4.B 1kBhc.'Series with positive and negative termskBhcAlternating Series.kBhcmfAn alternating series is a series in which the signs on the terms being added alternate between + and kBhc-kBhc;4. Here is perhaps the most famous alternating series muS(kBhckBhc n = 1kBhc carFIGrapkBhc -1opxEkBhcnkBhcnkBhczThis series actually converges. Let us now show this. We have to start from first principles by examining the partial sum.kBhcSsbuSkBhcNkBhc  = muS(kBhcNkBhc n = 1kBhc carFIGrapkBhc -1opxEkBhcnkBhcnkBhcWhat is tmiLkBhc NϰkBhc  SsbuSkBhcNkBhcOH? We can get a sense for what happens by making a table of partial sums.lbaT$kBhcNkBhc1kBhc2kBhc3kBhc4kBhc5kBhc6kBhc7kBhc8kBhc9kBhc 10kBhc 11kBhc 12kBhc 13kBhc 14kBhc 15kBhc 16kBhc 17kBhc 18kBhc 19kBhc 20kBhcSsbuSkBhcNkBhc -1kBhc -0.5kBhc-0.83333333333333326kBhc-0.58333333333333326kBhc-0.78333333333333321kBhc-0.6166666666666667kBhc-0.7595238095238096kBhc-0.63452380952380938kBhc-0.74563492063492043kBhc-0.64563492063492056kBhc-0.7365440115440115kBhc-0.65321067821067802kBhc-0.73013375513375489kBhc-0.65870518370518349kBhc-0.72537185037185015kBhc-0.66287185037185015kBhc-0.72169537978361487kBhc-0.66613982422805929kBhc-0.71877140317542765kBhc-0.66877140317542783 kBhc\UThe table is not very illuminating, but if we plot the partial sums as a function of kBhcNkBhc4-, something rather dramatic jumps out at you. tlP2d4@NylbaTkBhcNkBhc1kBhc2kBhc3kBhc4kBhc5kBhc6kBhc7kBhc8kBhc9kBhc 10kBhc 11kBhc 12kBhc 13kBhc 14kBhc 15kBhc 16kBhc 17kBhc 18kBhc 19kBhc 20kBhcSkBhc -1kBhc -0.5kBhc-0.8333333333333333kBhc-0.5833333333333333kBhc-0.7833333333333332kBhc-0.6166666666666667kBhc-0.7595238095238096kBhc-0.6345238095238094kBhc-0.7456349206349204kBhc-0.6456349206349206kBhc-0.7365440115440115kBhc-0.653210678210678kBhc-0.7301337551337549kBhc-0.6587051837051835kBhc-0.7253718503718501kBhc-0.6628718503718501kBhc-0.7216953797836149kBhc-0.6661398242280593kBhc-0.7187714031754276kBhc-0.6687714031754278@2@4@6 @8$@10(@12,@140@162@184@20ɿ-0.2ٿ-0.4433333-0.6-0.8-1$4@$?@@@@@@NN @NN"@==$@ &@y4đ(@*@/*wA],@.@'?60@'?61@ 2@/vQ3@!,4@n$If0 kBhce^To try to understand this picture, let us take a closer look at the nature of the partial sum. kBhcSsbuSkBhcNkBhc  = muS(kBhcNkBhc n = 1kBhc carFIGrapkBhc -1opxEkBhcnkBhcn kBhc@9To start with, write down a typical partial sum for even kBhcNkBhcha. Grouping the terms in that partial sum in a particular way leads to an interesting observation.kBhcSsbuSkBhc6kBhc  = Grap@kBhc -1 + carF$kBhc1kBhc2kBhc  - GrapfcarF$kBhc1kBhc3kBhc  - carF$kBhc1kBhc4kBhc  - GrapfcarF$kBhc1kBhc5kBhc  - carF$kBhc1kBhc6kBhc  = carF%kBhc -1kBhc2kBhc  - GrapkBhc poskBhc  - GrapkBhc poskBhce^This tells us that the sequence of even partial sums is a decreasing sequence which starts at kBhc -1/2kBhc\U. This raises the question of whether that sequence of partial sums will decrease to kBhc -kBhc or if it will level out and have a limit. If we can somehow show that this sequence of partial sums is bounded below, then we will have shown that the sequence converges.kBhcWPSet aside that question for a moment and examine a typical partial sum with odd kBhcNkBhc^W. Once again, an appropriate version of the grouping trick shows something interesting.kBhcSsbuSkBhc7kBhc = -1 + GrapfcarF$kBhc1kBhc2kBhc  - carF$kBhc1kBhc3kBhc  + GrapfcarF$kBhc1kBhc4kBhc  - carF$kBhc1kBhc5kBhc  + GrapfcarF$kBhc1kBhc6kBhc  - carF$kBhc1kBhc7kBhc= -1 + GrapkBhc poskBhc  + GrapkBhc poskBhc  + GrapkBhc poskBhcThe odd partial sums form an increasing sequence of terms. The