Sequences and Series

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270 Chapter 11 Sequences and SeriesWe can compute the integral:∫ ∞N1x 2 dx = 1 N ,so N = 100 is a good starting point. Adding up the first 100 terms gives approximately1.634983900, and that plus 1/100 is 1.644983900, so approximating the series by the valuehalfwaybetweenthesewillbeatmost1/200 = 0.005inerror. Themidpointis1.639983900,but while this is correct to ±0.005, we can’t tell if the correct two-decimal approximationis 1.63 or 1.64. We need to make N big enough to reduce the guaranteed error, perhaps toaround 0.004 to be safe, so we would need 1/N ≈ 0.008, or N = 125. Now the sum of thefirst 125 terms is approximately 1.636965982, and that plus 0.008 is 1.644965982 and thepointhalfwaybetweenthemis1.640965982. Thetruevalueisthen1.640965982±0.004,andall numbers in this range round to 1.64, so 1.64 is correct to two decimal places. We havementioned that the true value of this series can be shown to be π 2 /6 ≈ 1.644934068 whichrounds down to 1.64 (just barely) and is indeed below the upper bound of 1.644965982,again just barely. Frequently approximations will be even better than the “guaranteed”accuracy, but not always, as this example demonstrates.Exercises 11.3.Determine whether each series converges or diverges.∞∑ 11.n ⇒2. ∑ ∞nπ/4 n 2 +1 ⇒3.5.7.n=1∞∑n=1∞∑n=1∞∑n=2lnnn 2 ⇒1e n ⇒1nlnn ⇒9. Find an N so that10. Find an N so that11. Find an N so that12. Find an N so that∞∑n=1∞∑n=0∞∑n=1∞∑n=21n is between ∑ N 4n=11e is between ∑ N nn=0lnnn 2 is between N∑n=11n(lnn) is between ∑ N 2n=14.∞∑n=16.∞∑n=18.∞∑n=21n and ∑ N 4n=11e and ∑ N nn=01n 2 +1 ⇒ne n ⇒lnnn 2 and N∑n=21n(lnn) 2 ⇒1n 4 +0.005. ⇒1e n +10−4 . ⇒n=11n(lnn) and ∑ N 2lnnn 2 +0.005. ⇒n=21n(lnn) 2 +0.005. ⇒
½½º ÐØÖÒØÒËÖ×11.4 Alternating Series 271Next we consider series with both positive and negative terms, but in a regular pattern:they alternate, as in the alternating harmonic series for example:∞∑ (−1) n−1n=1n= 1 1 + −12 + 1 3 + −14 +··· = 1 1 − 1 2 + 1 3 − 1 4 +···.In this series the sizes of the terms decrease, that is, |a n | forms a decreasing sequence,but this is not required in an alternating series. As with positive term series, however,when the terms do have decreasing sizes it is easier to analyze the series, much easier, infact, than positive term series. Consider pictorially what is going on in the alternatingharmonic series, shown in figure 11.4. Because the sizes of the terms a n are decreasing,the partial sums s 1 , s 3 , s 5 , and so on, form a decreasing sequence that is bounded belowby s 2 , so this sequence must converge. Likewise, the partial sums s 2 , s 4 , s 6 , and so on,form an increasing sequence that is bounded above by s 1 , so this sequence also converges.Since all the even numbered partial sums are less than all the odd numbered ones, andsince the “jumps” (that is, the a i terms) are getting smaller and smaller, the two sequencesmust converge to the same value, meaning the entire sequence of partial sums s 1 ,s 2 ,s 3 ,...converges as well.14s 2 = 1 2s 6...a 2 = − 1 2a 4a 6s 4 s 5 s 3a 5. .a 31 = s 1 = a 1Figure 11.4 The alternating harmonic series.There’s nothing special about the alternating harmonic series—the same argumentworks for any alternating sequence with decreasing size terms. The alternating series testis worth calling a theorem.THEOREM 11.29 Suppose that{a n } ∞ n=1 isanon-increasing sequence ofpositivenumbersand lim a n = 0. Then the alternating series (−1) n−1 a n converges.∞∑n→∞Proof. The odd numbered partial sums, s 1 , s 3 , s 5 , and so on, form a non-increasingsequence, because s 2k+3 = s 2k+1 − a 2k+2 + a 2k+3 ≤ s 2k+1 , since a 2k+2 ≥ a 2k+3 . Thisn=1
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Sequences and Series

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