Topics. Homework Problems. MATH 301 Introduction to Analysis Chapter Four Sequences. 1. Definition of convergence of sequences.

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1 MATH 301 Itroductio to Aalysis Chapter Four Sequeces Topics 1. Defiitio of covergece of sequeces. 2. Fidig ad provig the limit of sequeces. 3. Bouded covergece theorem: Theorem Theorems ad Limit theorems ad Ratio test: Theorem Defiitio of ifiite limits. Provig ifiite limits. 8. Theorems ad Defiitio of mootoe sequeces. 10. Mootoe covergece theorem: Theorem Theorem Defiitio of Cauchy sequeces. 13. Theorem : A sequece i R is coverget iff it is Cauchy. Homework Problems 1. Use the defiitio of limit ad prove the followig. (a) lim = 0 (b) lim = 3 (c) lim 2! = 0 (d) If x < 1 the lim x = Show that each of the sequece is diverget. (a) a = 2 (b) b = ( 1) (c) c = coπ 3 (d) d = ( ) 2

2 3. Fid a example of each of the followig. (a) A coverget sequece of ratioal umbers havig a irratioal limit. (b) A coverget sequece of irratioal umbers havig a ratioal limit. 4. Suppose that (a ), (b ), ad (c ) are sequeces such that a b c for all N. If lim a = lim c = L, prove that lim b = L as well. 5. Determie if the sequece coverges or diverges. Fid the limit if it exists. (a) = (b) = ( 1) +3 (c) = ( 1) 2 1 (d) = (e) = Determie if the sequece coverges or diverges. Fid the limit if it exists. (a) = (b) =! 2 (c) =! (d) = 2! 7. Prove or give a couterexample. (a) If ( ) ad (t ) are diverget sequeces, the ( + t ) is also diverget. (b) If ( ) ad (t ) are diverget sequeces, the ( t ) is also diverget. (c) If ( ) ad ( + t ) are coverget sequeces, the (t ) is also coverget. (d) If ( ) ad ( t ) are coverget sequeces, the (t ) is also coverget. 8. Give a example for each of the followig. (a) A coverget sequece of positive umbers ( ) such that lim +1 = 1. (b) A diverget sequece of positive umbers (t ) such that lim t +1 t = Prove the followig. (a) lim( + 1 ) = 0 (b) lim( ) = 0 (c) lim( 2 + ) = Prove the followig. (a) If + ad k > 0, the k +. (b) If + ad k < 0, the k.

3 (c) If + ad (t ) is a bouded sequece, the + t Prove that each sequece is mootoe ad bouded. Fid the limit. (a) s 1 = 1 ad +1 = 1 4 ( + 5) for all N (b) s 1 = 2 ad +1 = 1 4 ( + 5) for all N (c) s 1 = 2 ad +1 = for all N 12. Fid a example of a sequece i R for each descriptio. (a) Cauchy but ot mootoe. (b) Mootoe but ot Cauchy. (c) Bouded but ot Cauchy. 13. Let (a ) ad (b ) be mootoe sequeces. Prove or give a couterexample. (a) The sequece (a + b ) is mootoe. (b) The sequece (a b ) is mootoe. 14. Let s 1 = 6, s 2 = 6 + 6, s 3 = ,... ad i geeral +1 = 6 +. Prove that ( ) coverges, ad fid its limit. 15. Suppose x > 0. Defie a sequece ( ) by Prove that for ay k > 0, lim = x. s 1 = k, +1 = s2 + x 2 for N.

