4.3 Limit of a Sequence: Theorems


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1 4.3. LIMIT OF A SEQUENCE: THEOREMS Limit of a Sequence: Theorems These theorems fall in two categories. The first category deals with ways to combine sequences. Like numbers, sequences can be added, multiplied, divided,... Theorems from this category deal with the ways sequences can be combined and how the limit of the result can be obtained. If a sequence can be written as the combination of several "simpler" sequences, the idea is that it should be easier to find the limit of the "simpler" sequences. These theorems allow us to write a limit in terms of easier limits. however, we still have limits to evaluate. The second category of theorems deal with specific sequences and techniques applied to them. Usually, computing the limit of a sequence involves using theorems from both categories Limit Properties We begin with a few technical theorems. They do not play an important role in computing limits, but they play a role in proving certain results about limits. Theorem 30 Let x be a number such that ɛ > 0, x < ɛ, then x = 0. Proof. See problems at the end of the section. Theorem 3 If a sequence converges, then its limit is unique. Proof. We assume that a n L and a n L and show that L = L. Given ɛ > 0 choose N such that n N = a n L < ɛ. Similarly, choose N such that n N = a n L < ɛ. Let N = max (N, N ). If n N, then L L = L a n + a n L a n L + a n L < ɛ + ɛ = ɛ By theorem 30, it follows that L L = 0, that is L = L. Theorem 3 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 Theorem 33 If a sequence converges, then ɛ > 0 N : m, n N = a n a m < ɛ
2 6 CHAPTER 4. SEQUENCES AND LIMIT OF SEQUENCES Proof. Given ɛ > 0, we can choose N such that n, m N = a n L < ɛ and a m L < ɛ. 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 < ɛ + ɛ = ɛ 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, then so does a n+k where k is any natural number. The above theorem can be used to prove that a sequence does not converge by proving that the difference between two of its terms does not get smaller and smaller. Example 34 Find lim cos nπ We suspect the sequence diverges, as its values will oscillate between  and. We can actually prove it using theorem 33. 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. Theorem 35 Suppose that (a n ) converges. Then, any subsequence (a nk ) also converges and has the same limit. Proof. Suppose that a n L. Let b k = a nk be a subsequence. We need to prove that given ɛ > 0, there exists N such that k N = b k L < ɛ. Let ɛ > 0 be given. Choose N such that n k N = a nk L < ɛ. Now, if k N, then n k N therefore b k L = a nk L < ɛ 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 36 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. In the next two sections, we look at theorems which give us more tools to compute limits.
3 4.3. LIMIT OF A SEQUENCE: THEOREMS Limit Laws 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. Theorem 37 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 ) = (lim a n ) ± (lim b n ) = a ± b.. lim (a n b n ) = (lim a n ) (lim b n ) = ab. 3. if lim b n = b 0 then lim 4. lim a n = lim a n = a. 5. if a n 0 then lim a n 0. ( an b n 6. if a n b n then lim a n lim b n. ) = lim a n lim b n = a b. 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 ) = (lim a n )+(lim b n ). The proof of lim (a n b n ) = (lim a n ) (lim b n ) is left as an exercise. We need to prove that ɛ > 0, N : n N = a n + b n (a + b) < ɛ. Let ɛ > 0 be given, choose N such that n N = a n a < ɛ. Choose N such that n N = b n b < ɛ. Let N = max (N, N ). If n N, then a n + b n (a + b) = (a n a) + (b n b) a n a + b n b by the triangle inequality < ɛ + ɛ = ɛ. We need to prove that ɛ > 0, N : n N = a n b n ab < ɛ. Since a n converges, it is bounded, let M be the bound i.e. a n < M. Choose ɛ N such that n N = a n a < ( b + ). Choose N such that
4 8 CHAPTER 4. SEQUENCES AND LIMIT OF SEQUENCES n N = b n b < ɛ (M + ). Let N = max (N, N ). If n N then 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 ɛ < M (M + ) + b ɛ ( b + ) < ɛ + ɛ = ɛ 3. See problems 4. We need to prove that a n a that is ɛ > 0, N : n N = a n a < ɛ. Let ɛ > 0 be given, choose N such that n N = a n a < ɛ (since a n a). If n N, then we have: a n a < a n a by the triangle inequality < ɛ 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. We apply the results found in parts and 5 to the sequence a n b n. 7. See problems Remark 38 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 + =
5 4.3. LIMIT OF A SEQUENCE: THEOREMS 9 (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. (a) + and (b) 0 and 0 (c) and 0 0 Remark 39 When using theorems from this category, it is important to remember previous results since these theorems allow us to write a limit in terms of other limits, we hopefully know. The more limits we know, the better off we are. Example 30 If c 0, find lim c n. We know from an example in the previous section that lim n = 0. Therefore lim c n = c lim n = c 0 = 0 Example 3 Find lim n. In the previous section, we computed this limit using the definition. We can also do it as follows. lim ( n = lim n ) n ( = lim ) ( lim ) n n = 0 0 = 0 Remark 3 From the above example, we can see that if p is a natural number, lim n p = 0.
