7 Sequences of real numbers

Transcription

1 40 7 Sequeces of real umbers 7. Defiitios ad examples Defiitio 7... A sequece of real umbers is a real fuctio whose domai is the set N of atural umbers. Let s : N R be a sequece. The the values of s are s(),s(),s(3),...,s(),... It is customary to write s istead of s() i this case. Sometimes a sequece will be specified by listig its first few terms s, s, s 3, s 4,..., ad sometimes by listig of all its terms {s } N or {s } +. Oe way of specifyig a sequece is to give a formula, or recursio formula for its th term s. Notice that i this otatio s is the ame of the sequece ad is the variable. Some examples of sequeces follow. Example 7... (a), 0,, 0,, 0,,... ; (b), 0,,, 0,,,, 0,,,,, 0,,... ; (c),,,,,... ; (the costat sequece) (d), 3, 3, 4, 4, 3 4, 5, 5, 3 5, 4 5, 6, 6, 3 6, 4 6, 5 6, 7, 7, 3 7, 4 7, 5,... ; (What is the rage 7 of this sequece?) Recursively defied sequeces Example (a) x =, x + = + x 4, =,,3,...; (b) x =, x + = x + x, =,,3,...; (c) a =, a + = +a, =,,3,...; (d) s =, s + = +s, =,,3,...; (e) x = 9 0, x + = 9+x 0, =,,3,... (f) b =, b + =, =,,3,... b (g) f =, f + = (+)f, =,,3,... Some importat examples of sequeces are listed below.

2 4 b = c, c R. N, (7..) p = a, a R, N, (7..) ( x = + ), N, (7..3) ( y = + ) (+), N, (7..4) ( z = + ) a, a R, N, (7..5) f =, f + = f (+), N. (7..6) (The stadard otatio for the terms of the sequece {f } + is f =!, N.) q = a, a R, N, (7..7)! t =, t + = t +! N, (7..8) v = +a, v + = v + a, a R, N. (7..9)! Let {a } + be a arbitrary sequece. A importat sequece associated with {a } + followig sequece 7. Coverget sequeces is the S = a, S + = S +a +, N. (7..0) Defiitio 7... A sequece {s } + of real umbers coverges to the real umber L if for each ǫ > 0 there exists a umber N(ǫ) such that If {s } + coverges to L we will write N, > N(ǫ) s L < ǫ. lim s = L + or s L ( + ). The umber L is called the limit of the sequece {s } +. A sequece that does ot coverge to a real umber is said to diverge. Example 7... Let r be a real umber such that r <. Prove that lim + r = 0. Solutio. First ote that if r = 0, the r = 0 for all N, so the give sequece is a costat sequece. Therefore it coverges. Let ǫ > 0. We eed to solve r 0 < ǫ for. First simplify r 0 = r = r. Now solve r < ǫ by takig l of both sides of the iequality (ote that l is a icreasig fuctio) l r = l r < lǫ.

3 4 Sice r <, we coclude that l r < 0. Therefore the solutio is N(ǫ) = lǫ l r, the implicatio N, > N(ǫ) r 0 < ǫ > lǫ. Thus, with l r is valid. Example Prove that lim = +. Solutio. Let ǫ > 0 be give. We eed to solve Now ivet the BIN: Therefore the BIN is: = Solvig for is ow easy: < ǫ for. First simplify: = ( ) = = 3 = 3. 3 valid for N. 3 < ǫ. The solutio is > 3 ǫ. Thus, with N(ǫ) = 3, the implicatio ǫ > N(ǫ) < ǫ is valid. Usig the BIN, this implicatio should be easy to prove. This procedure is very similar to the procedure for provig limits as x approaches ifiity. I fact the followig two theorems are true. Theorem Let x f(x) be a fuctio which is defied for every x. Defie the sequece a : N R by a = f() for every N. If lim f(x) = L, the lim a = L. x + + Theorem Let x f(x) be a fuctio which is defied for every x (0,]. Defie the sequece a : N R by a = f(/) for every N. If lim x 0 f(x) = L, the lim + a = L.

