Lecture 12: The Ratio, Root, and Alternating Series Tests

18.100A: Complete Lecture Notes Lecture 12: The Ratio, Root, and Alternating Series Tests We continue our study of convergence tests. Theorem 1 (Ratio test) Suppose xn 6= 0 for all n and L = lim n→∞ |xn+1 | |xn | exists. Then, 1. if L < 1 then P xn converges absolutely. 2. if L > 1 then P xn diverges. Proof : We will first prove the second part of this theorem. 2) Suppose L > 1 and α ∈ (1, L). Then, there exists M0 ∈ N such that for all N ≥ M0 , for all n ≥ M0 , |xn+1 | ≥ |xn | =⇒ lim |xn | 6= 0. n→∞ P Therefore, xn diverges. |xn+1 | |xn | ≥ α ≥ 1. Thus, 1) Now suppose that L < 1. Let α ∈ (L, 1). Then, there exists M0 ∈ N such that ∀n ≥ M0 , Therefore,∀n ≥ M0 , |xn+1 | ≤ α|xn |. In other words, for all n ≥ M0 , |xn+1 | |xn | < α. |xn | ≤ α|xn−1 | ≤ α2 |xn−2 | ≤ · · · ≤ αn−M0 |xM0 |. Let m ∈ N. Then, m X |xn | = n=1 M 0 −1 X |xn | + n=1 ≤ M 0 −1 X m X ≤ = αn−M0 n=M0 |xn | + |xM0 | n=1 M 0 −1 X m X |xn | + |xM0 | n=1 M 0 −1 X |xn | n=M0 ∞ X α` `=0 |xn | + n=1 |xM0 | . 1−α Pm P ∞ Therefore, { n=1 |xn |}m=1 is bounded, and thus |xn | converges. Hence, xn is absolutely convergent. Let’s consider two examples where we can use the Ratio test. 1 Example 2 Show the series (−1)n n=1 n2 +1 P∞ converges absolutely. Proof : Notice 1 (−1)n 1 ≤ 2 < 2, n2 + 1 n +1 n and hence lim n→∞ (−1)n+1 (n+1)2 +1 (−1)n n2 +1 n2 = 1. n→∞ (n + 1)2 < lim  Example 3 Show that ∀x ∈ R, xn n=0 n! P∞ converges absolutely. Proof : This immediately follows from the Ratio test, noting that n! |x| |x|n+1 · n = lim = 0. n→∞ n + 1 n→∞ (n + 1)! |x| lim  Remark 4. As seen above, the Ratio test can be really helpful to use when we have a (−1)n or a factorial in the argument. Also note that if L = 1 then we the test doesn’t apply. Theorem 5 (Root test) P Let xn be a series and suppose that L = lim |xn |1/n n→∞ exists. Then, 1. if L < 1 then P xn converges absolutely. 2. if L > 1 then P xn diverges. Proof : 1. Suppose L < 1. Let L < r < 1. Then, since |xn |1/n → L, ∃M ∈ N such that ∀n ≥ M , |xn |1/n < r. Therefore, for all n ≥ M , |xn | ≤ rn . Thus, for all m ∈ N, m X |xn | = n=1 M −1 X |xn | + n=1 ≤ M −1 X ≤ |xn | + = n=1 m X |xn | + ∞ X rn n=M |xn | + rM . 1−r Pm P ∞ Thus, { n=1 |xn |}m=1 is bounded, and thus |xn | converges. 