6 0 CHAPTER 4. SEQUENCES AND LIMIT OF SEQUENCES Example 33 Find lim n +3n n +. This problem involves using a standard technique you should remember. We show all the steps, then we will draw a general conclusion. We begin by factoring the term of highest degree from both the numerator and denominator. lim n + 3n n + ) ( = lim n + 3 n n ( ) + n = lim + 3 n + n Now, and ( lim + 3 ) n lim ( + n ) = lim () + 3 lim n = = lim () + lim n = Since both the limit of the numerator and denominator exist and the limit of the denominator is not 0, we can write lim n + 3n n = lim ( ) + 3 n + lim ( ) + n = Remark 34 The same technique can be applied to every fraction for which the numerator and denominator are polynomials in n. We see that the limit of such a fraction will be the same as the limit of the quotient of the terms of highest degree. Let us look at some examples: Example 35 lim 3n + n 0 n + 5 = lim 3n n = lim 3 n = 3 lim n = Example 36 lim 5n3 n + 5n3 n 4 + 5n = lim n 4 = lim 5 n = 5 lim n = 0 Example 37 lim n3 n + n + n 3 + 0n More Theorems on Limits = lim n3 n 3 = lim = In example 304, we used an approximation to simplify the problem a little bit. In this particular example, the approximation was not really necessary, it was
7 4.3. LIMIT OF A SEQUENCE: THEOREMS 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 < ɛ, we usually simplify x n L to some expression involving n. Let E (n) denote this expression. This gives us the inequality E (n) < ɛ which we have to solve 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) < ɛ. E (n) should be such that solving the inequality E (n) < ɛ 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 38 (Bernoulli s inequality) If x, and n is a natural number, then ( + x) n + nx. Theorem 39 (binomial theorem) (a + b) n = a n +na n b+ n (n ) (n ) a n 3 b nab n + b n. 3! Corollary 330 ( + 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 or ( + x) n n (n ) (n ) x 3. 3! And so on. This is useful to get approximations on quantities like 3 n. We rewrite it as 3 n = ( + ) n Theorem 33 (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 ɛ > 0 N : n N = b n L < ɛ. Let ɛ > 0 be given. Choose N such that n N = a n L < ɛ or ɛ < a n L < ɛ. Similarly, choose N such that n N = ɛ < c n L < ɛ. Let N = max (N, N ). If n N then a n b n c n a n L b n L c n L ɛ < a n L b n L c n L < ɛ ɛ < b n L < ɛ b n L < ɛ Theorem 33 If 0 < a < then a n 0
8 CHAPTER 4. SEQUENCES AND LIMIT OF SEQUENCES Proof. Let x = a. Then, x > 0 and a =. For n, + x a n = ( + x) n by Bernoulli s inequality + nx < nx To show that a n 0, we need to show that a n < ɛ, for any ɛ whenever n N, that is a n < ɛ nx < ɛ n > xɛ So, given ɛ > 0, N, = xɛ will work. We look at a few more examples, to see how all these results come into play. Example 333 Find lim n + n using the definition + Of course, we can find this limit by using the theorems on limits. Here, we do it using the definition, as asked. We think the limit is. We want to show that for every ɛ > 0, there exists N such that n N = n + n + < ɛ. First, we simplify the absolute value. n + n + = n + = n + < n < n if n > So, we see that if n < ɛ, which happens when n >, then we will have ɛ n + n + < ɛ. So, N = will work. ɛ Example 334 Find lim n + n 4 We think the limit is 0. We need to prove that for every ɛ > 0, there exists N
