MIDTERM REVIEW FOR MATH The limit

Transcription

1 MIDTERM REVIEW FOR MATH 500 SHUANGLIN SHAO. The limit Define lim n a n = A: For any ε > 0, there exists N N such that for any n N, a n A < ε. The key in this definition is to realize that the choice of N depends on ε. We explain it by using an example. lim n n 3n + = 3. Proof. For any ε > 0, we need to find N N such that for any n N, a n 3 = n 3n + 3 < ε. Here n 3n + 3 = 6n 6n 3(3n + ) = 3(3n + ) < ε. That is to say 3n + > 3ε, n > 9ε 3. Roughly speaking, we do the following 9ε 3 is the choice of N. But N is a natural number, so if 9ε 3 0, take N = ; if 9ε 3 > 0, take N = [ 9ε 3 ] +, where [x] is the integer part of x, which denotes the greatest integer less than or equal to x. Having the choice of the N, we conclude, for any ε > 0, there exists N N, such that for any n N, a n 3 < ε.

2 .. Remark. One can apply the limit theorems to obtain the limit directly. n lim n 3n + = lim n 3 + = = n 3 + lim n 3. n But if you are asked to to evaluate the limit by using the ε N language, you can not apply the limit theorems directly.. The completeness Axiom and the least upper bound Let us first recall the definition of the least upper bound for a set A: we say L is the least upper bound of A if L is an upper bound of A, i.e., for any x A, x L. L is the least one among the upper bounds, i.e., for any positive number δ, if we lower L by δ to L δ, then L δ will not be an upper bound of A. The completeness Axiom says that, every nonempty subset of R attains its least upper bound. For instance, A = (, ). The least upper bound is. In class, we provide several examples and theorems to explain how we use the completeness Axiom establish some surprising properties of the real numbers, for instance, the Archimedean property of real numbers, and the approximation of any numbers in R by integers, and approximation of real numbers by rational numbers. Here we give an example to establish the monotone convergence theorem: every monotone increasing sequence which is bounded above has a limit. The key of using the completeness Axiom is to construct some set, then we show that this set is nonempty and bounded above. we can conclude that there exists the least upper bound of this set. Proof. Let A = {a n } satisfy the conditions, monotone increasing and bounded above. by the completeness Axiom, there exists the least upper bound L of A: for all n, a n L. For any ε > 0,, L ε is not an upper bound. From the second point, there exists N such that a N L ε. because {a n } is increasing, for any n N, a n a N L ε.

3 On the other hand, a n L. L ε a n L L + ε, for any n N. lim a n = L. n 3. The monotone convergence theorem and the sequence We just state the monotone convergence theorem, every monotone increasing sequence which is bounded above has a limit. The requirement in the test is how to apply this theorem. We give an example, which is from the homework and also the quiz 3. Question. Determine whether the sequence {a n }: a 0 = and a n+ = 3 + an converges or not. If it converges, then find the limit. Proof. We first show that it is monotone increasing. I have seen two ways: the first way is by using the mathematics induction. Since a 0 =, then a = 3 + a 0 = 5. So a 0 a. We assume that a n a n. 3 + an 3 + a n. a n a n+. The second way I talked about it in class. We look at a n+ a n = 3 + a n 3 + a n = (a n a n ) 3 + an a n. The denominator is always positive. We see that the sign of a n+ a n is determined by the previous difference, a n a n. Inductively, we see that it is determined by a a 0, which is positive. So a n+ a n, for all n. We then show that it is bounded. We guess the bound is 3. We also prove it by mathematics induction. a 0 = 3. We assume that a n 3. a n+ = 3 + a n = 3. 3

4 So a n 3 for all n. we apply the limit theorems to get the limit. A = 3 + A, i.e., A A 3 = 0. A = 3 or A =. But A > 0, so A = Series The convergence of a series k= a k is determined by its partial sum, n S n = a k = a + a + a n. k= If {S n } converges, then a k converges and lim n S n = a k. Otherwise, we say that the series diverges. The definition often provides a way to tell whether a series converges or diverges. Another way is to use the Cauchy criterion. a k converges if and only if, for any ε > 0, there exists N N such that for any n N, a n+ + a n+ + a n+m < ε, for any m N. Here I give one example to explain the definition and another example for the Cauchy criterion. Question. Show that k= n p converges if p >. Proof. We look at the partial sum, S n = + p + + n p. Firstly {S n } is increasing because S n+ S n. Secondly, {S n } is bounded. x dx. p S n = + p + + n p + The way is to compare it with the integral x p dx = + p <. So S n converges by the monotone convergence theorem. So a k converges. 4

5 Question. Use the Cauchy criterion to show that n p converges if p. Proof. For any ε > 0, we need to find N such that for any n N, We look at a n+ +a n+ + +a n+m = a n+ + a n+ + + a n+m < ε, for any m N. (n + ) p + + (n + ) p (n + ) + + (n + m). On the other hand, (n + ) (n + )n = n n +,, (n + m) (n + m)(n + m ) = n + m n + m. if we sum them together, a n+ + a n+ + + a n+m n + n + m n + n. we ask n < ε, i.e., ε < n. we may take N = [ ε ] +. Department of Mathematics, KU, Lawrence, KS address: slshao@math.ku.edu 5