One Variable Advanced Calculus

5.3. MORE TESTS FOR CONVERGENCE 851From Theorem 5.3.4, I need to find n such thatn 2 +1 ≤ 110and then n − 1 is thedesired value. Thus n = 3 and so∞∑(−1) n 12∑∣ n 2 + 1 − (−1) n 1n 2 + 1∣ ≤ 1 10n=1n=1Definition 5.3.6 A series ∑ a n is said to converge absolutely if ∑ |a n | converges.It is said to converge conditionally if ∑ |a n | fails to converge but ∑ a n converges.Thus the alternating series or more general Dirichlet test can determine convergenceof series which converge conditionally.5.3.2 Ratio And Root TestsA favorite test for convergence is the ratio test. This is discussed next. It is at theother extreme from the alternating series test, being completely oblivious to any sort ofcancellation. It only gives absolute convergence or spectacular divergence.Theorem 5.3.7 Suppose |a n | > 0 for all n and suppose|a n+1 |lim = r.n→∞ |a n |Then⎧∞∑ ⎨a n⎩n=1diverges if r > 1converges absolutely if r < 1test fails if r = 1.Proof: Suppose r < 1. Then there exists n 1 such that if n ≥ n 1 , then∣ 0 n, then |a m | < R m−n1 |a n1 | . By the comparison test and the theorem ongeometric series, ∑ |a n | converges. This proves the convergence part of the theorem.To verify the divergence part, note that if r > 1, then 5.7 can be turned around forsome R > 1. Showing lim n→∞ |a n | = ∞. Since the n th term fails to converge to 0, itfollows the series diverges.To see the test fails if r = 1, consider ∑ n −1 and ∑ n −2 . The first series divergeswhile the second one converges but in both cases, r = 1. (Be sure to check this lastclaim.) The ratio test is very useful for many different examples but it is somewhat unsatisfactorymathematically. One reason for this is the assumption that a n > 0, necessitatedby the need to divide by a n , and the other reason is the possibility that the limit mightnot exist. The next test, called the root test removes both of these objections. Beforepresenting this test, it is necessary to first prove the existence of the p th root of anypositive number. This was shown earlier in Theorem 2.10.2 but the following lemmagives an easier treatment of this issue based on theorems about sequences.