Exercise 2.3.1
(a) Let be arbitrary. Since , there exists such that for all we have Hence . Therefore, for all , we also haveThus we have .
(b) If by part (a) we are done. If , since for all , by Order Limit Theorem, we have . Since , we have .
Let be arbitrary. Since , there exists such that for all we have Therefore, for all , we have
Exercise 2.3.2
(a) Let be arbitrary. Since , there exists such that for all we have Therefore, for all , we have Therefore .
(b) Let be arbitrary. Since , there exists such that for all we have Since , we get Therefore, for all , we have Here we used that . Therefore .
Exercise 2.3.3
Let be arbitrary.
Since , there exists such that for all we have
Hence for all we have Similarly, since , there exists such that for all we have
Hence for all we have
Note that , therefore by we have for all .
By we have for all .
Combining and , we haveall . Therefore as well.
Exercise 2.3.4
Note that .
(a) By Algebraic Limit Theorem, we have Therefore using Algebraic Limit Theorem again,
(b) By Algebraic Limit Theorem, we have
(c) By Algebraic Limit Theorem and multiplying both numerator and denominator by , we have
Exercise 2.3.5
We first show “if” part. Set . Let be arbitrary. Since , there exists such that for all we have Similarly, since , there exists such that for all we have Let , we show thatfor all . If is odd, then . By we have If is even, then . By we have Therefore is convergent to the same limit of .
We then “show only if” part. Suppose . Let be arbitrary. There exists such that for all . Note that for all we also have and . Hence by we haveand Therefore .
Exercise 2.3.6
We are going to using the following useful formula,
By this formula, we can rewrite as follows,Then since , and by Exercise 2.3.1, we haveHence by the Algebraic Limit Theorem, we have
Exercise 2.3.7
(a) Here is an example. Let be the sequence (1,0,1,0,1,0,……). Let be the sequence (-1,0,-1,0,-1,0,……). Then both and are divergent. However, the sequence is given by . Therefore is convergent.
(b) This is impossible. Suppose and are convergent. Note that therefore by Algebraic Limit Theorem, we have is convergent.
(c) Let , then it is clear that . However, we haveSince is not bounded, therefore by Theorem 2.3.2, is divergent.
(d) This is impossible. Suppose is bounded, namely there exists such that for all , Since is convergent, by Theorem 2.3.2, is bounded, namely there exists such that for all , Hence for all , we haveThus is bounded.
(e) Take to be the sequence (0,0,0,0,……). Take to be the sequence (1,2,3,4,5,……). Clearly, and are convergent but is not since is not even bounded.
Exercise 2.3.8
(a) By the Algebraic Limit Theorem, we can show by induction that if , then for any . Let be a polynomial, where are real numbers. Then by the Algebraic Limit Theorem(b) Consider the function such that and Then take , it is clear that if , then
Exercise 2.3.9
(a) Let be arbitrary. Since is bounded, there exists some such that for all .
Since , there exists such that for all we haveTherefore it follows from and that for all We cannot use Algebraic Limit Theorem as the limit of may not exist.
(b) Note that hence it follows from parts (iii) and (iv) of Algebraic Limit Theorem that is convergent if and only if is convergent. Hence is very important.
(c) Since and exists, therefore is bounded. Then by part (a) we have that as . Note that in order to use part (a), the sequences and here correspond to the sequence and in part (a), respectively.
Exercise 2.3.10
(a) Counterexample. Let and be the same sequence . Then is the sequence . It is clear that . However the limits of and do not exist since they are not even bounded.
(b) We use the following inequality to prove it.
Let be arbitrary. Since , there exists such that for all we have Then by our inequality , for all we haveHence .
(c) Yes. Note thathence by Algebraic Limit Theorem we have
(d) Yes. Note that implies and Since , by Algebraic Limit Theorem we have . Now by the Sequeeze Theorem, we have Hence
Exercise 2.3.11
(a) Let be arbitrary. Since is convergent, by Theorem 2.3.2, is bounded. Hence there exists such that for all . Again since , there exists such that for all we have Now we take an integer , we are going to show that for all , we have . Indeed, for , we have Therefore Since , we have ThereforeHence as well.
(b) Take , then does not converge but .
Exercise 2.3.12
(a) True. Take any element . Since every is an upper bound of , we have for all . Therefore by the Order Limit Theorem, we have Hence . Since is chosen arbitrarily, we see that is an upper bound of .
(b) True. Suppose , then take , then and . Consider the -neighborhood , then. Recall the definition 2.2.3B, if , then there exists such that for all . This implies for all which contradicts our assumption. Hence our assumption that is impossible. Thus is in the complement of .
(c) False. We are going to use Theorem 1.4.3 to construct a counter example. Let which is irrational. By Theorem 1.4.3, for any we can find a rational number such thatThen we show that .
Let be arbitrary. Let , then . For all , we have Hence . By our choice all are rational, but the limit is which is irrational.
Exercise 2.3.13
(a) Note that , hence andSimilarly, , hence and
(b) Yes. It is cleat that . Similarly to part (a), we haveThe three limits are the same.
If , thenHowever does not exist. Suppose the limit exists. Taking , we have . Hence this sequences has infinitely many one-half’s. By the negation of Exercise 2.2.4(b), the limit has to be . Taking , then we have . Hence this sequences has infinitely many two-fifth’s. By the negation of Exercise 2.2.4(b), the limit has to be . But this limit should be unique, hence we get a contradiction and we are done.