8.5 Alternating infinite series


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1 Alteratig ifiite series I the previous two sectios we cosidered oly series with positive terms. I this sectio we cosider series with both positive ad egative terms which alterate: positive, egative, positive, etc. Such series are called alteratig series. For example = = = ++ Theorem 8.5. The Alteratig Series Test. If the alteratig series satisfies the followig two coditios: a a +a 3 a a + = i 0 < a + a for all =,,3,..., ii lim + a = 0, the the series is coverget a Proof. Assume that {a } + is a oicreasig sequece that is assume that i is true ad lim a = 0. + By the defiitio of covergece for each ǫ > 0 there exists Nǫ such that N, > Nǫ implies a 0 < ǫ. Sice a > 0, the last implicatio ca be simplified as N, > Nǫ implies a < ǫ We eed to show that the sequece of partial sums {S } +, S = a a a +a 3 a a, =,,3,4,..., is coverget. First cosider the sequece {S } + of eve partial sums. The S + S = a + a + 0, sice by i a + a +. Thus the sequece {S } + is odecreasig.
2 66 Next we compare a arbitrary eve partial sum S k with a arbitrary odd partial sum S j. Assume j k, the S k S j = a j +a j+ + aj+ +a j ak4 +a k3 + ak +a k ak. Each of the umbers i the parethesis is egative. Therefore the last sum is egative. That is S k S j for j k. Assume ow that j > k, the S j S k = a k+ a k+ + ak+3 a k aj5 a j4 + aj3 a j +aj. Each of the umbers i the parethesis is positive. Therefore the last sum is positive. That is S k S j for j > k. Thus we coclude that S k S j for all j,k =,,3, I particular meas that {S } + is bouded above ad that each S j, j =,,3,... is a upper boud. Sice the sequece {S } + is also odecreasig, the Mootoe Covergece Theorem, Theorem 7.3.4, implies that {S } + coverges to its least upper boud, call it S. Cosequetly S k S S j for all j,k =,,3, For each two cosecutive atural umbers, oe of them is eve ad oe is odd. Therefore the iequalities i imply that S S S S = a for all =,,3, Let ǫ > 0 be arbitrary. Let N be such that > Nǫ. The by we coclude that Combiig the iequalities ad we coclude that a < ǫ S S < ǫ. Thus we have proved that for each ǫ > 0 there exists Nǫ such that N, > Nǫ implies S S < ǫ. This proves that the sequece {S } + coverges ad therefore the alteratig series coverges. Example The series i 8.5. is called alteratig harmoic series. It coverges. Solutio. We verify two coditios of the Alteratig Series Test: a + a lim + sice + <, for all =,,3,..., = 0 is easy to prove by defiitio. Thus the Alteratig Series Test implies that the alteratig harmoic series coverges.
3 67 Remark The Alteratig Series Test does ot apply to the series i 8.5. sice the sequece of umbers,, 3,, 5, 3, 7, 4, 8, 5, 9, 6,..., ,... is ot oicreasig. Further exploratio of the series i 8.5. would show that it diverges. The Alteratig Series Test does ot apply to the series i sice this series does ot satisfy the coditio ii: + lim = 0. + Agai this series is diverget by the Test for Divergece. Exercise Determie whether the give series coverges or diverges. a cos π + b si π c si π d cos π + e l f si π g si π h i May of the exercises i the ext sectio use the Alteratig Series Test for covergece. Do those exercises as well. 8.6 Absolute ad Coditioal Covergece I the previous sectio we proved that the alteratig harmoic series = + coverges Later o we will see that the sum of this series is l. Talkig about ifiite series i class I have ofte used the aalogy with a ifiite colum i a spreadsheet ad fidig its sum. A series with positive ad egative terms oe ca iterpret as balacig a checkbook with ifiitely may deposits ad withdrawals. Lookig at the alteratig harmoic series 8.6. we see a sequece of alteratig deposits ad withdrawals, ifiitely may of them. What we proved i the previous sectio tells that uder two coditios o the deposits ad withdrawals, although it has ifiitely may trasactios, this checkbook ca be balaced. Now comes the first surprisig fact! Let s calculate how much has bee deposited to this accout: = +.
