Revision Flashcards
What is the Least Upper Bound Axiom?
Any non-empty set of real numbers that is bounded above has a least upper bound.
What is Bernoulli’s Inequality?
(1 + x)^k ≥ 1 + kx, x > −1
How do you prove there is a real number whose square is 2?
Consider the set of real numbers whose square is less than 2. Let r = Sup (S) and so either the square of r is <2, =2 or >2. Work to exclude the first and last possibility by showing that r cannot be the LUB, by taking a positive ε and finding r + ε to be in S for case 1 and r - ε to be an upper bound in case 3. This leaves only r^2 = 2.
How do rational and irrational add and multiply.
Given any irrational number x and any rational number p/q, x + (p/q) and xp/q are both irrational. To see this, we argue by contradiction: if x + p/q = p′/q′ , then x = p′/q′ − p/q = (p′q − pq′)/qq′ ∈ Q, contradicting the fact that x is irrational; and similarly if x(p/q) = p′/q′ , then x = p′q/pq′ ∈ Q.
What is the Archimedean Property?
For any real number r, there exists n ∈ N such that n > r.
What is the proof of the Archimedean Property?
If this does not hold then there is some real number r such that n < r for all n ∈ N, i.e. the set N of all natural numbers would be bounded above. In this case, as a non-empty subset of the real numbers that is bounded above, it would have a least upper bound R. Given any n ∈ N, n + 1 is also a natural number, so for every natural number n, n + 1 ≤ R. But this implies that n ≤ R−1 for every natural number n. So R−1 is an upper bound for N, contradicting the fact that R is the least upper bound.
How are rationals and irrationals interleaved?
Between any two real numbers a < b there is a rational number and an irrational number.
Proof that rationals and irrationals are interleaved.
First we find a rational number between a and b.
If a < 0 < b then we can take x = 0. If the result holds for 0 ≤ a < b then it also holds for a < b ≤ 0 (find x with −b < x < −a and then take −x with a < −x < b).
So we take 0 ≤ a < b with a, b ∈ R. Choose n ∈ N such that 1/n < b − a, and let m be the smallest integer such that m > an. Then m − 1 ≤ an and an < m, which means that a < m/n ≤ a + (1/n) < a + (b − a) = b, since we choose n to satisfy 1/n < b − a. So m/n is a rational number that lies between a and b. Now we show that there is an irrational number between a and b. We use the previous result to find a rational number x such that a/√ 2 < x < b/√ 2 . We showed above that if x is rational and y is irrational then xy is irrational, so x√ 2 is irrational and a < x√ 2 < b, as required.
What are the triangle inequalities?
For any two real numbers a and b,
(i) |a + b| ≤ |a| + |b| (triangle inequality);
(ii) |a − b| ≥ ||a| − |b|| (reverse triangle inequality).
What is the definition of a convergent sequence?
A sequence (an) converges to a limit l ∈ R if for every ε > 0 there exists N ∈ N such that |an − l| < ε for every n ≥ N.
To show that a sequence converges, you have to show that for any ε > 0 you can find an appropriate N so that (2.1) holds.
What is the Uniqueness of Limits Lemma?
A sequence can have at most one limit. This means that we can write limn→∞ an = l if an → l.
What is the proof of the Uniqueness of Limits Lemma?
We suppose that an → l1 and an → l2 and show that we must have l1 = l2. Assume that l1 ≠ l2 and deduce a contradiction. Suppose that l2 > l1. Take ε = (l2 − l1)/2. Since an→l1, there exists N1 such that |an − l1| < ε for every n ≥ N1; in particular, an > l1 − (l2 − l1)/2 = (l1 + l2)/2 for all n ≥ N1. Since an → l2, there exists N2 such that |an − l2| < ε for every n ≥ N2; in particular, an < l2 + (l2 − l1)/2 = (l1 + l2)/2 for all n ≥ N2. If we take n ≥ max(N1, N2) it follows that we must have both an > (l1 + l2)/2 and an < (l1 + l2)/2, which is impossible; so we must have l1 = l2.
What is the Shift Rule?
For any fixed k ∈ N, an → l as n → ∞ if and only if an+k → l as n → ∞.
Proof of the Shift Rule?
Given ε > 0, choose N such that |an − l| < ε for all n ≥ N. Then, if we take n > N, we have n + k > n > N, so |an+k − l| < ε also. Conversely, given ε > 0, find M such that |am+k − l| < ε for all m ≥ M. Then if we take N = M + k we have |an − l| < ε for all n ≥ N.
Prove that any convergent sequence is bounded?
Lemma 2.6. Any convergent sequence is bounded.
Proof. Take a sequence (an) such that an → a. Then, taking ε = 1 in the definition of convergence, there exists N such that |an − a| < 1 n ≥ N. So we have |an| = |(an − a) + a| ≤ 1 + |a| n ≥ N. Set A = max(|a1|, . . . , |aN−1|, |a| + 1) and then |an| ≤ A for all n ∈ N.
Prove that if an → a then |an| → |a|.
Proof. Using the reverse triangle inequality ||an| − |a|| ≤ |an − a|. Given any ε > 0 choose N such that |an − a| < ε for all n ≥ N; then ||an| − |a|| < ε for all n ≥ N too.
How does a sequence converge if it is bounded by another sequence?
Suppose that |an − l| ≤ |bn|, where bn → 0 as n → ∞. Then an → l.
What are the Algebra of Limits?
