Download Math 20B. Lecture Examples. Section 10.3. Convergence of series

(7/19/09)
Math 20B. Lecture Examples.
Section 10.3. Convergence of series with positive terms†
Theorem 1 (Convergence of series with positive terms) An infinite series with positive terms
either converges or diverges to ∞. The series converges if its partial sums are bounded and
diverges if its partial sums are not bounded.
Theorem 2 (The Integral Test for series with positive terms)
∞
X
Suppose that the series
an
(1)
n=n0
with positive terms is such that an = f (n) for integers n ≥ c with some c, where y = f (x) is
continuous on [c, ∞) and decreasing for x ≥ c. Then the infinite series (1) converges if and
only if the improper integral
Z
∞
f (x) dx
(2)
c
converges.
Figures 1 and 2 show why the series converges (1) if the integral (2) converges. The area of the
rectangles in Figure 1 is less than the area of the region in Figure 2. If the integral converges, then the
area of the region in Figure 2 is bounded as N → ∞, so that the partial sums of the series are bounded
and the series converges.
Area =
N
X
Area =
an
n=c+1
FIGURE 1
† Lecture
Z
N
f (x) dx
c
FIGURE 2
notes to accompany Section 10.3 of Calculus, Early Transcendentals by Rogawski.
1
Math 20B. Lecture Examples. (7/19/09)
Section 10.3, p. 2
Figures 3 and 4 show why the series (1) diverges if the integral (2) diverges. The area of the
rectangles in Figure 4 is greater than the area of the region in Figure 3. If the integral diverges, then the
area of the region in Figure 3 tends to ∞ as N → ∞, so that the partial sums of the series tend to ∞
and the series diverges.
Area =
Z
N +1
f (x) dx
Area =
c
Does
an
n=c
FIGURE 3
Example 1
N
X
∞
X
FIGURE 4
2
ne−n converge?
n=1
Answer:
Z
∞
2
xe−x dx =
1
series are plotted in Figure A1.)
1 −1
e
2
•
∞
X
2
ne−n converges by the Integral Test. (Some partial sums of the
n=1
y
y = sN
1.5
sN =
∞
X
n=1
1
−n2
ne
0.5
Figure A1
10
20
N
Section 10.3, p. 3
Example 2
Math 20B. Lecture Examples. (7/19/09)
Does
∞
X
n=2
Answer:
Z
∞
1
converge or diverge?
n ln n
1
x ln x
2
dx = ∞ •
plotted in Figure A2.)
∞
X
n=2
1
diverges by the Integral Test. (Some partial sums of the series are
n ln n
y
y = sN
2
sN =
N
X
n=2
1
1
n ln n
Figure A2
10
20
N
The Integral Test is used to establish the following result.
Theorem 3 (Convergence of the p-series)
The infinite series
∞
X
1
1
1
1
p = 1 + p + p + p + ···
n
2
3
4
n=1
converges if p > 1 and diverges if p ≤ 1.
Example 3
Does
∞
X
6
1.75
n
n=1
Answer:
∞
X
n=1
6
=6
n1.75
with p = 1.75 > 1.
Example 4
Does
Answer:
∞
X
n=1
1
converges because it is a constant multiplied by the p-series
n1.75
n=1
1
1.75
n
1
√ converge or diverge and why?
9
n
n=2
∞
X
1
√
9 n
1
< 1.
with p = 2
∞
X
Does
n=1
1
= 9
∞
X
n=2
1
n1/2
diverges because it is a constant multiplied by the p-series
∞
X
n=2
1
n1/2
7n
converge or diverge and why?
6n + 1
Answer: The series diverges because an =
Example 6
∞
X
∞
X
n=2
Example 5
converge or diverge and why?
7n
6n + 1
=
7
6 + 1/n
→
7
6
as n → ∞; the terms do not tend to 0.
∞
X
1
is called the harmonic series. Does it converge or diverge?
The series
n
n=1
Answer:
∞
X
n=1
1
n
diverges because it is the p-series with p = 1.
Math 20B. Lecture Examples. (7/19/09)
Section 10.3, p. 4
Theorem 4 (The Comparison Test with positive terms)
Suppose that
∞
X
an and
n=n0
(a) If
∞
X
n=n0
∞
X
∞
X
bn are series with positive terms.
n=n0
bn converges and there are constants M and N such that an ≤ M bn for n ≥ N,
an also converges.
then
n=n0
∞
X
(b) If
n=n0
bn dinverges and there are constants M > 0 and N such that an ≥ M bn for
n ≥ N, then
∞
X
an also diverges.
n=n0
Example 7
Does the series
∞
X
(0.6)n
converge or diverge and why?
n+1
n=0
Answer:
∞
X
(0.6)n
n=0
(0.6)n
n+1
≤
converges by the Comparison Test with the convergent geometric series
n+1
(0.6)n
∞
X
(0.6)n because
n=0
for n ≥ 0. (The partial sums of the first series in Figure A7a are less than the partial sums of the
second series in Figure A7b.)
y
y
2
2
1
1
10
20
y=
N
N
X
(0.6)n
n=0
n+1
Figure A7a
Example 8
Does
Answer:
n2 + 1
3
5n
20
y=
N
X
N
(0.6)n
n=0
Figure A7b
∞
X
n2 + 1
converge or diverge and why?
3
5n
n=1
∞
X
n2 + 1
n=1
2
>
10
n
3
5n
5n3
=
1
5
diverges by the Comparison Test with the divergent harmonic series
∞
X
n=1
1
n
is a positive constant.
for n ≥ 1, and 1
5
1
n
because
Section 10.3, p. 5
Math 20B. Lecture Examples. (7/19/09)
Theorem 5 (The Limit-Comparison Test with positive terms)
Suppose that
an
=L
b
n→∞ n
lim
(3)
where the an ’s and bn ’s are positive and L is either a nonnegative number or ∞.
(a) If L is finite and positive, then
∞
X
an converges if and only if
(c) If L = ∞ and
∞
X
bn converges, then
n=n0
∞
X
bn converges.
n=n0
n=j0
(b) If L = 0 and
∞
X
an converges.
n=n0
∞
X
bn diverges, then
n=n0
∞
X
an also diverges
j=n0
∞
X
10(2n )
Does
converge or diverge and why?
5n − 1
Example 9
n=1
Answer:
∞
X
n=1
n
10(2n )
n
converges by the Limit Comparison Test with the convergent geometric series
=
5 −1
∞
X
n=0
2 n
5
10(2 )
because
5n − 1
2
n
5n
2
n
5n
10(2n )
n
5 −1
=
5n
2
n
10(2n )
n
5 (1 − 1/5n )
=
10
1 − 1/ 5n
→ 10 as n → ∞, and 10 is a positive
number.
Example 10
Does the series
∞
X
n=1
Answer:
∞
X
n=1
n
6n2 + 1
n
2
6n + 1
converge or diverge and why?
diverges by the Limit Comparison Test with the divergent harmonic series
∞
X
n=1
1
n
because
n
6n2 + 1
1
=
n2
2
6n + 1
=
n2
2
n (6 + 1/n2 )
=
1
6 + 1/n2
→
1
6
as n → ∞, and 1
is a positive number.
6
n
Interactive Examples
Work the following Interactive Examples on Shenk’s web page, http//www.math.ucsd.edu/˜ashenk/:‡
Section 10.3: Examples 1–4
Section 10.4: Examples 1–4
‡ The
chapter and section numbers on Shenk’s web site refer to his calculus manuscript and not to the chapters and sections
of the textbook for the course.