- •Functions
- •The Concept of a Function
- •Trigonometric Functions
- •Inverse Trigonometric Functions
- •Logarithmic, Exponential and Hyperbolic Functions
- •Limits and Continuity
- •Introductory Examples
- •Continuity Examples
- •Linear Function Approximations
- •Limits and Sequences
- •Properties of Continuous Functions
- •The Derivative
- •The Chain Rule
- •Higher Order Derivatives
- •Mathematical Applications
- •Antidifferentiation
- •Linear Second Order Homogeneous Differential Equations
- •Linear Non-Homogeneous Second Order Differential Equations
- •Area Approximation
- •Integration by Substitution
- •Integration by Parts
- •Logarithmic, Exponential and Hyperbolic Functions
- •The Riemann Integral
- •Volumes of Revolution
- •Arc Length and Surface Area
- •Techniques of Integration
- •Integration by formulae
- •Integration by Substitution
- •Integration by Parts
- •Trigonometric Integrals
- •Trigonometric Substitutions
- •Integration by Partial Fractions
- •Fractional Power Substitutions
- •Numerical Integration
- •Integrals over Unbounded Intervals
- •Discontinuities at End Points
- •Improper Integrals
- •Sequences
- •Monotone Sequences
- •Infinite Series
- •Series with Positive Terms
- •Alternating Series
- •Power Series
- •Taylor Polynomials and Series
- •Applications
- •Parabola
- •Ellipse
- •Hyperbola
- •Polar Coordinates
- •Graphs in Polar Coordinates
- •Areas in Polar Coordinates
- •Parametric Equations
314CHAPTER 7. IMPROPER INTEGRALS AND INDETERMINATE FORMS
7.4Improper Integrals
1.Suppose that f is continuous on (−∞, ∞) and g0(x) = f(x). Then define each of the following improper integrals:
Chapter 8
Infinite Series
8.1Sequences
Definition 8.1.1 An infinite sequence (or sequence) is a function, say f, whose domain is the set of all integers greater than or equal to some integer m. If n is an integer greater than or equal to m and f(n) = an, then we express the sequence by writing its range in any of the following ways:
1.f(m), f(m + 1), f(m + 2), . . .
2.am, am+1, am+2, . . .
3.{f(n) : n ≥ m}
4.{f(n)}∞n=m
5.{an}∞n=m
Definition 8.1.2 A sequence {an}∞n=m is said to converge to a real number L (or has limit L) if for each > 0 there exists some positive integer M such that |an − L| < whenever n ≥ M. We write,
lim an = L or an → L as n → ∞.
n→∞
If the sequence does not converge to a finite number L, we say that it diverges.
315
316 |
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CHAPTER 8. |
INFINITE SERIES |
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Theorem 8.1.1 Suppose that c is a positive real number, |
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n}n=m |
and |
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are convergent sequences. Then |
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(i) |
lim (can) = c lim an |
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n→∞ |
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(ii) |
lim (an + bn) = lim an + lim bn |
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n→∞ |
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(iii) |
nlim (an − bn) = nlim an − nlim bn |
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(iv) |
lim |
(anbn) = lim an lim bn |
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n→∞ |
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(v) |
nlim |
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, if nlim bn 6= 0. |
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c
(vi) lim (an)c = |
lim an |
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(vii) lim (ean ) = elimn→∞ an
n→∞
(viii) Suppose that an ≤ bn ≤ cn for all n ≥ m and
lim an = lim cn = L.
n→∞ n→∞
Then
lim bn = L.
n→∞
Proof. Suppose that {an}∞n=m converges to a and {bn}∞n=m converges to b. Let 1 > 0 be given. Then there exist natural numbers N and M such that
|an − a| < 1 |
if n ≥ N, |
(1) |
|bn − b| < 1 |
if n ≥ M. |
(2) |
Part (i) Let > 0 be given and c =6 0. Let 1 = 2|c| and n ≥ N + M. Then by the inequalities (1) and (2), we get
|can − ca| = |c| |an − a|
<|c| 1
<.
8.1. SEQUENCES |
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317 |
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This completes the proof of Part (i). |
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Part (ii) Let > 0 be given and 1 = |
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≥ N + M. Then by the |
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inequalities (1) and (2), we get |
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|(an + bn) − (a + b)| = |(an − a) + (bn − b)| |
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≤ |an − a| + |bn − b| |
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< 1 + 1 |
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= . |
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This completes the proof of Part (ii). |
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Part (iii) |
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nlim (an − bn) = nlim |
(an + (−1)bn) |
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an + nlim [(−1)bn] |
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= nlim |
an + (−1) nlim bn |
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=a + (−1)b
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Part (iv) Let > 0 be given and 1 |
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1, |
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1 + a |
+ b |
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then by the inequalities (1) and (2) we have |
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|anbn − ab| = |[(an − a) + a][(bn − b) + b] − ab|
= |(an − a)(bn − b) + (an − a)b + a(bn − b| ≤ |an − a| |bn − b| + |b| |an − a| + |a| |bn − b|
< 21 + |b| 1 + |a| 1
= 1( 1 + |b| + |a|)
≤ 1(1 + |b| + |a|)
≤ .
Part (v) First we assume that b > 0 and prove that
lim 1 = 1.
n→∞ bn b
318 CHAPTER 8. INFINITE SERIES
1 |
b and using inequality (2) for n ≥ M, we get |
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Let > 0 be given. Choose 2 |
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If n ≥ N + M, then the inequalities (3) and (4) imply that |
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lim |
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8.1. SEQUENCES |
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319 |
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If b < 0, then |
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1 = (−a) −b
= ab .
This completes the proof of Part (v).
Part (vi) Since f(x) = xc is a continuous function,
c
lim (an)c = |
lim an = ac. |
n→∞ |
n→∞ |
Part (vii) Since f(x) = ex is a continuous function,
lim ean = elimn→∞ an = ea.
n→∞
Part (viii) Suppose that an ≤ bn ≤ cn for all n ≥ m and
lim an = L = lim cn = L.
n→∞ n→∞
Let > 0 be given. Then there exists natural numbers N and M such that
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n − |
L < |
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If n ≥ N + M, then n > N and n > M and, hence, |
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It follows that
lim bn = L.
n→∞
This completes the proof of this theorem.