10.6 ALTERNATING SERIES

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1 0.6 Alteratig Series Cotemporary Calculus 0.6 ALTERNATING SERIES I the last two sectios we cosidered tests for the covergece of series whose terms were all positive. I this sectio we examie series whose terms chage sigs i a special way: they alterate betwee positive ad egative. Ad we preset a very easy to use test to determie if these alteratig series coverge. A alteratig series is a series whose terms alterate betwee positive ad egative. For example, the followig are alteratig series: () ( )k+ k +... = ( ) k+ k k= (alteratig harmoic series) (2) ( )k k k = ( ) k k k+2 k= () ( )k 2k = k= ( ) k 2k+. Figures, 2 ad show graphs ad tables of values of several partial sums s for each of these series. As we move from left to right i each graph (as icreases), the partial sums alterately get larger ad smaller, a commo patter for the partial sums of alteratig series. The same patter appears i the tables s =! ( ) k+ k= k Fig. s s =! ( ) k k k+2 k= Fig. 2 s s =! ( ) k k= "2k s Fig

2 0.6 Alteratig Series Cotemporary Calculus 2 Alteratig Series Test The followig result provides a very easy way to determie that some alteratig series coverge. It says that if the absolute values of the terms decrease mootoically to 0 the the series coverges. This is the mai result for alteratig series. Alteratig Series Test If the umbers a satisfy the three coditios (i) (ii) a > 0 for all (each a is positive) a > a + (the terms a are mootoically decreasig) (iii) "# a = 0 the the alteratig series ( ) + a coverges. Proof: I order to show that the alteratig series coverges, we eed to show that the sequece of partial sums approaches a fiite it, ad we do so i this case by showig that the sequece of eve partial sums { s 2, s 4, s 6,... } ad the sequece of odd partial sums {s, s, s 5,...} each approach the same value. Eve partial sums: s 2 = a a 2 > 0 sice a > a 2 s 4 = a a 2 + a a 4 = s 2 + (a a 4 ) > s 2 sice a > a 4 s 6 = a a 2 + a a 4 + a 5 a 6 = s 4 + (a 5 a 6 ) > s 4 sice a 5 > a 6. I geeral, the sequece of eve partial sums is positive ad icreasig s 2+2 = s 2 + (a 2+ a 2+2 ) > s 2 > 0 sice a 2+ > a 2+2. Also, s 2 = a a 2 + a a 4 + a 5... a a 2 a 2 = a (a 2 a ) (a 4 a 5 )... (a 2 2 a 2 ) a 2 < a. so the sequece of eve partial sums is bouded above by a. Sice the sequece { s 2, s 4, s 6,... } of eve partial sums is icreasig ad bouded, the Mootoe Covergece Theorem of Sectio 0. tells us that sequece of eve partial sums coverges to some fiite it: "# s 2 = L.

3 0.6 Alteratig Series Cotemporary Calculus Odd partial sums: s 2+ = s 2 + a 2+ so "# s 2+ = "# s 2 + "# a 2+ = L + 0 = L. Sice the sequece of eve partial sums ad the sequece of odd partial sums both approach the same it L, we ca coclude that the it of the sequece of partial sums is L ad that the series ( ) + a coverges to L: ( ) + a = L. Example : Show that each of the three alteratig series satisfies the three coditios i the hypothesis of the Alteratig Series Test. The we ca coclude that each of them coverges. (a) = ( ) + (b) = ( ) + (c) 7 2. l(2) 7. l() l(4)... = =2 ( ) 7. l(). Solutio: (a) This series is called the alteratig harmoic series. (i) a = > 0 for (iii) all positive. (ii) Sice < +, the > + ad a > a +. "# a = "# = 0. Therefore, the alteratig harmoic series coverges. Fig. shows some partial sums for the alteratig harmoic series. (b) a = > 0. Sice < +, we have < + ad > +. "# a = "# = 0. The series (c) a =. > 0 for 2. Sice < + ad l() < l(+), we have l(). l() < (+). l(+) ad "# a = "# 7. l() > 7 (+). l(+). 7 $ l() = 0. The series =2 ( ) 7. l() +... coverges. coverges.

