Math 311 W08
Day 3
1. Read: 2.1 Page 25
2. Definition: A sequence is a real-valued function whose domain is the natural numbers.
3. Notation: Rather than using f(n) notation for sequence, we use and denote the entire sequence by {an}. And sometimes we just list out the elements of the range, like {1, 2, 3, 4, 5, …}
4. Examples:
{1/n} also known as {1, 1/2, 1/3, 1/4, 1/5, …}
{1 + (-1)n} ***
also known as {0, 2, 0, 2, …}
an = ***
also known as {, , , , …}
5. Definition: The sequence {an} is said to converge if there exists a real number a (called the limit of an) such that:
For every e>0, there exists a natural number N such that n³N Þ < e
(Picture on the board)
In this case, we often write an ® a.
Examples:
{1/n} converges to 0. Proof: By the Archimedean Property, for any e>0, there exists a natural number N s.t. 1/N <e. Thus if n ³N, = 1/n £ 1/N e. Ö
{1 + (-1)n} does not converge. Proof: Suppose the sequence converged to a ¹0. Then choose e = . Then given any natural number N, there is an odd number larger (e.g. 2N+1). So there is some n ³N such that = =2ee. Similarly, the sequence cannot converge to 0 since (with e =1) for any natural number there is a larger even number (so there is some n ³N where the value of the sequence is 2). Thus this term is not within epsilon of zero. So the sequence has no limit and thus does not converge.Ö
Now each of you try to prove the following: ***
A: and B: sin() does not converge.
Proof for A: Let e>0. By the Archimedean Property we can choose a natural number N such that 1/N < . Then if n ³N, = < <5=e.Ö
Proof for B: First note that the sequence cannot go to zero. Choosing e = ½, note that for any natural number, there is a bigger natural number of the form 4k + 1. For such an index the value of the sequence will be 1. Thus no matter how large we choose our N, we will find terms of the sequence that are not within e ( ½) of 0.
Now, for any other real number a, we choose e = . First note that 0 is not within epsilon of a. Then for any natural number there is a larger natural number of the form 4k. For such an index, the value of the sequence will be 0. Thus no matter how large we choose our N, will find terms that are not within e of a.
Thus no number can be the limit of this sequence, so it does not converge. Ö
6. Comparison Lemma: Suppose an ® a. Then the sequence {bn} converges to the number b if there is a non-negative number C and an index N1 such that £ C for all n ³N1
Proof: Let e>0. Let N2 be such that n ³N1Þ < .
Now let N = N1 + N2. Then if n ³N then £ C< Ce.Ö
7. Note: A lot of times people do not use the Comparison Lemma, but instead use the idea of the proof of the Comparison Lemma in other proofs. This was the case with our example A above. Our proof would have been slightly shorter if we had used the Comparison Lemma.
Proof: 1/n ®0, and for all natural numbers n, = so by the Comparison Lemma, . Ö
8. Definition: The sequence {an} is bounded if there exists some real number M such that £ M for every natural number n.
9. Theorem (Very Useful): Convergent sequences are bounded. ***
Proof: Suppose that an ® a. Choose e = 1. There exists a natural number N such that n ³N Þ < 1. Then by a homework problem (sort of), we know that + 1 for any n ³N. This gives us an upper bound for most of the guys in the sequence. But we don’t know about the first N-1 terms. But there are only finitely many of these things, so let’s just say M = Max {, , …, , + 1} and this is our bound! Ö
10. Theorem (The Bounded Away from Zero Theorem). Sequences that converge to nonzero elements are bounded away from zero in the following sense:
If an ® a and a¹0. Let b be any positive real number such that b <. Then there exists a natural number N such that n ³N Þ > b. (In other words at some point all remaining terms are more than b units away from zero.) ***
Proof: Suppose an ® a and a¹0. Let b >0. Choose e = -b. Then there is a natural number N s.t. n ³N Þ -b. But then + -b. Thus > b.
11. Theorem (Sum Property): Suppose that an ® a and bn ® b.
Then an + bn ® a + b. ***
Proof: Let e>0. Choose N1 such that n ³N1Þ < and choose N2 such that n ³N2Þ < . Let N = max {N1 , N2}. Then n ³NÞ
= £ ++=e. Ö
12. Theorem (Product Property): Suppose that an ® a and bn ® b.
Then an bn ® ab.
13. Questions about the proof:
a. Why did I choose a POSITIVE number M?
b. What is up with the “+1”?
c. Why does each step in the inequality chain work?
14. In the book the author proved a couple of cute little lemmas before proving the Product Property Theorem. Since we did the theorem without the little lemmas we can dismiss these with scorn and derision, beating them into submission with our sledgehammer of a theorem!
Lemma 2.11: If an → a then for any number α, αan → αa.
Lemma 2.12: If an and bn both converge to zero, then so does an bn.
For practice, try to prove these directly (they are in the book so try them without peeking). The book proves 2.11 using the Comparison Lemma. Try to also do it with just the definition of convergence.
15. Theorem (Flippy Property). Suppose that the {bn} is a sequence of nonzero numbers that converge to the nonzero number b. Then .
Proof: (The book uses the Comparison Lemma, I will do it without).
Let e>0. First note that since by the “bounded away from zero theorem”, there is a natural number N1 such that n ³N1 Þ > Þ
Now choose N2 such that n ³N2 Þ .
Then if we let N = Max{ N1 , N2 } then n ³N Þ
= = < < = e.Ö
16. Theorem (Quotient Property): If an → a and{bn} is a sequence of nonzero numbers that converge to the nonzero number b. Then .
Proof: This follows directly from the Flippy Property and the Product Property. You could also prove it directly by using the Bounded Away from Zero Theorem much like the previous proof (with the trick from the proof of the Product Property thrown in).