4 Solutios 1 1. (a) WLOG, assume ε < 1. It suffices to show that ε > 0, we ca fid N N, < ε < ε < ε > 1 ( ) 1 3 ε 1. So, choose N = ( ε 1) + 1. (b) It suffices to show that for all ε > 0, we ca fid N such that N, < ε. First choose large eough so that 2 > 3 ad > 5. (This is possible if > 5.) The = = ( 2 3) < = 10 2 = 10 < ε. So, > 10/ε, ad we shall choose N = max(5, 10/ε) + 1. (c) It suffices to show that for all ε > 0, we ca fid N such that N, 2! 0 < ε. First choose large eough so that < 2 ad 2 > 0, which is possible if > ! 0 = 2! = ( 1)! = 1 ( 2)! 1 < < ε. So, > 2 + 2/ε, ad we shall choose N = max(2, 2 + 2/ε) + 1. (d) If x = 0, it is obvious that lim x = 0. Otherwise, we wish to show that for all ε > 0, we ca fid N such that N, x 0 < ε. x 0 = x < ε l x < l ε > l ε l x So we choose N = l ε l x + 1. because x < 1, l x < For each of these sequeces, assume covergece ad prove by cotradictio. (a) Assume lim(2) = L. Let ε = 1. The we ca fid N such that N, 2 L < 1, 2 L < 1 2 < L + 1 < L + 1, 2 which meas L+1 2 is a upper boud for all N. This is a cotradictio because N is ubouded. (b) Assume lim( 1) = L. Let ε = 0.5. The we ca fid N such that N, ( 1) L < 0.5. Whe is eve, ( 1) = 1, so 1 L < 0.5, 0.5 < L < 1.5, or L (0.5, 1.5). Whe is odd, ( 1) = 1, so 1 L < 0.5, 1.5 < L < 0.5, or L ( 1.5, 0.5). This is a cotradictio because (0.5, 1.5) ( 1.5, 0.5) =. (c) Assume lim coπ 3 = L. Let ε = 0.5. The we ca fid N such that N, coπ 3 Whe is a multiple of 6, = 6m for some m N, coπ 3 L (0.5, 1.5). L < ε. = 1. So 1 L < 0.5, hece

5 Whe is 3 plus multiple of 6, = 3 + 6m for some m N, coπ 3 = 1. So 1 L < 0.5, hece L ( 1.5, 0.5). This is a cotradictio because (0.5, 1.5) ( 1.5, 0.5) =. (d) Assume lim( ) 2 = L. Let ε = 1. The we ca fid N such that N, ( ) 2 L < 1, ( ) 2 L < 1 2 < L + 1 < L + 1, which meas L + 1 is a upper boud for all N. This is a cotradictio because N is ubouded. 3. Cosider the floor fuctio x = sup{ : x}. The f (x) = 10 x /10 is the result of trucatig x after it-th decimal place. We costruct these examples as follows. (a) Cosider x = f (π). The x Q ad lim x = π. (Ca you prove the covergece? Use x x < 1/10.) So (x ) is a ratioal sequece with a irratioal limit. (b) Cosider y = f (π)/π. The y Q ad lim y = 1. So, (y ) is a irratioal sequece with a ratioal limit. 4. For ay ε > 0 we ca fid N 1 such that N, a L < ε because lim a = L. We ca also fid N 2 such that c L < ε because lim c = L. Let N = max(n 1, N 2 ), the for N, a L < ε ad c L < ε, i.e. L ε < a < L + ε ad L ε < c < L + ε. Sice a < b < c, it follows that L ε < b < L + ε, hece b L < ε, which implies lim b = L. 5. (a) lim = lim = 2+3 lim 1 1+lim 1 = 2. (b) lim ( 1) +3 = 0. You ca prove it by usig defiitio of limit, or by usig the squeeze theorem (Problem 4) with a = 1/( + 3) ad c = 1/( + 3). (c) The sequece diverges. It is obvious that s 2 1/2 ad s 2+1 1/2 a. You ca prove the divergece by cotradictio. (d) The sequece diverges to +, i.e. lim = +. You ca prove by usig defiitio. (e) lim = 0 by the ratio test (Theorem 4.2.7): lim(s k+1 /s k ) = 1/2 < (a) The sequece diverges. lim = + by ratio test: lim(s k+1 /s k ) = 3 > 1. (b) The sequece diverges. lim = + by ratio test: lim(s k+1 /s k ) = +. ( ) (c) lim = 0. By ratio test, +1 = (+1)! (+1) +1! = +1 = 1, which approaches 1 (1+ 1 ) e < 1 as. (d) lim = 0. By ratio test, +1 = (+1)2 (+1)!! 2 = (+1) , which approaches 0 a. 7. (a) The statemet is false. Couterexample: = ad t =. The + t = 0. So ( ) ad (t ) are diverget but ( + t ) is coverget. (b) The statemet is false. Couterexample: = t = ( 1). The t = 1. So ( ) ad (t ) are diverget but ( t ) is coverget. (c) The statemet is true. ( ) ad ( + t ) are coverget, so their differece (( + t ) ) = (t ) is coverget.