4 68 Applyig the Limit Compariso Test with the harmoic series it is easy to coclude this series diverges. Lookig at the withdrawals we see 4 6 =. Agai this is a diverget series. This is certaily a suspicious situatio: Dealig with a accout to which a ubouded amout of moey has bee deposited ad a ubouded amout of moey has bee withdraw. A simpler way to look at this is to look at the total amout of moey that wet through this accout oe ca call this amout the total activity of the accout: + = This is the harmoic series which is diverget. Sice we kow that a ubouded amout of moey has bee deposited to this accout we might wat to get i the spedig mood sooer ad do two withdrawals after each deposit, keepig the amouts the same: I ay real life checkig accout this might result i a occasioal low balace but if the deposits ad withdrawals are idetical, o mater how you arrage them they should result i the same fial balace. Amazigly this is ot always the case with ifiite series! This is the secod surprisig fact! The series i also coverges but to a differet umber the the series i To be specific deote the terms of the series by b, N. The b 3k = k, b 3k = 4k, b 3k = 4k, k N. It is clear that this series has the same terms as the alteratig harmoic series. The terms of the alteratig harmoic series has bee reordered. For k N, the term at the positios k odd terms i the alteratig harmoic series is at the positio 3k i the series 8.6.3, the term which is at the positio 4k a half of the eve terms i the alteratig harmoic series is at the positio 3k i the series ad the term which is at the positio 4k aother half of the eve terms i the alteratig harmoic series is at the positio 3k i the series The followig calculatio idicates that the sum of the series i is / of the sum of the alteratig harmoic series i Let us calculate the 3th partial sum of the series Sice this is a fiite sum we ca rearrage terms as we please. Here is the calculatio S 3 = = =
5 69 Hece, 3th partial sum of the series is idetical to oehalf of the th partial sum of the alteratig harmoic series. Sice the sum of the alteratig harmoic series is l we have Sice S 3+ = S lim S 3 = l +. ad S 3+ = S = S , we coclude that lim S 3+ = lim S 3+ = lim S 3 = l From the last three equalities oe ca prove rigorously that lim S = l +. This proves that the series coverges to l /. That is just rearragemet of terms chaged the sum. This is a remarkable observatio: a chage of order of summatio ca chage the sum of a ifiite series. This feature is closely related to the fact that the total activity of the accout expressed i 8.6. is a diverget series. This is a motivatio for the followig defiitio. Defiitio A coverget series of the absolute values of its terms Defiitio A series values of its terms a is called coditioally coverget if the series a is diverget. a is called absolutely coverget if the series of the absolute a is coverget. Example Prove that the series is absolutely coverget = + + Solutio. By the defiitio of absolute covergece we eed to determie the covergece of the series + = = This is a pseries with p =. Therefore this series coverges. Notice that at the begiig of Sectio 8.3 we proved that this series coverges by comparig it to a telescopig series.
6 70 Remark Oe ca iterpreted the series i Example as a checkig accout with ifiitely may alteratig deposits ad withdrawals. I this case the total activity of the accout is a coverget series. Cosequetly the total amout deposited ad the total amout withdraw = = + = are both coverget series. As we ca see, the total amout withdraw is /4 of the total activity of the accout. We metioed before that we ca ot prove it i this course Therefore = π + = = = 3 4 π 6 π 4 6 = π 6 = π Theorem If a series a is absolutely coverget, the it is coverget. Proof. Assume that The the algebra of coverget series the series we coclude that a is absolutely coverget, that is assume that 0 a + a a for all =,,3,... By the Compariso Test it follows that the series coverget series implies that the series is also coverget. a is coverget. a is coverget. Sice a a a, a + a a = a + a is coverget. The algebra of a The followig stroger versios of the Ratio ad the Root test ca be applied to ay series to determie whether a series coverges absolutely or it diverges.