4 43 The above two theorems are useful for provig limits of sequeces which are defied by a formula. For example you ca prove the followig limits by usig these two theorems ad what we proved i previous sectios. Exercise Fid ( ) the followig limits. Provide ( ) proofs. (a) lim si (b) lim + si + ( (d) lim l + ) ( ) (e) lim + cos + The Algebra of Limits Theorem holds for sequeces. (c) (f) ( lim l + lim + cos ) + ( ) Theorem Let {a } +, {b } + ad {c } +, be give sequeces. Let K ad L be real umbers. Assume that () lim x + a = K, () lim x + b = L. The the followig statemets hold. (A) If c = a +b, N, the lim x + c = K +L. (B) If c = a b, N, the lim x + c = KL. (C) If L 0 ad c = a b, N, the lim x + c = K L. Theorem Let {a } + ad {b } + be give sequeces. Let K ad L be real umbers. Assume that () lim x + a = K. () lim x + b = L. (3) There exists a atural umber 0 such that The K L. a b for all 0. Theorem Let {a } +, {b } + ad {s } + be give sequeces. Assume the followig. The sequece {a } + coverges to the limit L.. The sequece {b } + coverges to the limit L. 3. There exists a atural umber 0 such that The the sequece {s } + Prove this theorem. a s b for all > 0. coverges to the limit L.

5 Sufficiet coditios for covergece May limits of sequeces caot be foud usig theorems from the previous sectio. For example, the recursively defied sequeces (a), (b), (c), (d) ad(e) i Example 7..3 coverge but it caot be proved usig the theorems that we preseted so far. Defiitio Let {s } + be a sequece of real umbers.. If a real umber M satisfies s M for all N the M is called a upper boud of {s } + ad the sequece {s } + is said to be bouded above.. If a real umber m satisfies m s for all N, the m is called a lower boud of {s } + ad the sequece {s } + is said to be bouded below. 3. The sequece {s } + is said to be bouded if it is bouded above ad bouded below. Theorem If a sequece coverges, the it is bouded. Proof. Assume that a sequece {a } + coverges to L. By Defiitio 7.. this meas that for each ǫ > 0 there exists a umber N(ǫ) such that N, > N(ǫ) a L < ǫ. I particular for ǫ = > 0 there exists a umber N() such that N, > N() a L <. Let 0 be the largest atural umber which is N(). The 0 +, 0 +,... are all > N(). Therefore a L < for all > 0. This meas that L < a < L+ for all > 0. The umbers L ad L+ are ot lower ad upper bouds for the sequece sice we do ot kow how they relate to the first 0 terms of the sequece. Put Clearly m = mi{a,a,...,a 0,L } M = max{a,a,...,a 0,L+}. m a for all =,,..., 0 m L < a for all > 0.

6 45 Thus m is a lower boud for the sequece {a } +. Clearly a M for all =,,..., 0 a < L+ M for all > 0. Thus M is a upper boud for the sequece {a } +. Is the coverse of Theorem 7.3. true? The coverse is: If a sequece is bouded, the it coverges. Clearly a couterexample to the last implicatio is the sequece ( ), N. This sequece is bouded but it is ot coverget. The ext questio is whether boudedess ad a additioal property of a sequece ca guaratee covergece. It turs out that such a property is mootoicity defied i the followig defiitio. Defiitio A sequece {s } + of real umbers is said to be o-decreasig if s s + for all N, strictly icreasig if s < s + for all N, o-icreasig if s s + for all N. strictly decreasig if s > s + for all N. A sequece with either of these four properties is said to be mootoic. The followig two theorems give powerful tools for establishig covergece of a sequece. Theorem If {s } + is o-decreasig ad bouded above, the {s } + coverges. Theorem If {s } + is o-icreasig ad bouded below, the {s } + coverges. To prove these theorems we have to resort to the most importat property of the set of real umbers: the Completeess Axiom. The Completeess Axiom. If A ad B are oempty subsets of R such that for every a A ad for every b B we have a b, the there exists c R such that a c b for all a A ad all b B. Proof of Theorem Assumethat{s } + is a o-decreasig sequece ad that it is bouded above. Sice {s } + is o-decreasig we kow that s s s 3 s s s +. (7.3.) Let A be the rage of the sequece {s } +. That is A = { s : N }. Clearly A. Let B be the set of all upper bouds of the sequece {s } +. Sice the sequece {s } + is bouded above, the set B is ot empty. Let b B be arbitrary. The b is a upper boud for {s } +. Therefore s b for all N.

