11.3-11.4 Integral and Comparison Tests The Integral Test: Suppose a function f(x) is continuous, positive, and decreasing on [1; 1). Let an 1 P R 1 be defined by an = f(n). Then, the series an and the improper integral 1 f(x) dx either BOTH n=1 CONVERGE OR BOTH DIVERGE. Notes: • For the integral test, when we say that f must be decreasing, it is actually enough that f is EVENTUALLY ALWAYS DECREASING. In other words, as long as f is always decreasing after a certain point, the \decreasing" requirement is satisfied. • If the improper integral converges to a value A, this does NOT mean the sum of the series is A. Why? The integral of a function will give us all the area under a continuous curve, while the series is a sum of distinct, separate terms. • The index and interval do not always need to start with 1. Examples: Determine whether the following series converge or diverge. 1 n2 • X n2 + 9 n=1 1 2 • X n2 + 9 n=3 1 1 n • X n2 + 1 n=1 1 ln n • X n n=2 Z 1 1 p-series: We saw in Section 8.9 that the integral p dx converges if p > 1 and diverges if p ≤ 1. So, by 1 x 1 1 the Integral Test, the p-series X converges if p > 1 and diverges if p ≤ 1. np n=1 Notes: 1 1 • When p = 1, the series X is called the harmonic series. n n=1 • Any constant multiple of a convergent p-series is also convergent. Example: 1 2017 • X p n n n=1 2 1 1 P P Comparison Test: Suppose an and bn are two series with positive terms. n=1 n=1 1 1 P P • If an ≤ bn for all n and if bn converges, then an also converges. Just like with improper integrals, n=1 n=1 if the larger converges, so does the smaller. 1 1 P P • If an ≥ bn for all n and if bn diverges, then an also diverges. Just like with improper integrals, if n=1 n=1 the smaller diverges, so does the larger. Examples: Determine whether the following series converge or diverge. 1 n2 • X p 9 3 n=1 n + n + 10 1 6n + 1 • X n2 − n − 3 n=3 1 n3 − 2 • X 8n4 + 6n n=5 1 1 • X 7n + n n=2 3 1 5n • X 6n − 3 n=1 1 sin2 n + arctan n + e−n • X p 9=5 3 n=1 n + n 1 cos n + 4 + e−n • X p n n=1 4 If a correct inequality cannot be established for the Comparison Test, the following test may be more helpful. 1 1 P P Limit Comparison Test: Suppose an and bn are two series with positive terms. n=1 n=1 a If lim n = L where L > 0 (and is finite), then either BOTH SERIES DIVERGE OR BOTH n!1 bn SERIES CONVERGE. If the limit is 0 or 1, no conclusion can be made. 1 n3 − 2 • X 8n4 + 6n n=5 p 1 n + 3 • X n2 + n − 1 n=1 **The Limit Comparison Test tells us that for rational/algebraic fractions, we can just look at the ratio of the highest powers from the numerator and denominator to determine whether the series converges or diverges.** It is often easier to use than the Comparison Test, especially if you don't want to think about inequalities. 5 1 arctan n • X p n n=1 1 5n • X 6n − 3 n=1 1 1 • X sin n n=1 1 1 • X sin n2 n=1 6 1 P Remainder Estimate for the Integral Test: Suppose we know that the series an converges by using the n=1 Integral Test where f(n) = an. This means we know that the series has a finite sum S. If we use the nth partial sum, sn, to approximate the actual sum S, then the remainder of this estimate is Rn = S − sn, and it satisfies the following inequalities: Z 1 Z 1 f(x) dx ≤ Rn ≤ f(x) dx n+1 n So, if we find that A ≤ Rn ≤ B, then we can say that the actual sum of the series S satisfies sn + A ≤ S ≤ sn + B 1 5 Example: Consider the series X , which is a convergent p-series. n5 n=1 • Find s10 and the upper and lower bounds on the remainder. • Find upper and lower bounds on the actual sum S. • How many terms would be needed to guarantee accuracy to within 0.00001? 7 1 1 Example: Consider the series X . It can be shown this series converges by the Integral Test. n(1 + ln n)2 n=1 • Estimate the error in using s6 to approximate the sum of the series. • How many terms are needed to guarantee the approximation is accurate to within 0.001? 1 sin n + 2 Example: Consider the series X . n3 n=1 • Show the series converges. • Estimate the error in using s5 to approximate the sum of the series. 8 11.5-11.6 Other Convergence Tests In the previous section, the convergence tests we used were for series with positive terms. However, some series do not have all positive terms. An alternating series is a series whose terms alternate positive and negative. Alternating Series Test: 1 P n Suppose you have an alternating series (−1) bn where the bn's are just the positive parts (the absolute n=1 values) of each term. This alternating series converges if: (1) lim bn = 0 n!1 (2) bn+1 ≤ bn for all n. In other words, the absolute values of the terms are decreasing. Notes: • Criteria (1) above is basically just the Test for Divergence. If lim bn is NOT 0, then the limit of the n!1 alternating terms themselves DNE and the series diverges by the Test for Divergence. • As with the Integral Test, criteria (2) just has to be always true eventually. As long as bn+1 ≤ bn is always true after some finite point, we're good to go. Examples: Determine whether the following series converge or diverge. 1 n X (−1) n n=1 1 n−1 2 X (−1) 5n 7n2 + n n=1 9 1 n+1 2 X (−1) n n3 + 4 n=1 1 P n Alternating Series Estimation Theorem: If an alternating series (−1) bn converges, then the error (re- n=1 mainder) in using the nth partial sum sn to approximate the sum of the series satisfies the inequality jRnj ≤ bn+1 Example: 1 (−1)n Consider the convergent alternating series X p . n=1 n + 3 • Estimate the error in using s10 to approximate the sum of the series. • How many terms are needed to approximate the sum to within 0.01? 10 1 (−1)n+1 Example: Approximate the sum of the series X to within 0.00005. 3nn! n=1 P P Absolute Convergence: A series an is said to be absolutely convergent if the series janj is convergent. In other words, if the series of just the absolute values of the terms converges, the series is absolutely convergent. Notes: • It is possible for a series to be convergent, but not absolutely convergent. This is called conditional convergence. • **If a series is absolutely convergent, then it is convergent.** Examples: Determine whether the following series are absolutely convergent, convergent, or divergent. 1 X 1 n3 n=2 1 n X (−1) n n=2 11 1 n X (−1) n n3 + 9 n=3 1 n−1 X (−1) n ln n n=2 1 X cos n 3=2 n=1 n 12 The following is a VERY VERY USEFUL test for absolute convergence, and thus convergence. 1 P an+1 The Ratio Test: For a series an, let lim = L. n=1 n!1 an (i) If L < 1, the series is absolutely convergent, and thus convergent. (ii) If L > 1 or if the limit is infinite, the series is divergent. (iii) If L = 1, the Ratio Test is inconclusive. You cannot say whether the series converges or diverges. Use another test. Notes: • Use the Ratio Test when the series involves factorials and/or exponentials. • The Ratio Test will NOT work if the series involves only algebraic expressions involving powers of n. • The Ratio Test can be used on positive series and/or alternating series. Examples: Determine if the following seres are absolutely convergent, convergent, or divergent. 1 n X (−5) n7n+1 n=1 1 X n! 62n+1 n=1 13 1 n 2 n−1 X (−1) n 8 (3n + 1)! n=1 1 n X (−1) (n + 2) (n + 1)3 n=1 1 X 5 · 9 · 13 ··· (4n + 1) (−1)n+1n4(n + 2)! n=1 14.