7 7 Theorem The Ratio Test. Let a If R <, the the series coverges absolutely. b If R >, the the series diverges. Theorem The Root Test. Let a If R <, the the series coverges absolutely. b If R >, the the series diverges. a + a be a series for which lim = R. The + a a be a series for which lim a = R. The + Noticethatiftherootortheratiotestapplytoaseries, theseries eithercoverges absolutely or diverges. This implies that if a series coverges coditioally, the either a + a + lim = or lim + a + a does ot exist, ad also lim + a = or lim + a does ot exist. I other words, the root ad the ratio test caot lead to a coclusio that a series coverges coditioally. It turs out that our oly tool which ca be used to coclude coditioal covergece is the alteratig series test. Exercise Determie whether the give series coverges coditioally, coverges absolutely or diverges. cosπ siπ/ + + a b c d + + e + p + f +e/ g +! h + + i j +l k + e / l l = I probleme determie all the values of p for which the series coverges absolutely, coverges coditioally ad diverges. + l + Exercise Determie whether the give series coverges coditioally, coverges absolutely or diverges. a +si b c + cos d + si
8 7 9 Series of fuctios 9. Power Series The most importat series is the geometric series: a+ar +ar +ar 3 + +ar + = If < r < the geometric series coverges. Moreover, we proved ar = a+ar+ar +ar 3 + +ar + = a r ar. Replacig r by x ad lettig a = we ca rewrite the formula i 9.. as x = +x+x +x 3 + +x + = x The formula 9.. ca be viewed as a represetatio of the fuctio fx =, < x <, x as a ifiite series of powers of x: = x 0,x,x,x 3,...: x = +x+x +x 3 + +x + = for < r <. 9.. for < x <. 9.. x for < x <. You will agree that the oegative iteger powers of x are very simple fuctios. Therefore, it is atural to explore the followig questio: Q: Which fuctios ca be represeted as ifiite series of costat multiples of oegative iteger powers of x? I other words: Which fuctios x fx ca be represeted as fx = a 0 +a x+a x +a 3 x 3 + +a x + = The ifiite series a 0 +a x+a x +a 3 x 3 + +a x + = is called a power series. The first questio to aswer about a power series is: a x for? < x <?. a x 9..3
9 73 Q: For which real umbers x does the power series coverge? Sice we are workig with the powers of x ad sice there is o restrictio o the sigs of a ad x, we ca use Theorems ad the ratio ad root test to determie the absolute covergece of the power series To apply Theorem we calculate Assume that a + x + a + x a + lim = lim = x lim. + a x + a + a a + lim = L a If L = 0, the Theorem implies that the series 9..3 coverges for all real umbers x. If L > 0, the Theorem implies that the series 9..3 coverges absolutely for x L <, that is for L < x < L diverges for x L >, that is for x < L or x > L If the limit i 9..4 does ot exist, the o coclusio about the covergece or divergece ca be deduced. To apply Theorem we calculate a x = x lim a. Assume that lim + lim + + a = L If L = 0, the Theorem implies that the series 9..3 coverges for all real umbers x. If L > 0, the Theorem implies that the series 9..3 coverges absolutely for x L <, that is for L < x < L diverges for x L >, that is for x < L or x > L If the limit i 9..5 does ot exist, the o coclusio about the covergece or divergece ca be deduced. Example 9... Cosider the power series 0! +! x+! x + 3! x3 + +! x + = I this example a = /!, = 0,,,... We calculate a + L = lim + a = lim + +!!! x. = lim + + = 0. Cosequetly the give power series coverges absolutely for every x R.
10 74 Example 9... Cosider the power series +x+3x +4x x + = Here a = +, = 0,,,... ad we calculate a + L = lim + a +x. + = lim + + =. Cosequetly the give power series coverges absolutely for every x,. Clearly the series diverges for x = ad for x =. Example Cosider the power series x x + 3 x3 4 x x + = Here a 0 = 0 ad a = + /, =,,... We calculate a + L = lim = lim + a x. = lim + + =. Cosequetly the give power series coverges absolutely for every x,. Clearly the series diverges for x = ad coverges coditioally for x =. Example Cosider the power series + x+ x + 3 x3 + + x + = Here a =, = 0,,,... We calculate a + L = lim = lim + a + + = lim + =. x Cosequetly the give power series coverges absolutely for every x,. Clearly the series diverges for x = ad for x =. Notice that we ca actually calculate the sum of this series usig the followig substitutio or you ca call this a trick. Substitute u = x/ i The 9..6 becomes +u+u +u 3 + +u + = u We kow that the sum of the series i 9..7 is /u for u,, that is, +u+u +u 3 + +u + = u = u, u,. Substitutig back u = x/ we get: + x+ x + 3 x3 + + x + = x = x, x,.