Suppose that an → a and bn → b. Then:
(i) an + bn → a + b;
(ii) anbn → ab;
(iii) if b ≠ 0 then an/bn → a/b.
Prove that if an ≤ bn for all n, an → a and bn → b, then a ≤ b.
Suppose that a > b. Choose ε = (a − b)/2; then there exists N such that for all n ≥ N, |an − a| < (a − b)/2 and |bn − b| < (a − b)/2 . So then a − an ≤ |an − a| < (a − b)/2 ⇒ an > a − (a − b)/2 = (a + b)/2 and bn − b ≤ |bn − b| < (a − b)/2 ⇒ bn < b + (a − b)/2 = (a + b)/2 . This contradicts the assumption that an ≤ bn for all n.
What is the Sandwich Rule?
Suppose that an ≤ bn ≤ cn, where an → l and cn → l. Then bn → l.
Prove the Sandwich Rule
Take ε > 0. Since an → l there exists N1 such that for all n ≥ N1, l − ε < an < l + ε; since cn → l there exists N2 such that for all n ≥ N2, l − ε < cn < l + ε. If we take n ≥ N := max(N1, N2) then l − ε < an ≤ bn ≤ cn < l + ε, i.e. |bn − l| < ε, which shows that bn → l.
Prove that for any x > 0, x 1/n → 1 as n → ∞.
First we take x > 1 and write x^(1/n) = 1 + hn. Since x^(1/n) > 1 when x > 1, we know that hn > 0. So we have x^(1/n) = (1 + hn)^n ≥ 1 + nhn > nhn ⇒ hn < x/n . (2.3) Since we have 0 < hn < x n , it follows by the Sandwich Rule that hn → 0 as n → ∞, and so (by sum rule for limits, since x^(1/n) = 1 + hn) x^(1/n) → 1 as n → ∞. For x < 1 we can write x^(1/n) = (1/(1/x))^(1/n), and use the fact that (1/x)^(1/n) → 1 as n → ∞ and the quotient rule for limits.
How does n^(1/n) converge?
n^(1/n) → 1 as n → ∞.
What does it mean for a sequence to tend to infinity?
We say that an → ∞ as n → ∞ if for every R ∈ R there exists N such that an > R for every n ≥ N.
Prove that if an → ∞ as n → ∞ then 1/an → 0 as n → ∞.
Given ε > 0, since an → ∞ there exists N such that an > 1/ε for all n ≥ N. Then 0 ≤ 1/an < ε for all n ≥ N, so in particular |1/an| < ε for all n ≥ N and so 1/an → 0.
What is the Comparison Test for Limits?
If an ≥ bn and bn → ∞ as n → ∞, then an → ∞ as n → ∞.
What is the Ratio Test for Limits?
Suppose that (an) is a sequence of positive terms, and that limn→∞ (an+1)/an = r. If r < 1 then an → 0, and if r > 1 then an → ∞. If r = 1 then we cannot deduce anything: consider the three sequences an = 1/n for each n, an = 1 for all n, and an = n for each n. In all three cases an+1/an → 1 as n → ∞. (We will see similar issues later.)
What are the known/example limits for the Ratio Test?
We have the following limits as n → ∞:
(i) |x^n → 0 if |x| < 1 and |x|^n → ∞ if |x| > 1;
(ii) for α ∈ Q, n^α → ∞ if α > 0 and n^α → 0 if α < 0;
(iii) x^n/n^k → ∞ if x > 1 (and converges to 0 if 0 ≤ x ≤ 1);
(iv) x^n/n! → 0; and
(v) n!/n^n → 0
What does it mean for a sequence to be increasing, decreasing and monotonic?
A sequence (an) is increasing if a1 ≤ a2 ≤ a3 ≤ · · · (in general an+1 ≥ an) - the terms increase or stay the same. [If an+1 > an we say that the sequence is strictly increasing.] A sequence (an) is decreasing if a1 ≥ a2 ≥ a3 ≥ · · · (i.e. an+1 ≤ an); the terms decrease or stay the same. [If an+1 < an for all n then we say that the sequence is strictly decreasing.] A sequence is (strictly) monotonic if it is (strictly) increasing or (strictly) decreasing. [Note that a constant sequence is both ‘increasing’ and ‘decreasing’.]
What does it mean for a sequence to be bounded?
A sequence (an) is bounded above if there exists M ∈ R such that an ≤ M for all n, and bounded below if there exists K ∈ R such that an ≥ K for all n. [This is the same as the set {an} being bounded above or below.]
What is the theorem for convergence of an increasing sequence when it is/is not bounded?
If (an) is increasing then:
(i) if it is bounded above then it converges
(ii) if it is not bounded above then an → ∞.
Prove the theorem for convergence of an increasing sequence when it is/is not bounded?
Consider the set A = {an : n ∈ N}. This is non-empty and bounded above (by assumption), so (by the Least Upper Bound Axiom) it has a least upper bound; we set l = sup A. We will show that an → l as n → ∞. Since l is an upper bound for all elements of A, an ≤ l for every n ∈ N. (3.1) Now take ε > 0. Since l is the least upper bound, there exists an element a of A such that a > l − ε, i.e. there exists N such that aN > l − ε. Since (an) is increasing, we know that if n ≥ N then an ≥ aN > l − ε. Combining this with (3.1) it follows that for all n ≥ N we have |an − l| < ε, which shows that an → l as n → ∞. (ii) Choose R ∈ R. Since (an) is not bounded above there exists N such that aN > R. Now since (an) is increasing, it follows that an > R for all n ≥ N; so an → ∞.