4 0.6 Alteratig Series Cotemporary Calculus 4 Practice : Example 2: Show that these two alteratig series satisfy the three coditios i the hypothesis of the Alteratig Series Test. The we ca coclude that each of them coverges. (a) = ( ) + 2 (b) l(2) l() + l(4) l(5) +... = Does ( ) + +2 coverge? ( ) l() =2 Solutio: a = +2 > 0, but "# a = "# +2 = 0. Sice the terms do ot approach 0, we ca coclude from the N th Term Test For Divergece (Sectio 0.2) that the series diverges. Fig. 2 shows some of the partial sums for this series. You should otice that the eve ad the odd partial sums i Fig. 2 are approachig two differet values. Practice 2: (a) Does ( ) + coverge? (b) Does ( ) + 2+ coverge? Example Of A Diverget Alteratig Series If the terms of a series, ay series, do ot approach 0, the the series must diverge (N th Term Test For Divergece). If the terms do approach 0 the series may coverge or may diverge. There are diverget alteratig series whose terms approach 0 (but the approach to 0 is ot mootoic). For example, is a alteratig series whose terms approach 0, but the series diverges. The eve partial sums of our ew series are s 2 = 2 2 =, s 4 = = ( 2 2 ) + ( 4 4 ) = + 2, s 6 = = ( 2 2 ) + ( 4 4 ) +( 6 6 ) = + 2 +, ad s 2 = You should recogize that these partial sums are the partial sums of the harmoic series, a diverget series, so the partial sums of our ew series diverge ad our ew series is diverget. If the terms of a alteratig series approach 0, but ot mootoically, the the Alteratig Series Test does ot apply, ad the series may coverge or it may diverge.

5 0.6 Alteratig Series Cotemporary Calculus 5 Approximatig the Sum of a Alteratig Series If we kow that a series coverges ad if we add the first "may" terms together, the we expect that the resultig partial sum is close to the value S obtaied by addig all of the terms together. Geerally, however, we do ot kow how close the partial sum is to S. The situatio with may alteratig series is much icer. The ext result says that for some alteratig series (those that satisfy the three coditios i the ext box), the differece betwee S ad the th partial sum of the alteratig series, S s, is less tha the magitude of the ext term i the series, a +. Estimatio Boud for Alteratig Series If S is the sum of a alteratig series ( ) + a that satisfies the three coditios (i) a > 0 for all (each a is positive), (ii) a > a + (the terms are mootoically decreasig), ad (iii) "# a = 0 (the terms approach 0), the the th partial sum s is withi a + of the sum S: s a + < S < s + a + ad approximatio "error" usig s as a estimate of S = S s < a +. Note: This Estimatio Boud oly applies to alteratig series. It is temptig, but wrog, to use it with other types of series. Geometric idea behid the Estimatio Boud: If we have a alteratig series that satisfies the hypothesis of the Estimatio Boud, the the graph of the sequece {s } of partial sums is "trumpet shaped" or "fuel shaped" (Fig. 4). The partial sums are alterately above ad below the value S, ad they "squeeze" i o the value S. Sice the distace from s to S is less tha the distace betwee the successive terms s ad s + (Fig. 5), the S s < s s + = a +. Proof of the Estimatio Boud for Alteratig Series:

6 0.6 Alteratig Series Cotemporary Calculus 6 S s = ( a a 2 + a a ( ) + a + ( ) +2 a +...) ( a a 2 + a a ( ) + a ) = ( ) +2 a + + ( ) + a +2 + ( ) +4 a = ( ) +2 ( a + a +2 + a +... ). The 0 S s = ( ) +2 ( a + a +2 + a + a +4 + a ) = a + a +2 + a + a +4 + a sice ( ) +2 = ad the rest is positive = a + (a +2 a + ) (a +4 a +5 )... < a +. The Estimatio Boud is typically used i two differet ways. Sometimes we kow the value of, ad we wat to kow how close s is to S. Sometimes we kow how close we wat s to be to S, ad we wat to fid a value of to esure that level of closeess. The ext two Examples illustrate these two differet uses of the Estimatio Boud. Example : 4 How close is ( ) + 2 = to the sum ( ) + 2? Solutio: a = 4 2 ad s 4 = ( ) + 2 = so, by the Estimatio Boud, we ca coclude that S s 4 < a 5 = 25 = 0.04 : is withi 0.04 of the exact value S. The < S < ad < S < Similarly, s 9 = is withi a 0 = 00 = 0.0 of the exact value of S, ad s is withi a 00 = 00 2 = of the exact value of S. The < S < ad < S < Practice : Evaluate s 4 ad s 9 for the alteratig series ( ) + for S s 4 ad S s 9. ad determie bouds Example 4: Fid the umber of terms eeded so that s is withi 0.00 of the exact value of ( ) +! ad evaluate s. Solutio: We kow S s < a + so we wat to fid so that a = 000. With a little umerical experimetatio o a calculator, we see that 6! = 720 ad /720 is ot small eough, but