7 46 By the defiitio of A this meas Sice b B was arbitrary we have a b for all a A. a b for all a A ad for all b B. By the Completeess Axiom there exists c R such that s c b for all N ad for all b B. (7.3.) Thus c is a upper boud for {s } + ad also c b for all upper bouds b of the sequece {s } +. Therefore, for a arbitrary ǫ > 0 the umber c ǫ (which is < c) is ot a upper boud of the sequece {s } +. Cosequetly, there exists a atural umber N(ǫ) such that c ǫ < s N(ǫ). (7.3.3) Let N be ay atural umber which is > N(ǫ). The the iequalities (7.3.) imply that s N(ǫ) s. (7.3.4) By (7.3.) the umber c is a upper boud of {s } +. Hece we have s c for all N. (7.3.5) Puttig together the iequalities (7.3.3), (7.3.4) ad (7.3.5) we coclude that c ǫ < s c for all N such that > N(ǫ). (7.3.6) The relatioship (7.3.6) shows that for N such that > N(ǫ) the distace betwee umbers s ad c is < ǫ. I other words N, > N(ǫ) implies s c < ǫ. This is exactly the implicatio i Defiitio 7... Thus, we proved that Example Prove that the sequece coverges. lim s = c. + t = l, =,,3,..., Solutio. Use the defiitio of l as the itegral to prove that for > ( t > Floor(x) ) dx. x Deduce that t > 0. Represet t t + = ( l(+) l ) + as a area (or a differece of two areas). Coclude that t t + > 0. Now use oe of the precedig theorems.

8 47 8 Ifiite series of real umbers 8. Defiitio ad basic examples The most importat applicatio of sequeces is the defiitio of covergece of a ifiite series. From the elemetary school you have bee dealig with additio of umbers. As you kow the additio gets harder as you add more ad more umbers. For example it would take some time to add S 00 = It gets much easier if you add two of these sums, ad pair the umbers i a special way: S 00 = A straightforward observatio that each colum o the right adds to 0 ad that there are 00 such colums yields that S 00 = 0 00, that is S 00 = 0 00 This ca be geeralized to ay atural umber to get the formula = S = ( )+ = (+). This procedure idicates that it would be impossible to fid the sum where the last set of idicates that we cotiue to add atural umbers. The situatio is quite differet if we cosider the sequece, 4, 8, 6,...,,... ad start addig more ad more cosecutive terms of this sequece: = = = 4 = 3 4 = 8 = 7 8 = 6 = 5 6 = 3 = 3 3 = 64 = 63 64

9 48 These sums are icely illustrated by the followig pictures I this example it seems atural to say that the sum of ifiitely may umbers,,,... equals 4 8 : = Why does this make sese? This makes sese sice we have see above that as we add more ad more terms of the sequece we are gettig closer ad closer to. Ideed, ad, 4, 8, 6,...,, = lim ( ) =. + This reasoig leads to the defiitio of covergece of a ifiite series: Defiitio 8... Let {a } + be a give sequece. The the expressio a + a + a a +

10 49 is called a ifiite series. We ofte abbreviate it by writig a + a + a a + = For each atural umber we calculate the (fiite) sum of the first terms of the series We call S a partial sum of the ifiite series S = a + a + a a. If the sequece {S } + coverges ad if lim + S = S, the the ifiite series + a is called coverget ad we write a + a + a a + = S a. a. (Notice that {S } + is a ew sequece.) or a = S. The umber S is called the sum of the series. If the sequece {S } + does ot coverge, the the series is called diverget. I the example above we have Therefore we say that the series a = = ( ), S = = ( ) =. lim = + coverges ad its sum is. We write I our startig example =. a =, S = = (+) (+) lim does ot exist. +

