Download MAT 371 - Test 1 Solution

MAT 371 - Test 1 Solution
1. Define the following terms:
(a) A sequence {an }∞
n=1 converges to A
For all > 0, there exists N ∈ Z+ such that for all n ≥ N , |an − A| < .
(b) If A ⊆ R, the supremum or sup of the set A
A number s is sup(A) if s is an upper bound for A, and if for every upper bound
b of A we have s ≤ b.
(c) Cauchy sequence
+
A sequence {an }∞
n=1 is Cauchy if for all > 0 there exists N ∈ Z such that for
all n, m ∈ Z such that n ≥ N and m ≥ N , |an − am | < .
2. State the following:
(a) The least upper bound property
Every non-empty set of real numbers which is bounded above has a least upper
bound.
(b) Bolzano-Weierstrass theorem
Ever bounded infinite set of real numbers has at least one accumulation point.
3. Give an example of each of the following. You do not have to give proofs that your
examples work.
(a) An infinite set with no accumulation points
The integers Z
(b) An infinite set with exactly one accumulation point
{ n1 | n ∈ Z+ }
(c) A bounded sequence which does not converge
an = (−1)n
(d) (Extra credit) A bounded sequence whose range has exactly one accumulation point
but which does not converge.
The sequence
(
5 n is odd
an = 1
n is even
n
So the sequence starts 5, 21 , 5, 41 , 5, 16 , . . .. The key is that the sequence has to hit
one value infinitely often, and that value has to be different from the accumulation
point.
4. Suppose a, b ∈ R with a < b. Prove that the open interval (a, b) is a neighborhood of
every real number x ∈ (a, b).
Let x ∈ (a, b), so a < x < b. Then x−a > 0 and b−x > 0. Let = min(x−a, b−x) > 0.
Then (x − , x + ) is a neighborhood of x. If y ∈ (x − , x + ), then x − < y < x + .
But x − ≥ x − (x − a) = a and x + ≤ x + (b − x) = b. So a < y < b. Thus
(x − , x + ) ⊆ (a, b).
∞
5. Suppose {an }∞
n=1 and {bn }n=1 are sequences which converge to A and B respectively.
Prove {an + bn }∞
n=1 converges to A + B.
+
Let > 0. Since {an }∞
such that n ≥ N1
n=1 converges to A, there exists N1 ∈ Z
converges
to
B,
there
exists N2 ∈ Z+
implies |an − A| < /2. Similarly, since {bn }∞
n=1
such that n ≥ N2 implies |bn − B| < /2. Let N = max(N1 , N2 ). Then for all n ≥ N ,
|(an + bn ) − (A + B)| = |(an − A) + (bn − B)| ≤ |an − A| + |bn − B| < /2 + /2 = .
Thus, {an + bn }∞
n=1 converges to A + B.
6. Prove that the sequence given by
an =
3n3 + 2n
n3 − sin(n)
converges to 3. You may use any theorems from the course, but it is important that
the logic of your proof is sound.
First note that since n ∈ Z+ ,
an =
3 + 2/n2
3n3 + 2n
=
.
n3 − sin(n)
1 − sin(n)/n2
· n1 , so lim 1/n2 = 0 · 0 = 0. Since lim 2 = 2,
n→∞
n→∞
1
we again use limit of the product to get lim 2 · 2 = 2 · 0 = 0. So, in the numerator
n→∞
n
(using lim 3 = 3 and limit of a sum) we get lim 3 + 2/n2 = 3.
Now, we know lim 1/n = 0 and
n→∞
n→∞
1
n2
=
1
n
n→∞
Now | sin(x)| ≤ 1 for all x, so − sin(n) is a bounded sequence. We saw above that
1
{1/n2 }∞
n=1 converges to 0. So − sin(n) · n2 converges to 0. Using limit of a sum (and
that lim 1 = 1) we have lim 1 − sin(n)/n2 = 1. Finally, we apply limit of a quotient
n→∞
n→∞
3 + 2/n2
to get lim
= 3.
n→∞ 1 − sin(n)/n2
7. Working from the definition, prove that every Cauchy sequence is bounded.
Let {an }∞
n=1 be a Cauchy sequence. Using = 1 in the definition of Cauchy sequence,
there exists N ∈ Z+ such that for all m, n ≥ N , |an − am | < 1. In particular, for all
n ≥ N , |an − aN | < 1, and so (for all n ≥ N )
aN − 1 < an < aN + 1
There are only finitely many terms ai with 1 ≤ i < N , so those terms are bounded.
Combining with the bounds for an with n ≥ N , we get that {an }∞
n=1 is bounded.