C.7. Numerical series. Pag. 147 Proof of the converging criteria for series. Theorem 5.29 (Comparison test) Let a k and b k be positive-term series

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1 C.7 Numerical series Pag. 147 Proof of the converging criteria for series Theorem 5.29 (Comparison test) Let and be positive-term series such that 0, for any k 0. i) If the series converges, then also the series converges and ii) If. diverges, then diverges as well. Proof. i) Denote by {s n } and {t n } the sequences of partial sums of respectively. Since for all k, s n t n, n 0., By assumption, the series converges, so lim t n = t R. Proposition 5.28 implies the limit lim s n = s exists, finite or infinite. By the First comparison theorem (Theorem 4., p. 137) we have Therefore s R, and the series s = lim s n lim t n = t R. converges. Furthermore s t.

2 2 C.7 Numerical series ii) If the series converged, part i) of this proof would force to converge, too. Theorem 5.31 (Asymptotic comparison test) Let positive-term series and suppose the sequences { } k 0 and { } k 0 have the same order of magnitude for k. Then the series have the same behaviour. and Proof. Having the same order of magnitude for k is equivalent to Therefore the sequences { ak lim = l R \ {0}. k } { } bk and are both convergent, hence both k 0 bounded (Theorem 2., p. 137). There must exist constants M 1, M 2 > 0 such that M 1 and M 2 for any k > 0, i.e., k 0 M 1 and M 2. be Now it suffices to use Theorem 5.29 to finish the proof. Theorem 5.33 (Ratio test) Let have > 0, k 0. Assume the limit +1 lim = l k exists, finite or infinite. If l < 1 the series converges; if l > 1 it diverges. Proof. First take l finite. By definition of limit we know that for any ε > 0, there is an integer k ε 0 such that k > k ε +1 l < ε i.e., l ε < +1 < l + ε. Assume l < 1. Choose ε = 1 l 2 and set q = 1+l 2, so Repeating the argument we obtain 0 < +1 < l + ε = q, k > k ε. +1 < q < q 2 1 <... < q k kε ε+1

3 hence +1 < ε+1 q kε q k, k > k ε. C.7 Numerical series 3 The claim follows by Theorem 5.29 and from the fact that the geometric series, with q < 1, converges (Example 5.27). Now consider l > 1. Choose ε = l 1, and notice 1 = l ε < +1, k > k ε. Thus +1 > >... > ε+1 > 0, so the necessary condition for convergence fails, for lim 0. k Eventually, if l = +, we put A = 1 in the condition of limit, and there exists k A 0 with > 1, for any k > k A. Once again the necessary condition to have convergence does not hold. Theorem 5.34 (Root test) Given a series suppose lim k ak = l k with non-negative terms, exists, finite or infinite. If l < 1 the series converges, if l > 1 it diverges. Proof. Since this proof is essentially identical to the previous one, we leave it to the reader. Theorem 5.36 (Leibniz s alternating series test) An alternating series ( 1) k converges if the following conditions hold i) lim k = 0 ; ii) the sequence { } k 0 decreases monotonically. Denoting by s its sum, for all n 0 r n = s s n b n+1 and s 2n+1 s s 2n. Proof. As { } k 0 is a decreasing sequence, and s 2n = s 2n 2 b 2n 1 + b 2n = s 2n 2 (b 2n 1 b 2n ) s 2n 2 s 2n+1 = s 2n 1 + b 2n b 2n+1 s 2n 1. Thus the subsequence of partial sums made by the terms with even index decreases, whereas the subsequence of terms with odd index increases. For any n 0, moreover,

4 4 C.7 Numerical series s 2n = s 2n 1 +b 2n s 2n 1... s 1 and s 2n+1 = s 2n b 2n+1 s 2n... s 0. Thus {s 2n } n 0 is bounded from below and {s 2n+1 } n 0 from above. By Theorem 3.9 both sequences converge, so let us put lim s 2n = inf s 2n = s and lim s 2n+1 = sup s 2n+1 = s. n 0 Since s ( ) s = lim s2n s 2n+1 = lim b 2n+1 = 0, we conclude that the series ( 1) k has sum s = s = s. In addition, s 2n+1 s s 2n, n 0, in other words the sequence {s 2n } n 0 approximates s from above, while {s 2n+1 } n 0 approximates s from below. For any n 0 we have 0 s s 2n+1 s 2n+2 s 2n+1 = b 2n+2 and 0 s 2n s s 2n s 2n+1 = b 2n+1, n 0 i.e., r n = s s n b n+1. Theorem 5.40 (Absolute convergence test) If then it also converges and a k. converges absolutely Proof. This proof is analogous to the one of Theorem 10.7 (absolute convergence test for improper integrals). Let us introduce the sequences a + k = { ak if 0 0 if < 0 and a k = { 0 if ak 0 if < 0. Notice a + k, a k 0 for any k 0, and = a + k a k, = a + k + a k. Since 0 a + k, a k, for any k 0, the Comparison test (Theorem 5.29) says that the series and converge. Observing that a + k a k ( = a + k ) a k = a + k a k,

5 C.7 Numerical series 5 for any n 0, we deduce that also the series = a + k Finally, passing to the limit n the relation n n a k yields the desired inequality. a k converges.