MAT 127: Calculus C, Fall 2010Course Summary IIIExtremely Important: sequences vs. series (do not mix them or their convergence/divergencetests up!!!); what it means for a sequence or series to converge or diverge; power seriesVery Important: convergence/divergence tests for series; radius and interval of convergence forpower series; differentiation, integration, and limits of functions via power series; Taylor seriesImportant: estimating infinite sums by finite sums; finding radius and interval of convergence ofpower series; determining Taylor series of functions related to standard ones; applications of powerseries to computing sums of seriesJ: Series (cont’d)J.10 The Alternating Series Test applies to a narrow, but important, set of series with terms ofdifferent signs:if limn−→∞an= 0, |an|> |an+1|, and the signs of analternate (an> 0 for every n odd andan<0 for every n even, or the other way around), then the series∞Xn=1anconvergesThe alternating-sign condition is typically exhibited by the presence of (−1)nor (−1)n−1=−(−1)n;however, make sure to also check the first two conditions before concluding that the series converges.Typical examples are the series like∞Xn=1(−1)nn,∞Xn=1(−1)n−1(ln n)2n;both converge by the AST. The Alternating Series Test is a convergence test only: it states thata series converges if it meets 3 conditions. It can never be used to conclude that a series diverges;in this sense, it is the opposite of the most important divergence test, which can never be used toconclude that a series converges. If the first condition in the Alternating Series Test is not satisfied,the series does indeed diverge, but by the most important divergence test. However, there are lots ofseries that fail either the second or third condition (or both), but still converge; for example, thereare convergent series with only positive terms, that decay to zero, but are not strictly decreasing, e.g.∞Xn=12 + (−1)nn2.The Alternating Series Test is a consequence of the definition of convergence for series (convergenceof the sequence of partial sums) and the Monotonic Sequence Theorem.J.11 The substance of Absolute Convergence Test is that introducing some minus signs into aconvergent series with positive terms does not ruin the convergence:if the series∞Xn=1|an| converges, then so does the series∞Xn=1anThis test is useful when the signs are random, as opposed to strictly alternating as required forthe Alternating Convergence Te st. For example, the series∞Xn=1sin nn2converges by the ACT, becausethe series∞Xn=1sin nn2=∞Xn=1|sin n|n2converges since 0 ≤ |sin n|/n2≤ 1/n2and the series∞Xn=11n2converges (this argument uses 3 tests:Absolute Convergence, Comparison, and p-Series; Limit Comparison Test is less suitable in thiscase). The Alternating Series Test cannot be applied in this case, because the signs of sin n do notalternate:sin 1, sin 2, sin 3 > 0, sin 4, sin 5, sin 6 < 0;while the signs usually come in triples, occasionally there are four consecutive terms with the samesign. While the Absolute Convergence Test is less stringent about the alternating sign condition thanthe Alternating Series Test, the former is not a substitute of the latter. While either test can beused to conclude that the series∞Xn=1(−1)nn2converges, only the Alternating Series Test is applicableto the series∞Xn=1(−1)nnbecause the series∞Xn=1(−1)nn=∞Xn=11ndoes not converge. Neither of the two tests directly implies that the series∞Xn=1sin nnconverges1. Asthe Alternating Series Test, the Absolute Convergence Test is a convergence test only; it can neverbe used to conclude that a series diverges. The Absolute Convergence Test is a consequence of theComparison Test and the addition rule for series.J.12 The sum of a convergent series∞Xn=1ancan be estimated by a finite sub-sum: the sumsm=n=mXn=1an= a1+ a2+ . . . + amof the first m terms; this is the m-th partial sum. As m −→∞, smapproaches the sum of the series,so that∞Xn=m+1an=∞Xn=1an− sm−→ 0.In some cases, the above difference can be estimated:1this series does indeed converge b ecause of a more general version of the Alternating Series Test, called Dirichlet’sTest: if {bn} and {sn} are two sequences such that limn−→∞bn= 0, bn≥ bn+1, and there exists C > 0 such thatPn=mn=1sn≤ C for all m, then the seriesP∞n=1snbnconverges; in the case of the Alternating Series Test sn=±(−1)nis just the sign, and so C = 1 works2• if f =f(x) is p ositive, decreasing, and continuous on [1, ∞) andZ∞1f(x)dx converges, thenZ∞m+1f(x)dx <∞Xn=m+1an<Z∞mf(x)dx(J1)Note that increasing the lower limit (from m to m+1 here) makes the integral smaller becausef >0. In this case, the finite-sum estimate smis an under-estimate for the infinite sum becauselots of positive terms are dropped from the infinite series.• if limn−→∞an= 0, |an|> |an+1|, and the signs of analternate (an> 0 for every n odd and an< 0for every n even, or the other way around), then∞Xn=m+1an<am+1and the signs of∞Xn=m+1anand am+1are the same(J2)In this case, the finite-sum estimate smis an under-estimate for the infinite sum if am<0 andan over-estimate if am>0 (so determined by the last term used in the estimate).For example, let’s estimate the sum of the series∞Xn=11n2to within 1/5. Since f(x) = 1/x2> 0 iscontinuous and decreasing on [1, ∞), by (J1) we need to find the smallest integer m such thatZ∞mf(x)dx =Z∞m1x2dx =1m≤15.So m=5 and the required finite-sum estimate isn=5Xn=11n2=11+14+19+116+125=3600 + 900 + 400 + 225 + 1443600=52693600This is an under-estimate for the infinite sum, as only positive terms are dropped off from the latter.Let’s next estimate the sum of the series∞Xn=1(−1)n−1nto within 1/5. Since this series is alternating(odd terms > 0, even terms < 0), 1/n −→ 0, and 1/(n+1)>1/n, by (J2) we need to find the smallestinteger m such thatam+1=1m + 1≤15.So m=4 and the required finite-sum estimate isn=4Xn=1(−1)n−1n=11−12+13−14=12 − 6 + 4 − 312=712This is an under-estimate for the infinite sum, as the last term in the estimate is negative.Remark: The estimates (J1) and (J2) are closely tied to the Integral Test and the Alternating SeriesTest for convergence of series. In principle, there are estimates related to other convergence tests, inparticular the