4.3 Limit of a Sequence: Theorems
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1 4.3. LIMIT OF A SEQUENCE: THEOREMS Limit of a Sequence: Theorems 4.3. Elementary Theorems In example 76, we used an approximation to simplify the problem a little bit. In this particular example, the approximation was not really necessary, it was more to illustrate a point. Sometimes, if the problem is more complicated, it may be necessary to use such an approximation in order to be able to find the condition n has to satisfy. In other words, when we try to satisfy x n L <, we usually simplify x n L to some expression involving n. LetE (n) denote this expression. This gives us the inequality E (n) <whichwehavetosolve for n. If it is too hard, we then try to find a second expression we will call E (n) such that E (n) <E (n) <. E (n) should be such that solving the inequality E (n) <is feasible and easier. In order to achieve this, several tricks are used. We recall some useful results, as well as some theorems below. Theorem 78 (Bernoulli s inequality) If x, andn is a natural number, then ( + x) n +nx. Theorem 79 (binomial theorem) (a + b) n = a n +na n b+ n (n ) (n ) a n 3 b nab n + b n. 3! Corollary 80 ( + x) n =+nx + nx n + x n. n (n ) x + n (n ) a n b + n (n ) (n ) x ! In particular, when x 0, then( + x) n is greater than any part of the right hand side. For example, we obtain Bernoulli s inequality: ( + x) n +nx. We could also write ( + x) n n (n ) x. And so on. This is useful to get approximations on quantities like 3 n.werewriteitas 3 n =(+) n Theorem 8 Let x be a number such that >0, x <,thenx =0. Proof. See problems at the end of the section. Theorem 8 (squeeze theorem) If a n L, c n L and a n b n c n, then b n L Proof. We need to prove that >0 N : n N = b n L <. Let >0 be given. Choose N such that n N = a n L <or <a n L<. Similarly, choose N such that n N = < c n L <. Let N = max (N,N ).Ifn N then a n b n c n a n L b n L c n L <a n L b n L c n L< <b n L< b n L <
2 0 CHAPTER 4. SEQUENCES AND LIMITS OF SEQUENCES Theorem 83 If a sequence converges, then its limit is unique. Proof. We do a proof by contradiction. Assume that a n L and a n L. Given >0 choose N such that n N = a n L <. Similarly, choose N such that n N = a n L <. Let N =max(n,n ). If n N, then L L = L a n + a n L a n L + a n L < + = By theorem 8, it follows that L L =0, that is L = L. Theorem 84 If 0 <a< then a n 0 Proof. Let x =. Then,x>0 and a =.Forn, a +x a n = ( + x) n by Bernoulli s inequality +nx < nx To show that a n 0, we need to show that a n <, for any whenever n N, that is a n < nx < n> x So, given >0, N,= x will work. Theorem 85 If a sequence converges, then it is bounded, that is there exists a number M>0 such that a n M for all n. Proof. Choose N such that n N = a n L <. By the triangle inequality, we have a n L a n L < Thus, if n N, a n < + L. Let M =max( a, a,..., a N ). Let M = max (M, + L ). Then, clearly, a n <M
3 4.3. LIMIT OF A SEQUENCE: THEOREMS 03 Theorem 86 If a sequence converges, then > 0 N : m, n N = a n a m < Proof. Given >0, wecanchoosen such that n, m N = a n L < and a m L <. Now, a n a m = a n L + L a m = (a n L) (a m L) (a n L) + a m L by the triangle inequality < + = The above theorem simply says that if a sequence converges, then the difference between any two terms gets smaller and smaller. It should also be clear to the reader that if a n L, thensodoesa n+k where k is any natural number. We look at a few more examples, to see how all these results come into play. Example 87 Find lim n + n + We think the limit is. We want to show that for every >0, thereexistsn such that n N = n + n + <. First, we simplify the absolute value. n + n + = n + = n + < n < n if n> So, we see that if n <, which happens when n >, then we will have n + n + <.So,N = will work. Example 88 Find lim n +n 4 We think the limit is 0. We need to prove that for every >0, there exists N such that n N = n +n 4 <. We begin by noticing that n +n 4=n +(n ) >n if n>
4 04 CHAPTER 4. SEQUENCES AND LIMITS OF SEQUENCES Therefore, if n>, n +n 4 = n +n 4 < n ( If n>max, ),then < n n +n 4 (, <.So,N =max ) will work. Example 89 Find lim 4n n We suspect this sequence diverges to infinity. We need to prove that 4n n grows without bounds, that is for every M>0, there exists N such that n N = 4 n n >M.First,wenoticethat 4 n ( + 3)n = n n n (n ) 3 > by using the binomial theorem n 9(n ) > 4 We can see that 9(n ) >M n > 4M 4 9 4M +9 n> 9 4M +9 so, N = will work. 9 Example 90 Find lim cos nπ We suspect the sequence diverges, as its values will oscillate between - and. We can actually prove it using theorem 86. We notice that for any n, cos nπ cos ((n + )π) =. For the sequence to converge, this difference should approach 0. Hence, the sequence diverges Operations with Limits The theorems below are useful when finding the limit of a sequence. Finding the limit using the definition is a long process which we will try to avoid whenever possible. Since all limits are taken as n, in the theorems below, we will write lim a n for lim n a n.