7 0.6 Alteratig Series Cotemporary Calculus 7 7! = 5,040 >,000 so 7! = < Sice + = 7, s is the first partial sum guarateed to be withi 0.00 of S. I fact, s 6 is withi of S, so < S < Practice 4: Fid the umber of terms eeded so s is withi 0.00 of the value of ( ) ad evaluate s. The Estimatio Boud guaratees that s is withi a + of S. I fact, s is ofte much closer tha a + to S. The value a + is a upper boud o how far s ca be from S. Note : The first fiite umber of terms do ot affect the covergece or divergece of a series (they do effect the sum S) so we ca use the Alteratig Series Test ad the Estimatio Boud if the terms of a series "evetually" satisfy the coditios of the hypotheses of these results. By "evetually" we mea there is a value M so that for > M the series is a alteratig series. Note 2: If a series has some positive terms ad some egative terms ad if those terms do NOT "evetually" alterate sigs, the we ca NOT use the Alteratig Series Test it simply does ot apply to such series. PROBLEMS I problems 6 you are give the values of the first four terms of a series. (a) Calculate ad graph the first four partial sums for each series. (b) Which of the series are ot alteratig series?., 0.8, 0.6, ,.5, 0.7,., 2,, ,, 0.5, 0. 5., 0.6, 0.4, ,, 0.5, 0. I problems 7 2 you are give the values of the first five partial sums of a series. Which of the series are ot alteratig series. Why? 7. 2,,, 2, ,,.8,.4, ,, 2., 2.9, ,, 2.5,.5, 2.,, 0.8, 0.6, ,.6,.4,.8,.7. Fig. 6 shows the graphs of the partial sums of three series. Which is/are ot the partial sums of alteratig series? Why? 4. Fig. 7 shows the graphs of the partial sums of three series. Which is/are ot the partial sums of alteratig series? Why?

8 0.6 Alteratig Series Cotemporary Calculus 8 5. Fig. 8 shows the graphs of the partial sums of three series. Which is/are ot the partial sums of alteratig series? Why? 6. Fig. 9 shows the graphs of the partial sums of three series. Which is/are ot the partial sums of alteratig series? Why? I problems 7 determie whether the give series coverge or diverge ( ) ( ) l( ) = ( ) ( 0.99 ) = ( ) ( ) si( ) cos( π ) =4 = si( π ) + l( ) 25. ( ) + l( ) 26. ( ) l( ) = + l( ) 27. ( ) l( 0 ) = ( ) + 7 ( 2) ( ) 0. + ( 2) + ( ) +. cos( π ). si( π/ ) = I problems 2 40, (a) calculate s 4 for each series ad determie a upper boud for how far s 4 is from the exact value S of the ifiite series. The (b) use s 4 fid lower ad upper bouds o the value of S so that {lower boud} < S < {upper boud} ( ) + 6. ( ) l( + ) ( ) ( 0.8 ) 6. ( ) + 7. ( ) si( ) + 8. ( ) 2 9. ( ) cos( π ) l( )

9 0.6 Alteratig Series Cotemporary Calculus 9 I problems 4 50 fid the umber of terms eeded to guaratee that s is withi the specified distace D of the exact value S of the sum of the give series. ( ) , D = ( ) l( + ), D = ( ), D = ( 0.8 ), D = ( ), D = ( ) si( ), D = ( ) 2, D = ( ), D = cos( π ) l( ), D = 0.04 Problems ask you to use the series S(x), C(x), ad E(x) give below. For each problem, (a) substitute the give value for x i the series, (b) evaluate s, the sum of the first three terms of the series, (c) determie a upper boud o the error actual sum s = S s. (The partial sums of S(x), C(x), ad E(x) approximate si(x), cos(x), ad e x, respectively.) S(x) = x x x C(x) = x2 2 + x x ( ) x x E(x) = + x + x2 2 + x x ( ) x x! (2+)! (2)! x = 0.5 i S(x) 5. x = 0. i S(x) 52. x = i S(x) 5. x = 0. i S(x) 54. x = 0. i S(x) 55. x = i C(x) 56. x = 0.5 i C(x) 57. x = 0.2 i C(x) 58. x = 0. i C(x) 59. x = i E(x) 60. x = 0.5 i E(x) 6. x = 0.2 i E(x) Practice Aswers Practice : (a) (i) a = 2 > 0 for all. (ii) 2 < (+) 2 so 2 > (iii) (b) (i) a = (iii) "# a = "# 2 = 0. The series (+) 2 ad a > a +. ( ) + 2 coverges. l() > 0 for all 2. (ii) l() < l(+) so l() > l(+) ad a > a +. "# a = "# l() = 0. The series ( ) l() coverges.

10 0.6 Alteratig Series Cotemporary Calculus 0 Practice 2: (a) ( ) + = 0 so the terms ( ) + do NOT approach 0, ad the series diverges by the th Term Test for Divergece (Sectio 0.2). (b) (i) a = 2+ > 0 for all. (ii) 2+ < 2(+) + so 2+ < 2(+) + ad (iii) "# a = the series "# 2 + ( ) > 2(+) + so a > a +. = 0. Therefore, by the Alteratig Series Test, coverges. Practice : ( ) + : s 4 = S s 4 < a 5 = ( ) 6 5 = 25 = so s < S < s ad < S < s S s 9 < a 0 = ( ) 0 = 000 s < S < s ad < S < = 0.00 so Practice 4: We eed to fid a value for so a + < a + = ( ) +2 (+) + 5 = (+) + 5 a = = = = < 0.00 Sice a 0 < 0.00 we ca be sure that s 9 is withi 0.00 of S. 9 s 9 = ( ) = < S < so

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