9 4.3. LIMIT OF A SEQUENCE: THEOREMS 3 such that n N = n + n 4 < ɛ. We begin by noticing that n + n 4 = n + (n ) > n if n > Therefore, if n >, n + n 4 = n + n 4 < n ( If n > max, ), then ɛ < n n + n 4 (, < ɛ. So, N = max ) will work. ɛ Remark 335 The two examples above could have been done using the techniques discussed in the previous subsection, that is without using the definition. Example 336 Prove that lim 4n n! = 0 The trick we use is worth remembering. We will use the squeeze theorem. We note that 0 4n n! = n 43 3! 4 n since n But lim 43 3! 4 n = 0 hence the result follows by the squeeze theorem. Example 337 Find lim n 4 n We suspect the limit will be 0 since the denominator grows much faster than the numerator. So, we need to find a fraction slightly larger than n 4 n which we know converges to 0. Then, we will be able to invoke the squeeze theorem. Using the corollary of the binomial theorem, we note that 4 n = ( + 3) n 9n (n ) thus 0 n 4 n n 9n (n ) = 9n 9 0. lim n 4 n = 0. Hence, by the squeeze theorem, Example 338 Find lim n 4 n This is similar to the problem above. However, the numerator is a higher power
10 4 CHAPTER 4. SEQUENCES AND LIMIT OF SEQUENCES of n, so we need to write 4 n in terms of a higher power of n. Thus, 4 n ( + 3) n 33 n (n ) (n ) 3! 9n (n ) (n ) And lim 0 n 4 n n 9n (n ) (n ) n 9 (n ) (n ) n 9 (n ) (n ) = 0 (why?) hence by the squeeze theorem, lim n 4 n = 0. Example 339 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 > M. First, we notice that n We can see that so, N = 4M n ( + 3)n = n n n (n ) 3 > n 9 (n ) > 4 will work. 9 (n ) 4 by using the binomial theorem > M n > 4M 9 n > 4M Remark 340 We used the binomial theorem to deduce that ( + 3) n n (n ) 3. Often, when using this theorem, students do not know which term to keep, the term in x, or x, or x 3,... The answer is that it depends on what we are trying to achieve. Here, we wanted to show that 4n > M for any M, in other words, n
11 4.3. LIMIT OF A SEQUENCE: THEOREMS 5 4 n can be made as big as we want simply by taking n large enough. Since we n could not solve for n, we replaced 4n by a smaller term. That smaller term n should have similar properties, in other words, the smaller term should also get arbitrarily large. If we take the smaller term too small, it will not work. For example, it is true that 4n n > n. However, n cannot be made as large as we want, so this does not help us. If instead of using ( + 3) n n (n ) 3 (terms of second degree), we had used ( + 3) n + 3n, then we would have obtained + 3n n > M nm < + 3n nm 3n < n (M 3) < n < M 3 So, this does not work. This tells us that +3n n > M only for a few values of n. This is why we used terms of higher degree. Example 34 Find lim ( n + n ) lim ( n + n ) = lim ( n + n ) ( n + + n ) ( n + + n ) = lim n + + n = lim ( n + + n ) = 0 Remark 34 In many proofs or problems, diff erent versions of the triangle inequality are often used. As a reminder, here are the diff erent versions of the triangle inequality students should remember. a b a b a + b and a b a + b a + b Finally, we give a theorem which generalizes some of the examples we did above. Theorem 343 All limits are taken as n.. If p > 0 then lim n p = 0.