11 75 Example Cosider the power series We calculate x+ 4 x + 9 x3 + + x + = a + L = lim = lim + a + + x. = lim + + =. Cosequetly the give power series coverges absolutely for every x,. For x = we get the series. Therefore, for x = the give power series coverges. For x = we get the alteratig series which coverges absolutely. Therefore the give power series coverges absolutely o [, ]. The followig theorem aswers the questio Q above. Theorem Let a 0 +a x+a x +a 3 x 3 + +a x + = be a power series. The oe of the followig three cases holds. A The power series coverges absolutely for all x R. a x B There exists r > 0 such that the power series coverges absolutely for all x r,r ad diverges for all x such that x > r. C The power series diverges for all x 0. For x = 0 it is trivial that the power series coverges. The set o which a power series coverges is called the iterval of covergece. The umber r > 0 i Theorem 9..6 B is called the radius of covergece. I the case A i Theorem 9..6 we write r = +. I the case C i Theorem 9..6 we write r = 0. Remark I the case B i Theorem 9..6 the covergece of the power series at the poits x = r ad x = r must be determied by studyig the ifiite series a r ad a r. A review of the examples i this sectio shows that the iterval of covergece of a power series ca have ay of these four forms r,r, r,r], [r,r ad [r,r].
12 76 9. Fuctios Represeted as Power Series The followig theorem lists properties of fuctios defied by a power series. Theorem 9... Let I be the iterval of covergece of the power series a 0 +a x+a x +a 3 x 3 + +a x + = a x. Assume that I does ot cosist of a sigle poit. The the fuctio f defied o I by fx := a 0 +a x+a x +a 3 x 3 + +a x + = has the followig three properties. a The fuctio f is cotiuous o I. b The fuctio f is differetiable at all iterior poits of I. Moreover, f x = a +a x+3a 3 x + +a x ++a + x + = a x, x I, +a + x, for all x I excludig the edpoits if ay of I. c The fuctio f has derivatives of all orders,,3,..., at all iterior poits of I. I particular f0 = a 0, f 0 = a, f 0 = a, f 0 = 3 a 3,..., f 0 =!a, d If x I, the x 0 ftdt = a 0 x+ a x + a 3 x3 + + a x + a + x+ + = a x. Example 9... By 9.. we have x = +x+x +x 3 + +x + for < x <. 9.. Thus the fuctio fx = /x defied for x, ca be represeted by a power series. Applyig Theorem 9.. we get x = +x+3x +4x 3 + +x ++x + for < x <.
13 77 Example Substitutig x for x i 9.. we get +x = x+x x x + for < x < Thus the fuctio fx = /+x defied for x, ca be represeted by a power series. Applyig Theorem 9.. d we get l+x = x 0 +t dt = x x + 3 x3 4 x x + for < x <. For x = the above series is a alteratig harmoic series which coverges coditioally. Thus we foud a power series represetatio for the fuctio l + x o the iterval,]. By Theorem 9.. a this implies that the sum of the alteratig harmoic series is l. Example Substitutig x for x i 9..3 we get +x = x +x 4 x x + for < x <. Thus the fuctio fx = /+x defied for x, ca be represeted by a power series. Applyig Theorem 9.. d we get arctax = x 0 +t dt = x 3 x3 + 5 x5 7 x x + for < x <. For x = the above series is a coditioally coverget alteratig series. Moreover, π 4 = arcta = Thus we have a power series represetatio for the fuctio arctax o the iterval, ]. 9.3 Taylor series at 0 Maclauri series I the precedig sectio we foud power series represetatios for several well kow fuctios. It turs out that all well kow fuctios ca be represeted as power series. The key step i fidig the power series represetatio of elemetary fuctios are formulas 9.. which establish the relatioship betwee the coefficiets a, = 0,,,..., of a power series ad the derivatives of the fuctio f which is represeted by that power series. We rewrite formulas 9.. as a 0 = f0, a = f 0, a =! f 0, a 3 = 3! f3 0,..., a =! f 0, Let a > 0 ad let f be a fuctio defied o a,a. Assume that f has all derivatives o a, a. The the series power series f0+f 0x+! f 0x + 3! f3 0x 3 + +! f 0x + = +ifty! f 0x