5 4.3. LIMIT OF A SEQUENCE: THEOREMS 05 Theorem 9 Let (a n ) and (b n ) be two sequences such that a n a and b n b with a and b real numbers. Then, the following results hold:. lim (a n ± b n )=(lima n ) ± (lim b n )=a ± b. lim (a n b n ) = (lim a n ) (lim b n )=ab ( ) an 3. if lim b n = b 0then lim = lim a n = a lim b n b 4. lim a n = lim a n = a 5. if a n 0 then lim a n 0 b n 6. if a n b n then lim a n lim b n 7. if lim a n = a 0 then lim a n = lim a n = a Proof. We prove some of these items. The remaining ones will be assigned as problems at the end of the section.. We prove lim (a n + b n )=(lima n )+(lim b n ).Theproofoflim (a n b n )= (lim a n ) (lim b n ) is left as an exercise. We need to prove that >0, N : n N = a n + b n (a + b) <. Let >0 be given, choose N such that n N = a n a <. Choose N such that n N = b n b <.LetN =max(n,n ).Ifn N, then a n + b n (a + b) = (a n a)+(b n b) a n a + b n b by the triangle inequality < + =. We need to prove that >0, N : n N = a n b n ab <. Since a n converges, it is bounded, let M be the bound i.e. a n <M. Choose N such that n N = a n a < ( b +). Choose N such that n N = b n b < (M +).LetN = max (N,N ).Ifn N then 3. See problems a n b n ab = a n b n a n b + a n b ab = a n (b n b)+b (a n a) a n b n b + b a n a <M (M +) + b ( b +) < + =
6 06 CHAPTER 4. SEQUENCES AND LIMITS OF SEQUENCES 4. We need to prove that a n a that is > 0, N : n N = a n a <. Let > 0 be given, choose N such that n N = a n a <(since a n a).ifn N, thenwehave: a n a < a n a by the triangle inequality < 5. We prove it by contradiction. Assume that a n a<0. Choose N such that n N a n a < a.then, a n a< a a n < a a n < 0 which is a contradiction. 6. Weapplytheresultsfoundinpartsand5tothesequencea n b n. 7. See problems Remark 9 Parts, and 3 of the above theorem hold even when a and b are extended real numbers as long as the right hand side in each part is defined. You will recall the following rules when working with extended real numbers:. + = =( )( ) =. =( ) = ( ) = 3. If x is any real number, then (a) + x = x + = (b) + x = x = x (c) = x =0 (d) x { if x>0 0 = if x<0 { if x>0 (e) x = x = if x<0 { if x>0 (f) ( ) x = x ( ) = if x<0 4. However, the following are still indeterminate forms. Their behavior is unpredictable. Finding what they are equal to requires more advanced techniques such as l Hôpital s rule.