6 (d) This is false whe lim = 0. Couterexample: = 1/ 2 ad t =. The t = 1/. So ( ) ad ( t ) both coverges, but (t ) diverges. 8. (a) = 1/. The lim(+1 / ) = 1 ad lim = 0. (b) t =. The lim(t +1 /t ) = 1 ad lim t = (a) = + 1 = Sice lim( ) = +, lim = 0 by Theorem (b) = = Sice lim( ) = +, lim = 0 by Theorem (c) = 2 + = = 1, which approaches a. 10. (a) k > 0. If lim = + the for ay arbitrary M R, M/k is also a arbitrary umber, ad N N, > M/k. Multiply k to the iequality, the k > M. (b) k < 0. If lim = + the for ay arbitrary M R, M/k is also a arbitrary umber, ad N N, > M/k. Multiply k to the iequality, the k < M. (c) (t ) is bouded: Assume A t B for some A, B R. Sice lim = +, M, N 0 N 0, > M A. So + t > M A + t M. 11. (a) Let f(x) = (x + 5)/4. The +1 = f( ). Note that f(x) is a strictly icreasig fuctio because it is liear with slope 1/4. Sice s 1 = 1 < s 2 = 3/2, apply f to s 1 ad s 2 : f(s 1 ) < f(s 2 ), which leads to s 2 < s 3. Iductively, we ca prove that < +1 for all N. So ( ) is icreasig. To show ( ) is bouded, we ca easily prove by iductio that 2 for all N. Sice ( ) is icreasig ad bouded above, it has a limit. Assume lim +1 = lim = L. The L = (L + 5)/4, so L = 5/3. (b) Similar to (a), sice s 2 = 7/4 < s 1 = 2, we ca apply f to both, f(s 2 ) < f(s 1 ), so s 3 < s 2. Keep applyig f, ad we ca see iductively that decreasig. To show ( ) is bouded, observe easily that 0 for all N. > +1 for all N. So ( ) is Sice ( ) is decreasig ad bouded below, it has a limit. Assume lim +1 = lim = L. The L = (L + 5)/4, so L = 5/3. (c) f(x) = 2x + 1 is a icreasig fuctio, ad +1 = f( ). Sice s 1 = 2 < s 2 = 5, we ca show iductively that s 1 < s 2 < s 3 <... by repeatedly applyig f. So ( ) is icreasig. ( ) is bouded: We ca easily show that < 4, by iductio. Sice ( ) is icreasig ad bouded above, it has a limit. Assume lim +1 = lim = L. The L = 2L + 1, so L = Note that a sequece is Cauchy iff it is coverget. (a) = ( 1) = ( 1, 1 2, 1 3,..., ( 1),...). The () is coverget, hece Cauchy, but it is ot mootoe.

7 (b) =. The ( ) is icreasig but ot coverget, hece ot Cauchy. (c) s b = ( 1). The ( ) is bouded but ot coverget, hece ot Cauchy. 13. (a) The statemet is false. Cosider a = 2 ad b = 6. The both (a ) ad (b ) are mootoe. However iot mootoe. (a + b ) = ( 5, 8, 9, 8, 5, 0,...) (b) The statemet is false. Cosider a = b = 3. The (a ) = (b ) = ( 2, 1, 0, 1, 2,...) are mootoe. But iot mootoe. (a b ) = (4, 1, 0, 1, 4,...) 14. Let f(x) = 6 + x. The +1 = f( ). Also, f is a icreasig fuctio. Sice s 1 = 6 < s 2 = 6 + 6, f(s1 ) < f(s 2 ), so s 2 < s 3. Keep applyig f, ad by iductio, < +1. So ( ) is icreasig. For boudedess, we ca easily see (ad prove by iductio) that < 3. Sice ( ) is icreasig ad bouded above, it has a limit. Let lim = lim +1 = L, the L = 6 + L, L = View x as a costat. Let f(t) = t2 + x, f : R + R, the +1 = f( ). 2t f (t) = t2 x 2t 2 : f decreases o (0, x), icreases o ( x, ), ad has a miimum (= f( x) = x) at t = x. Let s 1 = k > 0, the s 2 = f(s 1 ) x because x is the miimum of f. By iductio, > x, so ( ) is bouded below. We wat to show that (s 2, s 3, s 4,...,,...) is decreasig: +1 = 1 2 ( x ) = x s Sice ( ) is decreasig ad bouded, it has a limit. Assume lim = L, the L = L2 + x 2L, so L = x.

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