14 78 is called Taylor series at 0 or Maclauri series of f. Usig formulas 9.3. it is ot difficult to calculate a Maclauri series for a give fuctio. The difficulties arise i provig that the fuctio defied by such power series is idetical to the give fuctio. Fortuately this is true for all well kow fuctios. Example Let fx = e x = expx, x R. The f x = e x for all = 0,,,... Therefore the coefficiets of the Maclauri series for the fuctio exp are a = /! ad it ca be proved that for all x R we have e x = +x+! x + 3! x3 + +! x +. Example Let fx = six, x R. The Cosequetly, f x = cosx, f x = six, f 3 x = cosx, f 4 x = six. f k 0 = 0, f k+ 0 = k, k = 0,,,... Therefore the coefficiets of the Maclauri series for the fuctio si are a k = 0, a k+ = k k +!, k = 0,,,... It ca be proved that for all x R we have six = x 3! x3 + 5! x5 7! x7 + + k k +! xk+ +. Example Let fx = cosx, x R. The Cosequetly, f x = six, f x = cosx, f 3 x = six, f 4 x = cosx. f k 0 = k, f k+ 0 = 0, k = 0,,,... Therefore the coefficiets of the Maclauri series for the fuctio cos are a k = k k!, a k+ = 0, k = 0,,,... It ca be proved that for all x R we have cosx =! x + 4! x4 6! x6 + + k k! xk +. Example The Biomial Series. Let α R. Let fx = +x α, x,. The f x = α+x α, f x = αα+x α, f 3 x = ααα+x α3,. f x = αα α++x α.
15 79 Therefore the coefficiets of the Maclauri series for the fuctio f are a 0 =, a = αα α+, N.! It ca be proved that for all x, we have +x α = + α! x+ αα! x + ααα 3! x αα α+ x +.! This series is called biomial series. The reaso for this ame is that for α N the biomial series becomes a polyomial: +x = + x +x = +x+ x +x 3 = +3x+ 3x + x 3 +x 4 = +4x+ 6x + 4x 3 + x 4 +x 5 = +5x+0x +0x 3 + 5x 4 + x 5 +x 6 = +6x+5x +0x 3 +5x 4 +6x 5 + x 6. +x m = m k=0 m x k, were m N ad k m = k m! k!mk! The last formula is called the biomial theorem. The coefficiets m m! mm mk + = = with m,k N, 0 k m, k k!mk! k! are called biomial coefficiets. With a geeral α R ad k N the coefficiets α αα αk + := k k! are called geeralized biomial coefficiets. By defiitio α 0 =. With this otatio the biomial series ca be writte as α +x α = x k for x, k k=0 Notice that formula 9.. is a special case of 9.3., sice k + = = k k! = k. k k! k! Notice also that differetiatig 9.. leads to +x = + k= k k +x k for < x <.
16 80 This is a biomial series with α =. To verify this we calculate = k 3 k + k! = k k +! k! = k k +. Thus For α = / the expressio / = k +x = + x! x + 3 = 3 k + k! 3 k3 k! = k 3 k 3 k k! 3 3! x ! x x 5 + for < x <. 5 5! Example Let fx = arcsix, x [,]. To calculate the Maclauri series for arcsi we otice that d arcsix = dx, x,. x Now calculate the Maclauri series for the last fuctio usig the biomial series with α = /. For α = / ad k N, we calculate Thus / = 3 x+ +x k = = 3 5 k k! k k! = k 3 k k k!! x ! x x 4 + for < x <, 4 4! that is, or usig the otatio of double factorials = + k 3 k +x k k! k= x k, = + kk!! x k. +x k!! k=
17 8 Substitutig x istead of x i the above formula we get = + + x k= k!! k!! x k, for < x <. Sice x itegratig the last power series we get arcsix = x+ k= 0 dt = arcsix, t k!! k +k!! xk+ = k=0 k k 4 k k + xk+, for < x < It is iterestig to ote that the above expasio holds at both edpoits x = ad x =. To prove this we eed to recall Theorem 9.. a ad prove that the series + k= k!! k +k!! coverges. This series coverges by The Compariso Test. Hit: Prove by mathematical k!! iductio that < k!! 3 for all k N. As a cosequece we obtai that k + k= + k!! k +k!! = k k 4 k k + = π. k=0
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