2 rn n=M n=1 M −1 X |xn | n=M n=1 M −1 X m X 2. Suppose L > 1. Then, since |xn |1/n → L > 1, there exists an M ∈ N such that for all n ≥ M , |xn |1/n > 1. P In other words, for all n ≥ M , |xn | > 1. Therefore, limn→∞ xn 6= 0, and thus xn diverges. Remark 6. Again, note that if L = 1 then the test doesn’t apply. Theorem 7 (Alternating Series test) Let {xn } be a monotone decreasing sequence such that xn → 0. Then, Proof : Let sm = Pm n n=1 (−1) xn . P (−1)n xn converges. Then, s2k = 2k X (−1)n xn n=1 = (x2 − x1 ) + (x4 − x3 ) + · · · + (x2k − x2k−1 ) ≥ (x2 − x1 ) + · · · + (x2k − x2k−1 ) + (x2k+2 − x2k+1 ) = s2(k+1) as {xn } is a monotone decreasing sequence. Thus, {s2k }∞ k=1 is monotone decreasing. Furthermore, s2k = −x1 + (x2 − x3 ) + (x4 − x5 ) + · · · + (x2k−2 − x2k−1 ) + x2k ≥ −x1 . In other words, {s2k } is a bounded below monotone decreasing sequence. Thus, {s2k }∞ k=1 converges. Let s = limk→∞ s2k . We now prove {sm }∞ converges to s. m=1 Let  > 0. Since s2k → s, ∃M0 ∈ N such that for all k ≥ M0 , |s2k − s| <  . 2 Since xn → 0, ∃M1 ∈ N such that ∀n ≥ M1 ,  . 2 Choose M = max{2M+ 0 + 1, M1 }. Suppose m ≥ M . If m is even, then |xn | < |sm − s| = |s2· m2 − s| < If m is odd, let k = m−1 2 m 2 ≥ M0 + 1/2 ≥ M0 . Therefore,  < . 2 so m = 2k + 1. Then, m ≥ M =⇒ k ≥ M0 and m ≥ M1 . Then, |sm − s| = |sm−1 + xm − s| ≤ |s2k − s + xm | ≤ |s2k − s| + |xm | < Thus, sm → s, and thus   + = . 2 2 P (−1)n xn converges. Corollary 8 We already showed that P (−1)n n does not absolutely converge. However, Proof : This follows immediately from the Alternating Series test. 3 P (−1)n n converges. Theorem 9 P P P Suppose xn converges absolutely and xn = x. Let σ : N → N be a bijective function. Then, xσ(n) P is absolutely convergent and xσ(n) = x. In other words, absolute convergence implies if we rearrange the sequence the new series will still converge to the same value of the original series. P Pm Proof : We first show |xσ(n) | converges, which is equivalent to showing the partial sums n=1 |xσ(n) | is bounded. P Since xn converges, ∃B ≥ 0 such that for all ` ∈ N, ` X |xn | ≤ B. n=1 Let m ∈ N. Then, σ({1, . . . , m}) is a finite subset of N. Thus, there exists an ` ∈ N such that σ({1, . . . , m}) ⊂ {1, . . . , `}. Thus, m X X |xσ(n) | = n=1 Therefore, P |xn | ≤ P∞ n=1 xn , and let  > 0. Then, ∃M0 ∈ N such that ∀m ≥ M0 , m X xn − x < n=1 Since P |xn | ≤ B. n=1 n∈σ({1,...,m}) |xσ(n) | converges. Let x = ` X  . 2 |xn | converges, ∃M1 ∈ N such that for all ` > m ≥ M1 , ` X |xn | < n=m+1  . 2 Let M2 = max{M0 , M1 }. Then, ∀` > m ≥ M2 , m X n=1 xn − x <  and 2 ` X |xn | < n=m+1 Since σ −1 ({1, . . . , M2 }) is a finite set, ∃M3 ∈ N such that {1, . . . , M2 } ⊂ σ({1, . . . , M3 }). 4  . 2 Choose M = M3 . Thus, if m0 ≥ M , 0 m X X xσ(n0 ) − x = n0 =1 xn − x n∈σ({1,...,m0 }) = M X ≤ max σ({1,...,m0 }) ≤ n=M +1 xn − x + n=1 < X xn − x + n=1 M X xn n∈σ({1,...,m0 })\{1,...,M } n=1 M X X xn − x + ` X n=M +1   + = . 2 2 5 |xn | |xn |