Chapter Infinite Series Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 8.2, Slide Section 8.2 Convergence Tests Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 8.2, Slide In this section we derive several tests that are useful in determining convergence Theorem 8.2.1 (Comparison Test) Let Σ an and Σ bn be infinite series of nonnegative terms That is, an ≥ and bn ≥ for all n Then (a) If Σ an converges and ≤ bn ≤ an for all n, then Σ bn converges (b) If Σ an = + ∞ and ≤ an ≤ bn for all n, then Σ bn = + ∞ Proof: Since bn ≥ for all n, the sequence (tn) of partial sums of Σ bn is an increasing sequence In part (a) this sequence is bounded above by the sum of the series Σ an , so (tn) converges by the monotone convergence theorem (4.3.3) Thus Σ bn converges In part (b) the sequence (tn) must be unbounded, for otherwise Σ an would have to converge But then lim tn = + ∞ by Theorem 4.3.8, so that Σ bn = + ∞ ♦ Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 8.2, Slide Example 8.2.2 Consider the series ∑ (n + 1)2 In Example 8.1.1 we saw that the series converges, ∑ (n +must1)also converge so For all n ∈ we have < 1 < n(n + 1) (n + 1) ∑ n(n + 1) Definition 8.2.4 If Σ | an | converges, then the series Σ an is said to converge absolutely (or to be absolutely convergent) If Σ an converges but Σ | an | diverges, then Σ an is said to converge conditionally (or be conditionally convergent) Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 8.2, Slide Theorem 8.2.5 If a series converges absolutely, then it converges Proof: Suppose that Σ an is absolutely convergent, so that Σ | an | converges By the Cauchy criterion for series (Theorem 8.1.6), given any ε > 0, there exists a natural number N such that n ≥ m ≥ N implies that | | am | + … + | a n | | < ε | am + L + an | ≤ | am | + L + | an | < ε But then by the triangle inequality, so Σ an also converges ♦ When the terms of a series are nonnegative, convergence and absolute convergence are really the same thing Thus the comparison test can be viewed as a test for absolute convergence If we have a series Σ bn with some negative terms, the corresponding series Σ| bn | will have only nonnegative terms, and we can use the comparison test on it Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 8.2, Slide If it happens that ≤ | bn | ≤ an for all n and if Σ an converges, then we can conclude that Σ| bn | converges That is, Σ bn converges absolutely In fact, changing the first few terms in a series will affect the value of the sum of the series, but it will not change whether or not the series is convergent So, given a nonnegative convergent series Σ an and a second series Σ bn, to conclude that Σ | bn | is convergent it suffices to show that ≤ | bn | ≤ an for all n greater than some N By using the comparison test and the geometric series, we can derive two more useful tests for convergence Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 8.2, Slide Theorem 8.2.7 (Ratio Test) Let Σ an be a series of nonzero terms (a) If lim sup | an + 1/an | < 1, then the series converges absolutely (b) If lim inf | an + 1/an | > 1, then the series diverges (c) Otherwise, lim inf | an + 1/an | ≤ ≤ lim sup | an + 1/an |, and the test gives no information about convergence or divergence Proof: If L < 1, then choose r so that L < r < (a) Let lim sup | an + 1/an | = L By Theorem 4.4.11(a) there exists N ∈ such that n ≥ N implies that | an + 1/an | ≤ r (If there were infinitely many terms | an + 1/an | greater than r, then a subsequence would converge to something greater than or equal to r, a contradiction to r being greater than the lim sup | an + 1/an |.) So, | aN + | ≤ r |aN |, and | aN + 2 | ≤ r |aN + | ≤ r |aN |, etc That is, n ≥ N implies | an + 1 | ≤ r |an | k It follows easily by induction that | aN + k | ≤ r |aN | for all k ∈ k Since < r < 1, the geometric series Σ r is convergent k Thus Σ | aN | r is also convergent, and Σ an is absolutely convergent by the comparison test Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 8.2, Slide Theorem 8.2.7 (Ratio Test) Let Σ an be a series of nonzero terms (a) If lim sup | an + 1/an | < 1, then the series converges absolutely (b) If lim inf | an + 1/an | > 1, then the series diverges (c) Otherwise, lim inf | an + 1/an | ≤ ≤ lim sup | an + 1/an |, and the test gives no information about convergence or divergence Proof: (b) If lim inf | an + 1/an | > 1, then it follows that | an + 1| > | an | for all n sufficiently large Thus the sequence (an) cannot converge to zero, and by Theorem 8.1.5 the series Σ an must diverge (c) This follows from the observation that the series Σ 1/n converges and the harmonic series Σ 1/n