1 =1+ k X 2−(p−1)(`−1) `=1 =1+ k−1 X 2−(p−1)` `=0 ≤1+ ∞ X 2−(p−1)` `=0 =1+ 1 1 − 2−(p−1) using the fact that p − 1 > 0, and using properties of geometric series. Thus, Claim 1 is proven. Pm Claim 2: {sm = n=1 n1p } is bounded. Proof: Let m ∈ N. Since 2m > m, we have that m sm m 2 X X 1 1 = ≤ n−p ≤ 1 + . p −(p−1) n 1 − 2 n=1 n=1 Hence, the partial sums are bounded, which implies {sm } converges. 4