11 50 Therefore we say that the series = diverges. Example 8.. (Geometric Series). Let a ad r be real umbers. The most importat ifiite series is a+ar +ar +ar 3 + +ar + = =0 ar (8..) This series is called a geometric series. To determie whether this series coverges or ot we eed to study its partial sums: S 0 = a, S = a+ar, S = a+ar +ar, S 3 = a+ar +ar +ar 3, S 4 = a+ar +ar +ar 3 +ar 4, S 5 = a+ar +ar +ar 3 +ar 4 +ar 5,. S = a+ar +ar + +ar +ar. Notice that we have already studied the special case whe a = ad r =. I this special case we foud a simple formula for S ad the we evaluated lim S. It turs out that we ca + fid a simple formula for S i the geeral case as well. First ote that the case a = 0 is ot iterestig, sice the all the terms of the geometric series are equal to 0 ad the series clearly coverges ad its sum is 0. Assume that a 0. If r = the S = a. Sice we assume that a 0, lim + a does ot exist. Thus for r = the series diverges. Assume that r. To fid a simple formula for S, multiply the log formula for S above by r to get: ow subtract, ad solve for S : S = a+ar +ar + +ar +ar, rs = ar+ar + +ar +ar + ; S rs = a ar +, S = a r+ r.

12 5 We already proved that if r <, the lim + r+ = 0. If r, the lim + r+ does ot exist. Therefore we coclude that I coclusio lim S = lim a r+ + + r = a r for r <, lim S does ot exist for r. + If r <, the the geometric series =0 ar coverges ad its sum is a r. If r, the the geometric series =0 ar diverges. The followig picture illustrates the sum of a geometric series with a > 0 ad 0 < r <. The width of the rectagle below is /( r) ad the height is a. The slopes of the lies show are ( r)a ad r( r)a. a ar 3 ar 5 ar 7 ar ar 6 ar 4 ar a ra a 0 I the picture above the terms of a geometric series are represeted as areas. As we ca see the areas of the terms fill i the rectagle whose area is a/( r). I the picture below we represet the terms of the geometric series by legths of horizotal lie segmets. The picture strogly idicates that the total legth of ifiitely may horizotal lie segmets is a/( r). The reaso for this is that by the costructio the slope of the hypotheuse of the right triagle i the picture below is ( r). Sice its vertical leg is a, its horizotal leg must be a( r).

13 5 a ar a ar ar ar 3 a ar ar ar 3 ar 4 ar 5 ar 6... Remark How to recogize whether a ifiite series is a geometric series? Cosider for example the ifiite series e. Here a = π+ e. Lookig at the formula (8..) we ote that the first term of the series is a ad that the ratio betwee ay two cosecutive terms is r. For a = π+ give above we calculate e a + a = π ++ e (+) π + e π + Sice a + a is costat, we coclude that the series = π+3 e e + π + = π e. π + is a geometric series with e a = a = π e ad r = π e for all =,,3,.... Sice r = π <, we coclude that the sum of this series is e π + π = e e π = π e e e e π = π e e π. Thus, to verify whether a give ifiite series is a geometric series calculate the ratio of the cosecutive terms ad see whether it is a costat: a for which a + a = r for all =,,3,... (8..) is a geometric series. I this case a = a (the first term of the series). Example 8..4 (Harmoic Series). Harmoic series is the series = +.

14 53 Agai, to explore the covergece of this series we have to study its partial sums: S =, S = +, S 3 = + + 3, S 4 = , S 5 = , S 6 = , S 7 = , S 8 = ,. S == Sice S + S = + > 0 the sequece {S } + is icreasig. Next we will prove that the sequece {S } + is ot bouded. We will cosider oly the atural umbers which are powers of :,4,8,..., k,... The followig iequalities hold: S = + + =+ S 4 = = =+ S 8 = = =+3 S 6 = = =+4 Cotiuig this reasoig we coclude that for each k =,,3,... the followig formula holds: S k = k k k k =+k k Thus S k +k for all k =,,3,... (8..3) This formula implies that the sequece {S } + is ot bouded. Namely, let M be a arbitrary real umber. We put j = max { Floor(M), }. The j Floor(M) > (M ).