7 4.3. LIMIT OF A SEQUENCE: THEOREMS 07 (a) + and (b) 0 and 0 (c) and 0 0 Theorem 93 Suppose that (a n ) converges. Then, any subsequence (a nk ) also converges and has the same limit. Proof. Suppose that a n L. Letb k = a nk be a subsequence. We need to prove that given >0, thereexistsn such that k N = b k L <. Let >0 be given. Choose N such that n k N = a nk L <. Now, if k N, then n k N therefore b k L = a nk L < This theorem is often used to show that a given sequence diverges. To do so, it is enough to find two subsequences which do not converge to the same limit. Alternatively, once can find a subsequence which diverges. Example 94 Study the convergence of cos nπ The subsequence cos nπ converges to, while the subsequence cos (n + )π converges to. Thus,cos nπ must diverge. Example 95 Find lim ( n + n ) lim ( n + n ) =lim ( n + n )( n ++ n ) ( n ++ n ) =lim n ++ n = lim ( n ++ n ) =0 Remark 96 In many proofs or problems, different versions of the triangle inequality are often used. As a reminder, here are the different versions of the triangle inequality students should remember. a b a b a + b and a b a + b a + b
8 08 CHAPTER 4. SEQUENCES AND LIMITS OF SEQUENCES Exercises. Prove that 5n n! 0.. Prove that n! n n Consider (x n ) a sequence of real numbers such that lim x n = x where n x>0, prove there exists an integer N>0such that n N = x n > Consider (x n ) a sequence of real numbers such that x n 0 for any n. If x is a partial limit of (x n ),provethatx 0. Prove the same result if x =limx n. Use this to show that if x n y n then lim x n lim y n. 5. Consider (x n ) a sequence of real numbers and let x R. Prove the two conditions below are equivalent. (a) x n x as n. (b) x n x 0 as n. 6. Consider (x n ) a sequence of real numbers and let x R. If lim x n = x, prove that lim x n = x. 7. Consider (x n ) a sequence of real numbers and let x R. Suppose that lim x n = x. Definey n = x n+p for some integer p. Provethatlim y n = x. 8. Given that a n b n for every n and that lim a n =, prove that lim b n =. 9. Suppose that (a n ) and (d n ) are sequences of real numbers such that a n b n < for every n. Prove that if is a partial limit of (a n ) then is also a partial limit of (b n ). 0. Two sequences are said to be eventually close if >0, N >0:n N = a n b n <. (a) Prove that if two sequences (a n ) and (b n ) are eventually close and if anumberx is the limit of the sequence (a n ) then x is also the limit of the sequence (b n ). (b) Prove that if two sequences (a n ) and (b n ) are eventually close and if anumberx is a partial limit of the sequence (a n ) then x is also a partial limit of the sequence (b n ).. Determine if the sequences below have limits. In each case, if a limit L exists, given >0, findn such that n N = x n L <. (a) n n +
9 4.3. LIMIT OF A SEQUENCE: THEOREMS 09 (b) 3n + n + (c) 5n3 + n 4 n 3 +3 n 3 (d) n + (e) 4n + n 3 + n (f) n + n n n ( (g) ( ) n ) (h) n ( )( 3 )( 4 ) (... ) n (i) 5n n (j) Prove that if a> then a >a> a> (k) Prove that n n n3 n4 0, 0, 0, n n n n 0. Prove theorem 8 3. Finish proving theorem 9 4. Let (x n ) and (y n ) be two sequences. Assume further that lim x n = x and y is a partial limit of (y n ).Provethatx+y is a partial limit for (x n + y n ). 5. State and prove similar results for subtraction, multiplication and division. 6. Give an example of two divergent sequences (x n ) and (y n ) such that (x n + y n ) is convergent. 7. Give an example of two sequences (x n ) and (y n ) such that x n 0, y n and: (a) x n y n 0 (b) x n y n c where 0 <c< (c) x n y n (d) (x n y n ) is bounded but has no limit. 8. Let (x n ) be a sequence of real numbers such that x n 0. Provethat x + x x n n 0
10 0 CHAPTER 4. SEQUENCES AND LIMITS OF SEQUENCES 9. Let (x n ) be a sequence of real numbers such that x n x for some real number x. Prove that x + x x n n x 0. Showbyexamplesthatif x n x then x n does not necessarily converge. If it does converge, it does not necessarily converge to x.. Prove that x n 0 x n 0.. Show by examples that if (x n ) and (y n ) are divergent sequences then (x n + y n ) is not necessarily divergent. Do the same for (x n y n ) 3. Show that if (x n ) 0 and (y n ) is bounded then (x n y n ) Prove that if (x n ) is a sequence in a set S R, then every finite partial limit of (x n ) must belong to S. 5. Prove that if S R and x S then there exists a sequence (x n ) in S such that x n x. 6. Let S R. Prove that the following conditions are equivalent: (a) S is unbounded above. (b) There exists a sequence (x n ) in S such that x n.
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