FINAL 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.
This definition is useful is when evaluating the limits; for instance, to show
2
2n
= .
3n + 1
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One can also apply the limit theorems to evaluate the limit as follows.
lim
n→∞
2
2n
= lim
n→∞ 3 +
n→∞ 3n + 1
lim
1
n
2
= .
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We talk about the limit theorem for limits. If limn→∞ an = A and limn→∞ bn =
B, then
lim an ± bn = A ± B
n→∞
lim an bn = AB
n→∞
lim
n→∞
an
A
= , if B 6= 0.
bn
B
The proof of this theorem is by use of the definition of limits.
There is also another useful theorem for evaluating the limits, “the squeezing
theorem”: If an ≤ bn ≤ cn and limn→∞ an = limn→∞ cn = A, then
lim bn = A.
n→∞
This theorem can be also proven by using the definition of limits. A typical
example is as follows.
sin n2
lim
=0
n→∞
n
Date: April 24, 2012.
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because
sin n2 1
≤ ,
0≤
n n
1
= 0.
n→∞ n
lim 0 = lim
n→∞
2. The monotone convergence theorem and the sequence
The monotone convergence theorem says that, if a sequence {an } is either monotone increasing and bounded above, or monotone decreasing and
bounded below, then {an } converges. We give an example to show how to
apply this theorem.
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
an+1 − an =
√
3 + 2an −
p
3 + 2an−1 = √
2(an − an−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.
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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.
3. 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.
There are several well-known examples of convergent or divergent examples:
the following series,
∞
∞
X
X
1
1
,
p
>
1,
and
, p > 1.
np
n(ln n)p
n=1
n=2
R∞
R∞
They are convergent by comparing to the integrals, 1 x1p dx and 1
both of which are finite because p > 1.
1
x(ln x)p dx,
On the other hand, the following series are divergent,
∞
∞
X
X
1
1
,
0
<
p
≤
1,
and
, 0 < p ≤ 1.
np
n(ln n)p
n=1
n=2
R∞
R∞
They are divergent by comparing to the integrals, 1 x1p dx and 1
both of which are infinity because 0 < p ≤ 1.
The geometric series is convergent.
∞
X
rn for |r| < 1.
n=1
3
1
x(ln x)p dx,
We can prove this by setting
Sn =
n
X
rk .
k=1
Then
(1 − r)Sn = (r + r2 + · · · rn ) − (r2 + · · · + rn + rn+1 ) = r − rn+1 .
Since |r| < 1,
Sn =
r − rn+1
r − limn→∞ rn+1
r
⇒ lim Sn =
=
.
n→∞
1−r
1−r
1−r
P∞ 1
We know that the harmonic series
n=1 n is divergent. However if we
n
P∞ (−1)n
1
is
add signs to each term n to turn it into (−1)
n=1
n , then the series
n
convergent. In this regard, there is a general theorem: Suppose the sequence
{an } satisfying
a1 ≥ a2 ≥ · · · ≥ an ≥ · · · ≥ 0 and limn→0 an = 0, then the
P∞
series n=1 (−1)n an is convergent.
P
A consequence that ∞
n=1 an is convergent is that the tail terms go to zero
as n goes to infinity, i.e.,
(1)
lim an = 0.
n→∞
This is a necessary
not sufficient condition for a P
convergent series, for exP
∞ 1
1
1
ample, for ∞
,
lim
=
0
but
the
series
n→∞ n
n=1 n
n=1 n is divergent.
However the equationP(1) can be used as a criterion toPtell whether a series
∞
converges or not:P
for ∞
n=1 an is divergent.
n=1 an , if limn→∞ an 6= 0, then
∞
n is divergent because lim
n = ∞. Another
2
2
An example is,
n→∞
n=1
P∞
1
1
example n=1 (1 + n ) is divergent because limn→∞ 1 + n = 0.
4. series of functions
In
the series
P∞the previous section, we talk about several theoremsPabout
∞
n
a
.
By
adding
a
variable
x,
the
series
of
functions
a
n=1 n
n=1 n x is also
n
a kind of series with terms an x . Obviously it is convergent if x = 0. In
general, there is a quantity associated to each series of functions, the radius
of convergence ρ, such that, for |x| < ρ, the series is absolutely convergent;
for |x| > ρ, it is divergent. However, for |x| = ρ, we need to study it case
by case. We usePthe ratio test to find ρ. For example, for which values of x,
xn
does the series
n converge?
Proof. For this example,
an =
4
1
.
n
Then
lim
n→∞
an+1
n
= 1, i.e., q = 1.