diverges In both series we have lim | an + 1/an | = ♦ Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 8.2, Slide Theorem 8.2.8 (Root Test) Given a series Σ an, let α = lim sup | an | 1/n (a) If α < 1, then the series converges absolutely (b) If α > 1, then the series diverges (c) Otherwise, α = 1, and the test gives no information about convergence or divergence Proof: (a) If α < 1, choose r so that α < r < By Theorem 4.4.11(a) we have | an | 1/n ≤ r for all n greater than some N n That is, | an | ≤ r for all n > N n Since < r < 1, the geometric series Σ r is convergent, so Σ an is absolutely convergent by the comparison test (b) If α > 1, then | an | 1/n ≥ for infinitely many indices n That is, | an | ≥ for infinitely many terms Thus the sequence (an) cannot converge to zero and, by Theorem 8.1.5, the series Σ an must diverge Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 8.2, Slide Theorem 8.2.8 (Root Test) Given a series Σ an, let α = lim sup | an | 1/n (a) If α < 1, then the series converges absolutely (b) If α > 1, then the series diverges (c) Otherwise, α = 1, and the test gives no information about convergence or divergence Proof: (c) follows from considering the convergent series Σ 1/n and the divergent series Σ 1/n 1/n In Example 4.1.11 we showed that lim n lim n Thus, Similarly, lim |1/n| 1/n = 1/ n = 1 = = = 1/ n 1/ n 2 lim (n ) lim ( n ) = 1, so the root test yields α = for both series Thus when α = we can draw no conclusion about the convergence or divergence of a given series ♦ Note: This is very useful in applying the root test to other sequences Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 8.2, Slide 10 (Integral Test) Theorem 8.2.13 Let f be a continuous function defined on [0, ∞), and suppose that f is positive and decreasing That is, if x1 Then the series Σ f (n) converges iff lim (∫ n exists as a real number n →∞ Proof: f ( x) dx ) n +1 ∫ n f ( x) dx Let an = f (n) and bn = Since f is decreasing, given any n ∈ we have f (n + 1) ≤ f (x) ≤ f (n) for all x ∈ [n, n + 1] f (n + 1) ≤ Since the length of each subinterval is 1, it follows that n +1 ∫n f ( x) dx ≤ f ( n) Geometrically, this means that the area under the curve y = f (x) from x = n to x = n + is between f (n + 1) and f (n) Thus < a n +1 ≤ b ≤ a for each n n n By the comparison test applied twice, Σ a converges iff Σ b converges n n y = f (x) But the partial sums of Σ b are the integrals n so Σ b converges precisely when n Area is Area is f (n) f (n + 1) n n ∫1 , lim n →∞ (∫ n f ( x) dx f ( x) dx exists as a real number ♦ n +1 Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 8.2, Slide 11 ) Example 8.2.14 (The p-series) p Any series of the form Σ 1/n , where p ∈ , is called a p-series It is easy to see that the ratio and root tests both fail to determine convergence For p ≠ 1, we have But, we can use the integral test n ∫1 dx = xp n −p x dx ∫ − p +1 n ( ) x = n1− p − = 1− p − p + 1 The limit of this as n → ∞ will be finite if p > and infinite if p < p Thus by the integral test Σ 1/n converges if p > and diverges if p < When p = 1, we get the harmonic series, which is divergent p So, the p-series Σ 1/n converges if p > and diverges if p ≤ This can be useful when applying the comparison test Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 8.2, Slide 12 If the terms in a series alternate between positive and negative values, the series is called an alternating series Our final test gives us a simple criterion for determining the convergence of an alternating series Theorem 8.2.16 (Alternating Series Test) n +1 If (a ) is a decreasing sequence of positive numbers and lim a = 0, then the series Σ (−1) a converges n n n Proof: The idea of the proof is to show that the sequence (sn) of partial sums converges by considering two subsequences: the subsequence (s2n) coming from the sum of an even number of terms, and the subsequence (s2n+1) coming from the sum of an odd number of terms In the text there are several examples of applying the various convergence tests to specific series Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 8.2, Slide 13 ... greater than or equal to r, a contradiction to r being greater than the lim sup | an + 1 /an |.) So, | aN + | ≤ r |aN |, and | aN + 2 | ≤ r |aN + | ≤ r |aN |, etc That is, n ≥ N implies | an + 1... Σ an and Σ bn be infinite series of nonnegative terms That is, an ≥ and bn ≥ for all n Then (a) If Σ an converges and ≤ bn ≤ an for all n, then Σ bn converges (b) If Σ an = + ∞ and ≤ an. .. Proof: (b) If lim inf | an + 1 /an | > 1, then it follows that | an + 1| > | an | for all n sufficiently large Thus the sequence (an) cannot converge to zero, and by Theorem