15 54 Therefore, + j > M. Together with the iequality (8..3) this implies that S j > M. Thus for a arbitrary real umber M there exists a atural umber = j such that S > M. This proves that the sequece {S } + is ot bouded ad therefore it is ot coverget. I coclusio: The harmoic series diverges. The ext example is a example of a series for which we ca fid a simple formula for the sequece of its partial sums ad easily explore the covergece of that sequece. Examples of this kid are called telescopig series. Example Prove that the series (+) coverges ad fid its sum. Solutio. We eed to examie the series of partial sums of this series: S = (+), =,,3,.... It turs out that it is easy to fid the sum S if we use the partial fractio decompositio for each of the terms of the series: k(k +) = k for all k =,,3,.... k + Now we calculate: S = (+) ( = ) ( + ) ( ) = +. Thus S = + Therefore the series ( ) ( + ) + for all =,,3,... Usig the algebra of limits we coclude that (+) ( lim S = lim ) = coverges ad its sum is : (+) =.

16 55 Exercise Determie whether the series is coverget or diverget. If it is coverget, fid its sum. ( ) ( ) +3 ( ) e +3 (a) 6 (b) (c) (d) π (e) (i) =0 π (cos) (f) (j) = 5 (g) (k) =0 =0 =0 (si) (ta) (h) (l) = ( l + ) A digit is a umber from the set {0,,,3,4,5,6,7,8,9}. A decimal umber with digits d,d,d 3,...,d,... is i fact a ifiite series: 0.d d d 3...d... = d 0. Therefore each decimal umber with digits that repeat leads to a geometric series. We use the followig abbreviatio: 0.d d d 3...d k = 0.d d d 3...d k d d d 3...d k d d d 3...d k d d d 3...d k... Exercise Express the umber as a ratio of itegers. (a) 0.9 = (b) 0.7 = (c) 0.7 (d) Basic properties of ifiite series A immediate cosequece of the defiitio of a coverget series is the followig theorem Theorem 8... If a series a coverges, the lim + a = 0. Proof. Assume that its sequece of partial sums {S } + lim S = S. Now usig the formula + ad the algebra of limits we coclude that a is a coverget series. By the defiitio of covergece of a series coverges to some umber S: lim + S = S. The also a = S S, for all =,3,4,..., lim + a = lim + S lim + S = S S = 0.

17 56 Warig: The precedig theorem caot be used to coclude that a particular series coverges. Notice that i this theorem it is assumed that a is a coverget. O a positive ote: Theorem 8.. ca be used to coclude that a give series diverges: If we kow that lim + a = 0 is ot true, the we ca coclude that the series a useful test for divergece. a diverges. This is Theorem 8.. (The Test for Divergece). If the sequece {a } + does ot coverge to 0, the the series a diverges. Example Determie whether the ifiite series Solutio. Just perform the divergece test: ( ) lim cos = 0. + Therefore the series cos ( ) diverges. cos ( ) coverges or diverges. Example Determie whether the ifiite series ( ) + coverges or diverges. Solutio. Cosider the sequece { ( ) + } + :, 3, 3 4, 4 5, 5 6, 6 7, 7 8, 8 9, 9 0, 0,, 3,..., (k ) k, k k +,... (8..) Without givig a formal proof we ca tell that this sequece diverges. I my iformal laguage the sequece (8..) is ot costatish sice it ca ot decide whether to be close to 0 or. Therefore the series ( ) + diverges. ( ) Remark The divergece test ca ot be used to aswer whether the series si ( ) coverges or diverges. It is clear that lim si = 0. Thus we ca ot use the test for + divergece.

18 57 Theorem 8..6 (The Algebra of Coverget Ifiite Series). Assume that coverget series. Let c be a real umber. The the series a ad b are ca, ( ) + ( ) a +b, ad a b, are coverget series ad the followig formulas hold Remark The fact that we write ca = c a, ( ) + a +b = a + ( ) + a b = a b, ad b. b does ot ecessarily mea that b is a geuie ifiite series. For example, let m be a atural umber ad assume that b = 0 for all > m. The m b = b. I this case the series b is clearly coverget. If a is a coverget (geuie) ifiite series, the Theorem 8..6 implies that the ifiite series coverget ad ( ) + a +b = a + m b. ( a + b ) is This i particular meas that the ature of covergece of a ifiite series ca ot be chaged by chagig fiitely may terms of the series. For example, let m be a atural umber. The: The series a coverges if ad oly if the series a m+k k= coverges. Moreover, if + a coverges, the the followig formula holds a = m a j + j= k= a m+k.

