121616949-math.264 - sequence because s 2 k 3 = s 2 k 1-a 2...

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250 Chapter 10 Sequences and Series 1 4 1 = s 1 = a 1 a 2 = - 1 2 s 2 = 1 2 a 3 s 3 a 4 s 4 a 5 s 5 a 6 s 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .... ................................................................................................... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .... ............................................................ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .... Figure 10.4 The alternating harmonic series. since the “jumps” (that is, the a i terms) are getting smaller and smaller, the two sequences must converge to the same value, meaning the entire sequence of partial sums s 1 , s 2 , s 3 , . . . converges as well. There’s nothing special about the alternating harmonic series—the same argument works for any alternating sequence with decreasing size terms. The alternating series test is worth calling a theorem. THEOREM 10.29 Suppose that { a n } n =0 is a non-increasing sequence of positive num- bers and lim n →∞ a n = 0. Then the alternating series X n =0 ( - 1) n a n converges. Proof. The odd numbered partial sums, s 1 , s 3 , s 5 , and so on, form a non-increasing
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Unformatted text preview: sequence, because s 2 k +3 = s 2 k +1-a 2 k +2 + a 2 k +3 ≤ s 2 k +1 , since a 2 k +2 ≥ a 2 k +3 . This sequence is bounded below by s 2 , so it must converge, say lim k →∞ s 2 k +1 = L . Likewise, the partial sums s 2 , s 4 , s 6 , and so on, form a non-decreasing sequence that is bounded above by s 1 , so this sequence also converges, say lim k →∞ s 2 k = M . Since lim n →∞ a n = 0 and s 2 k +1 = s 2 k + a 2 k +1 , L = lim k →∞ s 2 k +1 = lim k →∞ ( s 2 k + a 2 k +1 ) = lim k →∞ s 2 k + lim k →∞ a 2 k +1 = M + 0 = M, so L = M , the two sequences of partial sums converge to the same limit, and this means the entire sequence of partial sums also converges to L . Another useful fact is implicit in this discussion. Suppose that L = ∞ X n =0 (-1) n a n...
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