Download MIDTERM REVIEW FOR MATH 500 1. The limit Define limn→∞ an

MIDTERM REVIEW FOR MATH 500
SHUANGLIN SHAO
1. The limit
Define limn→∞ an = A: For any ε > 0, there exists N ∈ N such that for
any n ≥ N ,
|an − A| < ε.
The key in this definition is to realize that the choice of N depends on ε.
We explain it by using an example.
lim
n→∞
2n
2
= .
3n + 1
3
Proof. For any ε > 0, we need to find N ∈ N such that for any n ≥ N ,
2
2n
2
|an − | = |
− | < ε.
3
3n + 1 3
Here
2n
6n − 6n − 2 2
2
3n + 1 − 3 = 3(3n + 1) = 3(3n + 1) < ε.
That is to say
2
2
1
3n + 1 > , ⇒ n >
− .
3ε
9ε 3
2
Roughly speaking, 9ε
− 13 is the choice of N . But N is a natural number, so
we do the following
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• if 9ε
− 13 ≤ 0, take N = 1;
2
2
• if 9ε − 13 > 0, take N = [ 9ε
− 13 ] + 1, where [x] is the integer part of
x, which denotes the greatest integer less than or equal to x.
Having the choice of the N , we conclude,
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for any ε > 0, there exists N ∈ N , such that for any n ≥ N, |an − | < ε.
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1.1. Remark. One can apply the limit theorems to obtain the limit directly.
2n
2
2
2
lim
=
= lim
1 = 3.
n→∞ 3n + 1
n→∞ 3 + 1
3 + limn→∞ n
n
But if you are asked to to evaluate the limit by using the ε—N language,
you can not apply the limit theorems directly.
2. The completeness Axiom and the least upper bound
Let us first recall the definition of the least upper bound for a set A: we say
L is the least upper bound of A if
• L is an upper bound of A, i.e., for any x ∈ A, x ≤ L.
• L is the least one among the upper bounds, i.e., for any positive
number δ, if we lower L by δ to L − δ, then L − δ will not be an
upper bound of A.
The completeness Axiom says that, every√nonempty
subset of R attains its
√
2,
2).
The
least upper bound is
least
upper
bound.
For
instance,
A
=
(−
√
2.
In class, we provide several examples and theorems to explain how we use
the completeness Axiom establish some surprising properties of the real
numbers, for instance, the Archimedean property of real numbers, and the
“approximation” of any numbers in R by integers, and approximation of
real numbers by rational numbers.
Here we give an example to establish the monotone convergence theorem:
every monotone increasing sequence which is bounded above has a limit.
The key of using the completeness Axiom is to construct some set, then we
show that this set is nonempty and bounded above. Then we can conclude
that there exists the least upper bound of this set.
Proof. Let A = {an } satisfy the conditions, monotone increasing and bounded
above. Then by the completeness Axiom, there exists the least upper bound
L of A:
• for all n, an ≤ L.
• For any ε > 0,, L − ε is not an upper bound.
From the second point, there exists N such that aN ≥ L − ε. Then because
{an } is increasing, for any n ≥ N ,
an ≥ aN ≥ L − ε.
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On the other hand, an ≤ L. Then
L − ε ≤ an ≤ L ≤ L + ε, for any n ≥ N.
Then
lim an = L.
n→∞
3. The monotone convergence theorem and the sequence
We just state the monotone convergence theorem, every monotone increasing
sequence which is bounded above has a limit. The requirement in the test is
how to apply this theorem. We give an example, which is from the homework
and also the quiz 3.
Question.
Determine whether the sequence {an }: a0 = 1 and an+1 =
√
3 + 2an converges or not. If it converges, then find the limit.
Proof. We first show that it is monotone increasing. I have seen two ways:
the first
√ using the mathematics induction. Since a0 = 1, then
√ way is by
a1 = 3 + 2a0 = 5. So
a0 ≤ a1 .
We assume that an−1 ≤ an . Then
p
√
3 + 2an−1 ≤ 3 + 2an .
Then
an ≤ an+1 .
The second way I talked about it in class. We look at
p
√
2(an − an−1 )
.
an+1 − an = 3 + 2an − 3 + 2an−1 = √
√
3 + 2an + 3 + 2an−1
The denominator is always positive. We see that the sign of an+1 − an is
determined by the previous difference, an − an−1 . Inductively, we see that
it is determined by a1 − a0 , which is positive. So
an+1 ≥ an , for all n.
We then show that it is bounded. We guess the bound is 3. We also prove
it by mathematics induction.
• a0 = 1 ≤ 3.
• We assume that an ≤ 3. Then
√
√
an+1 = 3 + 2an ≤ 3 + 2 × 3 = 3.
3
So an ≤ 3 for all n.
Then we apply the limit theorems to get the limit.
√
A = 3 + 2A, i.e., A2 − 2A − 3 = 0.
Then
A = 3 or A = −1.
But A > 0, so
A = 3.
4. Series
The convergence of a series
Sn =
P∞
k=1 ak
n
X
is determined by its partial sum,
ak = a1 + a2 + · · · an .
k=1
P
P
If {Sn } converges, then
ak converges and limn→∞ Sn =
ak . Otherwise,
we say that the series diverges. The definition often provides a way to tell
whether a series converges or diverges.
P
Another way is to use the Cauchy criterion.
ak converges if and only if,
for any ε > 0, there exists N ∈ N such that for any n ≥ N ,
|an+1 + an+2 + · · · an+m | < ε, for any m ∈ N .
Here I give one example to explain the definition and another example for
the Cauchy criterion.
Question. Show that
P∞
1
k=1 np
converges if p > 1.
Proof. We look at the partial sum,
1
1
+ ··· + p.
p
2
n
Firstly {Sn } is increasing because Sn+1 ≥ Sn .
Sn = 1 +
Secondly,
{Sn } is bounded. The way is to compare it with the integral
R∞ 1
1 xp dx.
Z ∞
1
1
1
1
Sn = 1 + p + · · · + p ≤ 1 +
dx = 1 +
< ∞.
p
2
n
x
p−1
1
So Sn converges by the monotone convergence theorem. So
4
P
ak converges.
Question. Use the Cauchy criterion to show that
P
1
np
converges if p ≥ 2.
Proof. For any ε > 0, we need to find N such that for any n ≥ N ,
|an+1 + an+2 + · · · + an+m | < ε, for any m ∈ N .
We look at
|an+1 +an+2 +· · ·+an+m | =
1
2
1
1
+· · ·+
≤
+· · ·+
.
p
p
2
(n + 1)
(n + 1)
(n + 1)
(n + m)2
On the other hand,
1
1
1
1
1
1
1
1
≤
= −
,··· ,
≤
=
−
.
(n + 1)2
(n + 1)n
n n+1
(n + m)2
(n + m)(n + m − 1)
n+m−1 n+m
Then if we sum them together,
1
1
1
1
−
≤
≤ .
|an+1 + an+2 + · · · + an+m | ≤
n+1 n+m
n+1
n
Then we ask
1
1
< ε, i.e., < n.
n
ε
Then we may take
1
N = [ ] + 1.
ε
Department of Mathematics, KU, Lawrence, KS 66045
E-mail address: [email protected]
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