19 58 Example Prove that the series ( π (+) ) coverges ad fid its sum. Exercise Determie whether the series is coverget or diverget. If a series is coverget fid its sum. 3 + ( 3 (a) (b) arcta (c) (d) π ) e (e) =0 e +π (f) si ( ) (g) =0 =0 (+) + (h) = =0 ((0.9) +(0.) ) Exercise Express the followig sums as ratios of itegers ad as repeatig decimal umbers. (a) (b) (c) Compariso Theorems Warig: All series i the ext two sectios have positive terms! Do ot use the tests from these sectios for series with some egative terms. The covergece of the series i Examples 8.. ad 8..5 was established by calculatig the limits of their partial sums. This is ot possible for most series. For example we will soo prove that the series coverges. To uderstad why the sum of this series is exactly π you eed to take a class about 6 Fourier series, Math 430. I hope that you have doe your homework ad that you proved that the series = coverges ad that you foud its sum. If you did t here is a way to do it: (It turs out that this is a telescopig series.) Let S = Sice S + S = (+) > 0 the sequece {S } + = is icreasig. For each k =,3,4,... we have the followig partial fractios decompositio k = (k )(k+) = ( k ). k +

20 59 Next we use this formula to simplify the formula for the -th partial sum ( S = k = k ) = ( k + k ) k + k= k= k= = (( ) ( + 3 ) ( ) ( ) ( + )) + = ( + ) + = ( 3 + ) = 3 (+) 4 + (+). Usig the algebra of limits we calculate lim + + (+) = lim + + (+) = lim + Therefore, usig the algebra of limits agai, we calculate Clearly S < 3 for all =,3,... 4 Now cosider the series Let The fact that T + T = Sice we coclude that lim S = = = T = = 0+0 = 0. (+) > 0 implies that the sequece {T } + is icreasig. 4 < 3, 9 < 8, 6 < 5,..., <, T = < = +S < Thus T < 7 4 for all =,3,4,... Sice the sequece {T } + is icreasig ad bouded above it coverges by Theorem Thus the series coverges ad its sum is < 7 4. The priciple demostrated i the above example is the core of the followig compariso theorem.

21 60 Theorem 8.3. (The Compariso Test). Let terms. Assume that a ad a b for all =,,3,.... b be ifiite series with positive (a) If b coverges, the a coverges ad a b. (b) If a diverges, the b diverges. Sometimes the followig compariso theorem is easier to use. Theorem 8.3. (The Limit Compariso Test). Let positive terms. Assume that If diverges. b coverges, the Example Determie whether the series a lim = L. + b a ad a coverges. Or, equivaletly, if b be ifiite series with a diverges, the coverges or diverges. b Solutio. The domiat term i the umerator is ad the domiat term i the deomiator is 6 = 3. This suggests that this series behaves as the coverget series. Sice we are tryig to prove covergece we will take a = i the Limit Compariso Test. Now calculate: lim (+) = lim = lim ad b = (+) I the last step we used the algebra of limits ad the fact that lim + + = 6 which eeds a proof by defiitio. 3 + = lim =.

22 Sice we proved that lim Limit Compariso Test implies that the series = ad sice we kow that coverges. 6 is coverget, the I the ext theorem we compare a ifiite series with a improper itegral of a positive fuctio. Here it is presumed that we kow how to determie the covergece or divergece of the improper itegral ivolved. Theorem (The Itegral Test). Suppose that x f(x) is a cotiuous positive, decreasig fuctio defied o the iterval (0,+ ). Assume that a = f() for all =,,... The the followig statemets are equivalet (a) The itegral + f(x) dx coverges. (b) The series a coverges. At this poit we assume that you are familiar with improper itegrals ad that you kow how to decide whether a improper itegral coverges or diverges. We will use this test i two differet forms: Prove that the itegral Prove that the itegral + + f(x) dx coverges. Coclude that the series f(x) dx diverges. Coclude that the series a coverges. a diverges. Example (Covergece of p-series). Let p be a real umber. The p-series coverget if p > ad diverget if p. p is Solutio. Let >. The the fuctio x x is a icreasig fuctio. Therefore, if p <, the p <. Cosequetly, >, for all > ad p <. p Sice the series all p. + diverges, the Compariso Test implies that the series diverges for p

