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Unformatted text preview: Proposition 3 The geometric series âˆž summationdisplay m =0 q m = 1 + q + q 2 + Â·Â·Â· (783) converges with the sum 1 / (1 âˆ’ q ) if  q  < 1 and diverges if  q  â‰¥ 1 . Proof: For  q  â‰¥ 1 we have  q m  â‰¥ 1, m â‰¥ 0, and Thm. 14 implies divergence. Let  q  < 1. The nth partial sum is given by s n = 1 + q + Â·Â·Â· + q n , n âˆˆ N , (784) so that qs n = q + Â·Â·Â· + q n + q n +1 . (785) After subtraction, we are left with s n âˆ’ qs n = (1 âˆ’ q ) s n = 1 âˆ’ q n +1 . (786) Since q negationslash = 1, we have 1 âˆ’ q negationslash = 0, and we may solve (786) for s n : s n = 1 âˆ’ q n +1 1 âˆ’ q = 1 1 âˆ’ q âˆ’ q n +1 1 âˆ’ q â†’ 1 1 âˆ’ q , n â†’ âˆž , (787) since  q  < 1. square In the following sections, we will mostly use the ratio test to establish convergence of a series. Theorem 17 If a series z 1 + z 2 + Â·Â·Â· with z n negationslash = 0 , n âˆˆ N , has the property that for some N âˆˆ N and for some q < 1 vextendsingle vextendsingle vextendsingle vextendsingle z n +1 z n vextendsingle vextendsingle vextendsingle vextendsingle â‰¤ q, âˆ€ n > N, (788) then this series converges absolutely. If vextendsingle vextendsingle vextendsingle vextendsingle z n +1 z n vextendsingle vextendsingle vextendsingle vextendsingle â‰¥ 1 , âˆ€ n > N, (789) then the series z 1 + z 2 + Â·Â·Â· diverges. 134 Remark: q = 1 is not sufficient (cf. harmonic series)! Proof: From (789), we have that  z n +1  â‰¥  z n  , n > N , so divergence of the series follows from Thm. 14. If (788) holds, then  z n +1  â‰¤ q  z n  , n > N . In particular,  z N + p  â‰¤ q  z N + p 1  â‰¤ q 2  z N + p 2  â‰¤ Â·Â·Â· â‰¤ q p 1  z N +1  , p âˆˆ N . With Prop. 3, we find that  z N +1  +  z N +2  + Â·Â·Â· â‰¤  z N +1  ( 1 + q + q 2 + Â·Â·Â· ) â‰¤  z N +1  1 1 âˆ’ q . (790) The comparison test (Thm. 16) yields absolute convergence of the series z 1 + z 2 + Â·Â·Â· . square If the sequence of ratios in Thm. 17 converges, we get the more convenient Theorem 18 If a series z 1 + z 2 + Â·Â·Â· with z n negationslash = 0 , n âˆˆ N , is such that lim n â†’âˆž vextendsingle vextendsingle vextendsingle vextendsingle z n +1 z n vextendsingle vextendsingle vextendsingle vextendsingle = L, (791) then a) If L < 1 , the series converges absolutely. b) If L > 1 , the series diverges. c) If L = 1 , the series may converge or diverge, i. e. the test fails and permits no conclusion. Proof: a) We define k n :=  z n +1 /z n  and let L = 1 âˆ’ b < 1. By definition of the limit, k n must eventually get close to 1 âˆ’ b , say, k n â‰¤ q := 1 âˆ’ b/ 2 < 1, n > N , for some N . Convergence of z 1 + z 2 + Â·Â·Â· now follows from Thm. 17. b) For L = 1 + c > 1 we have k n â‰¥ 1 + c/ 2 > 1, n > N * , for some N * sufficiently largte, which implies divergence of z 1 + z 2 + Â·Â·Â· by Thm. 17....
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 Spring '08
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 Math, Geometric Series, Power Series, Mathematical Series, lim

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