1 More concise proof of part (a) of the monotone convergence theorem.

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1 Math 0450 Honors intro to analysis Spring, 009 More concise proof of part (a) of the monotone convergence theorem. Theorem If (x n ) is a monotone and bounded sequence, then lim (x n ) exists. Proof. (a) Suppose that (x n ) is increasing. Let S = fx n : n N g and let x = sup S, which exists because S is a bounded set. I claim that lim (x n ) = x: To prove this, suppose that " > 0: Then x " is not an upper bound for S; by the denition of supremum. Hence there is an n N such that x " < x n. But x is an upper bound for S; so x " < x n x. Hence jx n xj < ". Let K (") = n : If n K ("), then x n x n x, since we assumed that (x n ) is increasing. Hence jx n xj jx n xj < ". This proves that lim (x n ) = x. Remark: I haven't stated specically that the x n are real numbers; that is understood. Sometimes I wrote \n N ", or n N because it was easy to t in and didn't disrupt the ow. But I avoided the phrase \n is a natural number", because that is awkward. Usually we understand from the context that x is a real number, n a positive integer. If we want to make it clear that negative integers are also included, we could write n Z. If we want only non-negative integers, we could write n 0: Generally, again depending on context, x 0 would be understood to mean real numbers, not just integers. Innite series Denition: Suppose that (x n ) is a sequence of real numbers. Then P x n denotes another sequence, called the \sequence of partial sums" of (x n ), and dened by s = x s n = s n x n for n : P This sequence is also called an \innite series", and may be denoted as well by x n. The series is said to \converge" if the sequence (s n ) converges, according to the usual denition of convergence of a sequence. n=

2 Often, it is assumed that the sequence (x n ) starts with n = 0; and to make this P clear, the series is written as x i. Hence, Example: (x n ) = n, for n 0. Then s n = nx x i = nx i = 4 : n : () s = ; s = 3 ; s 3 = 7 4 : By a formula from chapter, the n th partial sum is s n = n From this it is easy to see that lim (s n ) = : Often this fact is written as X i = : Again, this must be understood as saying that the sequence of partial sums, (s n ), converges to. Example: (pg. 90, Example 3.7.(c) ) Consider the series P convergence or divergence. : n= : Discuss its n(n) I will do this more formally than the text, by using the denition of the sequence of partial sums. For this example, s = s = 6 s 3 = 6 and s n = s n (n ) (n ) :

3 The key is the partial fractions formula Thus, for example, s 3 = n (n ) = n n 3 3 = 4 4 : () (This is a \collapsing sum", or a \telescoping sum".) From this we have a lemma: Lemma s n = n. Proof. Since s = ; the lemma is true for n = : Suppose it is true for n = k: Then, s k = s k (k ) (k ) = s k : k k From the induction hypothesis, s k = k This proves the lemma. k Now the result is obvious: lim (s n ) = : = k k :. The harmonic series. This is the series X n : It is a very important result that this series diverges. sequence of partial sums, then lim (s n ) = (See section 3.6.) Proof: We consider the subsequence More precisely, if (s n ) is the s ( n ) = ; ; 3 4 ; ; :: 3

4 It helps to see what is going on if we group the terms appropriately, so that the last term is s 3 = > : Lemma: s n n. Proof: s = 3 = that s k = s k : Now suppose the formula is true for n = k: Then I claim X k i= k i k k = k k : This is because k k = k ; so that s k has k more terms than s k, and each of those terms is greater than or equal to the last term, which is : This proves k the lemma. The divergence of the series follows because a subsequence is unbounded. Since the sequence of partial sums is monotone increasing, this proves that lim (s n ) =. 3 About the exam The exam will cover through page 90. A problem like the examples on page 90, or like problems 8 and 9 below may be chosen for the exam. Also, problems 3(a), (4) and (5) would be possibilities. The harmonic series is example 3.5.6(c), pg. 84, and so fair game, though I do it dierently above. Regarding earlier material, if you are asked to prove something using the properties of real numbers from chapter, you will be given a list of those properties. If, in chapter 3, you are asked to prove that a sequence converges or diverges \using the denition of convergence," then you must nd K (") or show that no K (") exists. You cannot use the theorems in section 3. or later. 4 Homework This assignment is due on Tuesday, Feb. 7, in advance of the exam on Feb. 8. 4

5 Page 80: # 7(b), 8(b), pg. 86, # 5, 8, 3 pg. 88, # 8. Prove that the sequence (s n ) of partial sums for the sequence (x n ) = n! is a Cauchy sequence, using the denition of Cauchy sequence. Hint: Use problem 4 on page Determine if P n= p n converges. 5