23 6 Now assume that p >. Cosider the fuctio f(x) = xp, x > 0. This fuctio is a cotiuous, decreasig, positive fuctio. Let me calculate the improper itegral ivolved i the Itegral Test for covergece: + t dx = lim dx = lim xp t + xp = p lim t + ( t t + p p ) x p = p ( ) = p Thus this improper itegral coverges. Notice that the coditio p > was essetial to coclude that lim t + t = 0. Sice = f() for all =,,3,..., the Itegral Test implies that the p p series coverges for p >. p Remark Wehave otproved this forall p > thefuctio f(x) = xp, x > 0, iscotiuous. Oe way to prove that for a arbitrary a R the fuctio x x a, x > 0 is cotiuous is to use the idetity x a = e alx, x > 0. This idetity shows that the fuctio x x a, x > 0 is a compositio of the fuctio exp(x) = e x, x R ad the fuctio x a lx, x > 0. The later fuctio is cotiuous by the algebra of cotiuous fuctios: It is a product of a costat a ad a cotiuous fuctio l. We proved that exp is cotiuous. By Theorem 6..3 a compositio of cotiuous fuctio is cotiuous. Cosequetly x x a, x > 0 is cotiuous. Exercise Determie whether the series is coverget or diverget. (a) (b) e (c) (d) l (e) (i) = (l) b cos( ) (f) (j) =0! π +e e+π (g) = si ( ) For the series i (e) fid all umbers b for which the series coverges. (k)! t (h) (l) = =0 l si( ) Exercise A digit is a umber from the set {0,,,3,4,5,6,7,8,9}. A decimal umber with digits d,d,d 3,...,d,... is i fact a ifiite series: 0.d d d 3...d... = d 0. Use a theorem from this sectio to prove that the series above always coverges.

24 Ratio ad root tests Warig: All series i this sectio have positive terms! Do ot use the tests from this sectio for series with egative terms. I Remark 8..3 we poited out (see (8..)) that a series a for which a + a = r for all =,,3,... is a geometric series. Cosequetly, if r < this series is coverget, ad it is diverget if r. Testig the series usig this criteria leads to the ratio 3 + =0 ( ) ) ( ) ( = 3 + = 3 ( ) + ) = ( 3 ( ) which certaily is ot costat, but it is costatish. I propose that series for which the ratio a + /a is ot costat but costatish, should be called geometrish. The followig theorem tells that covergece ad divergece of these series is determied similarly to geometric series. Theorem 8.4. (The Ratio Test). Assume that The (a) If R <, the the series coverges. (b) If R >, the the series diverges. a + lim = R. + a Aother way to recogize a geometric series is: A series a for which a+ a is a series with positive terms ad that a = r for all =,,3,... is a geometric series. Cosequetly, if r < this series is coverget, ad it is diverget if r. ( ) + Testig the series usig this criteria leads to the root + =0 ( ) + = = + which certaily is ot costat, but it is costatish. +

25 64 Theorem 8.4. (The Root Test). Assume that a is a series with positive terms ad that lim + a = R. The (a) If R <, the the series coverges. (b) If R >, the the series diverges. Remark Notice that i both the ratio test ad the root test if the limit R = we ca coclude either divergece or covergece. I this case the test is icoclusive. Exercise Determie whether the series is coverget or diverget. ( ) + 4 (a) (b) (c) (d) 3 3 (e) (i) = 3! (!) ()! (f) (j) e! (g) (3 +) (k) e / a (h) 3 3 (l)! 3 5 ( ) 4 6 () 3 5 ( ) (arcta) (+) (m) () (o) (p)! For some of the problems you might eed to use tests from previous sectios. 3 5 ( ) 4 6 ()