12 6 CHAPTER 4. SEQUENCES AND LIMIT OF SEQUENCES. If p > 0 then lim n p =. 3. lim n n =. 4. If p > and α R then lim nα p n = If p < then lim p n = 0 6. p R, lim pn n! = 0. Proof. We prove each part separately.. We can use the definition. Let ɛ > 0 be given. We show there exists N > 0 : n N = n p 0 < ɛ. We begin with what we want to achieve. n p 0 < ɛ n p < ɛ n p > ɛ n > p ɛ So, we see that if N is the smallest ineteger larger than follow. p, the result will ɛ. See problems 3. Let x n = n n. Proving the result amounts to proving that lim x n = 0. From x n = n n, we can write n n = xn + n = (x n + ) n Using the binomial theorem, we see that if n n n (n ) x n Thus x n n It follows that for all n we have 0 x n n Since lim n = 0, it follows using the squeeze theorem that lim x n = 0.
13 4.3. LIMIT OF A SEQUENCE: THEOREMS 7 4. Here again, we will use the binomial theorem. Since p >, we can write p = + q with q > 0. Therefore, if k is a positive integer such that k > α, we have p n = ( + q) n > n (n )... (n k + ) q k k! Now, if n > k, then k < n. It follows that n k + > n + > n. It follows that and therefore Since n (n )... (n k + ) k! k k! lim n 0 nα p n k k! q k n k q k n k = k k! q k = 0 > nk k k! lim n It follows from the squeeze theorem that lim nα p n = 0 n k 5. See problems (hint: write p = ± q α = 0). and use part 4 of the theorem with 6. See problems Exercises. Prove that 5n n! 0.. Prove that n! n n Finish proving theorem Consider (x n ) a sequence of real numbers such that lim n x n = x where x > 0, prove there exists an integer N > 0 such 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 ), prove that x 0. Prove the same result if x = lim x n. Use this to show that if x n y n then lim x n lim y n. 6. Consider (x n ) a sequence of real numbers and let x R. Prove the two conditions below are equivalent.
14 8 CHAPTER 4. SEQUENCES AND LIMIT OF SEQUENCES (a) x n x as n. (b) x n x 0 as n. 7. Consider (x n ) a sequence of real numbers and let x R. If lim x n = x, prove that lim x n = x. 8. Show by examples that if x n x then x n does not necessarily converge. If it does converge, it does not necessarily converge to x. 9. Consider (x n ) a sequence of real numbers and let x R. Suppose that lim x n = x. Define y n = x n+p for some integer p. Prove that lim y n = x. 0. Given that a n b n for every n and that lim a n =, prove that lim b n =.. Suppose that (a n ) and (b 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 ).. Two sequences are said to be eventually close if ɛ > 0, N > 0 : n N = a n b n < ɛ. (a) Prove that if two sequences (a n ) and (b n ) are eventually close and if a number x 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 a number x is a partial limit of the sequence (a n ) then x is also a partial limit of the sequence (b n ). 3. Determine if the sequences below have limits. In each case, if a limit L exists, given ɛ > 0, find N such that n N = x n L < ɛ. (a) n n + (b) 3n + n + (c) 5n3 + n 4 n n 3 (d) n + (e) 4n + n 3 + n (f) n + n n n ( (g) ( ) n ) n
15 4.3. LIMIT OF A SEQUENCE: THEOREMS 9 (h) ( ) ( ) ( ) (... ) 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 4. Prove theorem Finish proving theorem 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 ). Prove that x+y is a partial limit for (x n + y n ). 7. State and prove similar results for subtraction, multiplication and division. 8. Give an example of two divergent sequences (x n ) and (y n ) such that (x n + y n ) is convergent. 9. 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. 0. Let (x n ) be a sequence of real numbers such that x n 0. Prove that x + x x n n 0. 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. Prove that x n 0 x n 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 ) 4. 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.
16 30 CHAPTER 4. SEQUENCES AND LIMIT OF SEQUENCES 6. Prove that if S R and x S then there exists a sequence (x n ) in S such that x n x. 7. 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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