JB

Question 1

Use sign analysis to find the solution set for x^2 > 9

Answer 1

Re arrange, to =0. Then make a table of f(x) and factors of f(x) (LHS) and on the top have important values of x.
Then depending on whether f(x) is =ve,-ve or 0 and what the equality is you’ll have your solution set.

Question 2

|xy| =?

Answer 2

|x|.|y|

Question 3

What’s the triangle inequality?

Answer 3

For any a,b ∈ R,

|a+b| ≤ |a|+|b|.

Question 4

Prove the triangle inequality

Answer 4

For a,b ∈ R,
ab≤|ab|, so that ab≤|a||b|
Thus a^2 + 2ab + b^2 ≤ a^2 + 2|a||b| + b^2 = |a|^2 + 2|a||b| + |b|^2 
Factored = |a+b|^2 ≤ (|a|+|b|)^2
Taking the positive square root 
|a+b| ≤ |a|+|b|
QED

Question 5

What’s the definition for a sequence of real numbers tending to infinity?

Answer 5

Given any real number A>0 there exists N∈Nat such that an > A for all n>N.
or
∀A > 0 ∃N ∈ N s.t. an > A ∀n > N.

Question 6

Prove that the sequence (an) given by an = √n tends to infinity.

Answer 6

Let A > 0 be given. We are required to find an N ∈ Nat s.t √n > A for all n > N.
Notice that if n > A^2 then √n > A. This suggests choosing N to be any natural
number greater than or equal to A^2. With this choice of N we have
n > N =⇒ n > A^2 =⇒ √n > A,
as required.

Question 7

Prove that the sequence (an), given by
an = n^2 + n/(n + 5)
tends to infinity.

Answer 7

(†) an = n2 + n/(n + 5) ≥ n^2/(n + 5) ≥n^2/(n + 5n) = n/6
for all n ∈ Nat.
Let A > 0 be given, and choose N to be any nat ≥
6A. With this choice of N we have
n > N =⇒ n > 6A =⇒ n/6 > A =⇒ an > A,
as required. Here in the last implication we have used (†).

Question 8

Thm 1.3 - Suppose N0 ∈ N and that (an) and (bn) are sequences of real numbers satisfying
an ≥ bn for all n ≥ N0. ?
If bn → inf, what can we conclude?

Answer 8

an → inf

Question 9

What’s the proof for Thm 1.3 (if an>bn and b->inf)?

Answer 9

Let A > 0. Since bn → inf there exists N′ ∈ N such that
bn > A for all n > N′.
Now let N = max{N′,N0} and observe that by the hypotheses of the theorem,
an ≥ bn > A for all n > N.
Hence an → inf.

Question 10

What is the null sequence test?

Answer 10

If a series ∑(inf,n=1)an converges then an->0 as n->0

Question 11

What is the Comparison test?

Answer 11

Suppose that 0≤an≤bn ∀ n∈ |N.

(1) If ∑(inf,n=1)bn converges then ∑(inf,n=1)an converges.
(2) If ∑(inf,n=1) an diverges then ∑(inf,n=1)bn diverges.

Question 12

What is the Ratio Test?

Answer 12

Let (an) be a sequence of non-negative real numbers s.t
lim(n->inf) a(n+1)/an = r, then the series ∑(inf,n=1)an
(1) converges if r<1
(2) diverges if r>1
if r=1, the ratio test tells us nothing

Question 13

What is the Integral test?

Answer 13

Let f: |R->|R be cts,decreasing, and positive on[0,inf). Then ∑(inf,n=1) f(n) converges iff the sequence of integrals ∫(n,1) f(x)dx converges as n->inf

Question 14

What is the Alternating Series Test?

Answer 14

Suppose the non-negative sequence (an)
(i) is decreasing,
(ii) and converges to 0.
Then a1 - a2 +a3 - a4 +… = ∑(inf,n=1) (-1)^(n+1) an converges.

Question 15

What is the absolute convergence test?

Answer 15

If ∑(inf,n=1)an is absolutely convergent then ∑(inf,n=1)an is convergent in the usual sense.

Question 16

What is the definition of a increasing sequence of real numbers?

Answer 16

if a(n+1)≥a(n) ∀ n∈ |N.

Question 17

What is the definition of a decreasing sequence of real numbers?

Answer 17

if a(n+1) ≤ a(n) ∀ n∈ |N.

Question 18

What is the definition of a strictly increasing sequence of real numbers?

Answer 18

if a(n+1) > a(n) ∀ n∈ |N.

Question 19

What is the definition of a strictly decreasing sequence of real numbers?

Answer 19

if a(n+1) < a(n) ∀ n∈ |N.

Question 20

State MCT

Answer 20

(i) If (an) is increasing and bounded above, then (an) converges.
(ii) If (an) is decreasing and bounded below, then (an) converges.

Question 21

What does it mean for a sequence to be bounded from below?

Answer 21

an is bounded below if there exists some M ∈ |R such that an ≥ M for all n ∈ N.

Question 22

What does it mean for a sequence to be bounded from above and below?

Answer 22

(an) is bounded if it is both bounded above and below; i.e. there exist M1,M2 ∈ |R such
that M1 ≤ an ≤ M2 for all n ∈ N.