= lim
n→∞ (n + 1)
an
Hence the radius of convergence ρ = 1q = 1. That is to say, for |x| < 1, the
P
P∞
n
n
series ∞
n=1 an x is absolutely convergent. For |x| > 1, the series
n=1 an x
is divergent.
P1
For |x| = 1, x = 1 or x = −1. For x = 1, the series becomes
, which
P∞ (−1)n n
diverges. However for x = −1, the series becomes
n=1
n , which is
convergent because it is a series of alternating terms.
To conclude, the series is convergent on [−1, 1), and divergent for |x| > 1
and x = 1.
5. Limits of functions, continuity and uniform continuity
Let f : D → R, and a ∈ R. We say that limx→a f (x) = A, if for every
sequence {xn } ∈ D satisfying xn → a,
lim f (xn ) = A.
n→∞
Note that the point a may not be in the domain D. If there exists two
sequences {xn } and {yn } both converging to a, and limits limn→a f (xn ) and
limn→∞ f (yn ) are not equal, then limx→a f (x) does not exist. This is a
typical way to show that a limit does not exist. An example,
x
: R \ {0} → R, in this case, lim f (x) does not exist.
f (x) =
x→0
|x|
The continuity of functions can be defined in a similar way. We say that f
is continuous at x = a, if for every sequence {xn } ∈ D satisfying xn → a,
lim f (xn ) = f (a).
n→∞
Note that a is in the domain D. If f is continuous at every point of D, then
f is continuous on D.
We have learned two theorems about continuous functions. One is the intermediate value theorem, and the other is the maximum principle. In the
final, you are not required to know the proofs of these two theorems, but
you need to know how to apply them to solve the problems.
There is another way to phrase the continuity of functions, which we call the
“ − δ” formulation of continuity of functions. We say that f is continuous
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at x = a: for any > 0, there exists δ > 0 such that for |x − a| < δ,
|f (x) − f (a)| < .
Here δ may depend on and a.
The second definition has the advantage of generalizing to uniform continuity
of functions. We say that f is uniformly continuous on D: for any > 0,
there exists δ > 0 such that for |x − y| < δ,
|f (x) − f (y)| < .
In general, the uniform continuity of functions implies the continuity of
functions, but the converse is not true. For instance, f (x) = x2 : R → R is
continuous on R but not uniformly continuous. However on a closed interval,
say [a, b], they are equivalent.
6. differentiability
Having seen continuity of functions, we talk about another property of functions, differentiability. We say that f is differentiable at a point a, if the
following limit exists,
f (a + h) − f (a)
.
lim
h→0
h
If the limit exists, we denote
f (a + h) − f (a)
.
h→0
h
f 0 (a) = lim
The definition is important when we want to say whether a function is
differentiable at a point; also it will gives the formula of the derivatives.
For instance, f (x) = xn with n 6= −1, is differentiable at every point x
and f 0 (x) = nxn−1 . Another example is, for f (x) = sin x, f 0 (x) = cos x.
The most common example of non-differentiable functions is f (x) = |x|; for
this function, although it is continuous at x = 0, we know that it is not
differentiable at x = 0.
Related to differentiability, there are two important theorems, Rolle’s theorem and the mean value theorem. The proof of Rolle’s theorem makes
use of the maximum principle we just mentioned. Roughly speaking, for
continuous functions over [a, b], there exists c and d in [a, b] such that
f (c) = max f (x),
f (d) = min f (x).
a≤x≤b
a≤x≤b
At these points, one can show that f 0 = 0. The Rolle Theorem can be used
to prove the mean value theorem.
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7. integrability
Let f be a bounded real function. We say that f is Riemann integrable on
[a, b], if
L(f ) = U (f ),
where,
L(f ) = sup{LP (f )}, U (f ) = inf {UP (f )}.
P
P
Here P is a partition of [a, b].
We have shown that continuous functions or monotone functions over [a, b]
are integrable. The Dirichlet function f is not Riemann integrable, where
(
1, for irrational x;
f (x) =
0, for rational x.
Related to Riemann integrals, the fundamental theorem of calculus is important. Let f be a continuous function on [a, b]. Then
Z x
F (x) =
f (t)dt
a
is differentiable at x ∈ (a, b) and
Z
0
b
f (t)dt = F (b) − F (a).
F (x) = f (x), and
a
Note that F is called the anti-derivative of f . The last equation in the
Rb
Fundament Theorem of Calculus tells a way to compute integrals a f (t)dt:
we just need to find what are the antiderivative of f . For example,
Z π
sin xdx = (− cos π) + cos 0 = 2.
0
This is because the antiderivative of sin x is − cos x.
Department of Mathematics, KU, Lawrence, KS 66045
E-mail address: [email protected]
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