final question to answer is what is the relationship between the even partial sums and the odd partial sums?kBhcHow does, for example, kBhcSsbuSkBhc7kBhcC< compare to all of the even partial sums that come after it?kBhcSsbuSkBhc 2 kkBhc  = SsbuSkBhc7kBhc  + muS)kBhc 2kkBhc n = 8kBhc carFIGrapkBhc -1opxEkBhcnkBhcn muS)kBhc 2kkBhc n = 8kBhc carFIGrapkBhc -1opxEkBhcnkBhcnkBhc  = GrapfcarF$kBhc1kBhc8kBhc  - carF$kBhc1kBhc9kBhc  + GraphcarF%kBhc1kBhc 10kBhc  - carF%kBhc1kBhc 11kBhc + + carF&kBhc1kBhc 2 kkBhc  > 0kBhcSsbuSkBhc 2 kkBhc  = SsbuSkBhc7kBhc  + kBhc poskBhcSsbuSkBhc 2 kkBhc  > SsbuSkBhc7kBhcWhat we have now is that the sequence of even partial sums is a strictly decreasing sequence which is bounded below, and therefore must converge.kBhcBy similar reasoning, the story with the odd partial sums is that they are strictly increasing and bounded above. Hence, they must converge.kBhcThere is just one question left to resolve, and here the picture is not much help. We have established that separately the even partial sums and the odd partial sums both form convergent sequences. The only thing left to prove is that these two limits arekBhc in fact the same limit. We can do that pretty easily by examining the difference between the even and odd partial sums in the limit. tmiLkBhc kϰVsbAWkBhcSsbuSkBhc 2kkBhc  - SsbuSkBhc 2k+1kBhc  = tmiLkBhc kϰVsbAucarFikBhc-GrapkBhc -1opxEkBhc 2k+1GrapkBhc 2k + 1kBhc  = tmiLkBhc kϰcarF*kBhc1kBhc2 k + 1kBhc  = 0kBhcThis finally shows that  muS(kBhckBhc n = 1kBhc carFIGrapkBhc -1opxEkBhcnkBhcnkBhc converges. kBhcIf you go back and take a close look at what we have just shown, you will see that the entire proof hangs on one essential fact about carFIGrapkBhc -1opxEkBhcnkBhcnkBhc: in absolute value each term is smaller than the term that came before it. Therefore, if we have any series with exactly alternating signs and strictly decreasing terms, we can apply exactly the same arguments to it to show that the series converges.! kBhcTheoremkBhc=6 (Alternating series test) If the terms of the series muS(kBhckBhc n = 1kBhc GrapkBhc -1opxEkBhcnkBhc  asbuSkBhcnkBhc*# have the property that all of the kBhcasbuSkBhcnkBhc terms are positive and"kBhcasbuSkBhc n+1kBhc  < asbuSkBhcn#kBhcfor all kBhcnkBhc#, then the series converges.$kBhcAbsolute Convergence%kBhcAlternating series are very nice. To prove that an alternating series converges, we only have to prove that the terms in the series decrease in absolute value as kBhcnkBhc increases. A practical problem that sometimes occurs in practice is that you may have a series that has both positive and negative terms in it but the positive and negative terms do not follow the strict alternating pattern. An example of this kind is thkBhce series& muS(kBhckBhc n = 1carF@kBhc cos nkBhcnopxEkBhc2'kBhcThis series has both positive and negative terms, but the terms don't form a strict alternation. In cases like this, the only thing we can do is the following.(kBhc DefinitionkBhc A series) muS(kBhckBhc n = 1kBhc  asbuSkBhcn*kBhc is kBhcabsolutely convergentkBhc if the series+ muS(kBhckBhc n = 1kBhc VsbA,kBhcasbuSkBhcnkBhcconverges. A series is kBhcconditionally convergentkBhc4- if the series converges, but not absolutely.kBhc The series, muS(kBhckBhc n = 1carF@kBhc cos nkBhcnopxEkBhc2kBhc' is absolutely convergent because- muS(kBhckBhc n = 1VsbALcarF@kBhc cos nkBhcnopxEkBhc2kBhc  < muS(kBhckBhc n = 1carF<kBhc1kBhcnopxEkBhc2kBhc2+and the latter series is known to converge.