Advanced Calculus (I)
WEN-CHINGLIEN
Department of Mathematics National Cheng Kung University
6.2 Series with nonnegative terms
Theorem
Suppose that ak ≥ 0 for k ≥ N. Then
∞
P
k =1
ak converges if and only if its sequence of partial sums {sn} is bounded, i.e., if and only if there exists a finite number M > 0 such
that
n
X
k =1
ak
≤ M for all n ∈ N.
6.2 Series with nonnegative terms
Theorem
Suppose that ak ≥ 0 for k ≥ N. Then
∞
P
k =1
ak converges if and only if its sequence of partial sums {sn} is bounded, i.e., if and only if there exists a finite number M > 0 such
that
n
X
k =1
ak
≤ M for all n ∈ N.
Proof:
Set sn =
n
P
k =1
ak for n ∈N. If
∞
P
k =1
ak converges, then sn
converges as n → ∞. Since every convergent sequence is bounded (Theorem 2.8),
∞
P
k =1
ak has bounded partial sums.
Conversely, suppose that |sn| ≤ M for n ∈ N. Since ak ≥ 0 for k ≥ N, snis an increasing sequence when n ≥ N.
Hence by the Monotone Convergence Theorem (Theorem 2.19), sn converges. 2
Proof:
Set sn =
n
P
k =1
ak for n ∈N. If
∞
P
k =1
ak converges, then sn
converges as n → ∞. Since every convergent sequence is bounded (Theorem 2.8),
∞
P
k =1
ak has bounded partial sums.
Conversely, suppose that |sn| ≤ M for n ∈ N. Since ak ≥ 0 for k ≥ N, snis an increasing sequence when n ≥ N.
Hence by the Monotone Convergence Theorem (Theorem 2.19), sn converges. 2
Proof:
Set sn =
n
P
k =1
ak for n ∈N. If
∞
P
k =1
ak converges, then sn
converges as n → ∞. Since every convergent sequence is bounded (Theorem 2.8),
∞
P
k =1
ak has bounded partial sums.
Conversely, suppose that |sn| ≤ M for n ∈ N. Since ak ≥ 0 for k ≥ N, snis an increasing sequence when n ≥ N.
Hence by the Monotone Convergence Theorem (Theorem 2.19), sn converges. 2
Proof:
Set sn =
n
P
k =1
ak for n ∈N. If
∞
P
k =1
ak converges, then sn
converges as n → ∞. Since every convergent sequence is bounded (Theorem 2.8),
∞
P
k =1
ak has bounded partial sums.
Conversely, suppose that |sn| ≤ M for n ∈ N.Since ak ≥ 0 for k ≥ N, snis an increasing sequence when n ≥ N.
Hence by the Monotone Convergence Theorem (Theorem 2.19), sn converges. 2
Proof:
Set sn =
n
P
k =1
ak for n ∈N. If
∞
P
k =1
ak converges, then sn
converges as n → ∞. Since every convergent sequence is bounded (Theorem 2.8),
∞
P
k =1
ak has bounded partial sums.
Conversely, suppose that |sn| ≤ M for n ∈ N. Since ak ≥ 0 for k ≥ N,snis an increasing sequence when n ≥ N.
Hence by the Monotone Convergence Theorem (Theorem 2.19), sn converges. 2
Proof:
Set sn =
n
P
k =1
ak for n ∈N. If
∞
P
k =1
ak converges, then sn
converges as n → ∞. Since every convergent sequence is bounded (Theorem 2.8),
∞
P
k =1
ak has bounded partial sums.
Conversely, suppose that |sn| ≤ M for n ∈ N.Since ak ≥ 0 for k ≥ N, snis an increasing sequence when n ≥ N.
Hence by the Monotone Convergence Theorem (Theorem 2.19), sn converges. 2
Proof:
Set sn =
n
P
k =1
ak for n ∈N. If
∞
P
k =1
ak converges, then sn
converges as n → ∞. Since every convergent sequence is bounded (Theorem 2.8),
∞
P
k =1
ak has bounded partial sums.
Conversely, suppose that |sn| ≤ M for n ∈ N. Since ak ≥ 0 for k ≥ N,snis an increasing sequence when n ≥ N.
Hence by the Monotone Convergence Theorem (Theorem 2.19),sn converges. 2
Proof:
Set sn =
n
P
k =1
ak for n ∈N. If
∞
P
k =1
ak converges, then sn
converges as n → ∞. Since every convergent sequence is bounded (Theorem 2.8),
∞
P
k =1
ak has bounded partial sums.
Conversely, suppose that |sn| ≤ M for n ∈ N. Since ak ≥ 0 for k ≥ N, snis an increasing sequence when n ≥ N.
Hence by the Monotone Convergence Theorem (Theorem 2.19), sn converges. 2
Proof:
Set sn =
n
P
k =1
ak for n ∈N. If
∞
P
k =1
ak converges, then sn
converges as n → ∞. Since every convergent sequence is bounded (Theorem 2.8),
∞
P
k =1
ak has bounded partial sums.
Conversely, suppose that |sn| ≤ M for n ∈ N. Since ak ≥ 0 for k ≥ N, snis an increasing sequence when n ≥ N.
Hence by the Monotone Convergence Theorem (Theorem 2.19),sn converges. 2
Proof:
Set sn =
n
P
k =1
ak for n ∈N. If
∞
P
k =1
ak converges, then sn
converges as n → ∞. Since every convergent sequence is bounded (Theorem 2.8),
∞
P
k =1
ak has bounded partial sums.
Conversely, suppose that |sn| ≤ M for n ∈ N. Since ak ≥ 0 for k ≥ N, snis an increasing sequence when n ≥ N.
Hence by the Monotone Convergence Theorem (Theorem 2.19), sn converges. 2
Theorem (Integral Test)
Suppose that f : [1, ∞) →R is positive and decreasing on [1, ∞). Then
∞
P
k =1
f (k ) converges if and only if f is improperly integrable on [1, ∞), i.e., if and only if
Z ∞ 1
f (x )dx < ∞.
Theorem (Integral Test)
Suppose that f : [1, ∞) →R is positive and decreasing on [1, ∞). Then
∞
P
k =1
f (k ) converges if and only if f is improperly integrable on [1, ∞), i.e., if and only if
Z ∞ 1
f (x )dx < ∞.
Corollary:
[p-Series Test]The series
(4)
∞
X
k =1
1 kp converges if and only if p > 1.
Corollary:
[p-Series Test]The series
(4)
∞
X
k =1
1 kp converges if and only if p > 1.
Proof:
If p = 1 or p ≤ 0, the series diverges. If p > 0 and p 6= 1, set f (x ) = x−p and observe that f0(x ) = −px−p−1 <0 for all x ∈ [1, ∞). Hence, f is nonnegative and decreasing on [1, ∞). Since,
Z ∞ 1
dx
xp = lim
x →∞
x1−p 1 − p
n
1
= lim
n→∞
n1−p− 1 1 − p
has a finite limit if and only if 1 − p < 0, it follows from the Integral Test that (4) converges if and only if p > 1.2
Proof:
If p = 1 or p ≤ 0,the series diverges. If p > 0 and p 6= 1, set f (x ) = x−p and observe that f0(x ) = −px−p−1 <0 for all x ∈ [1, ∞). Hence, f is nonnegative and decreasing on [1, ∞). Since,
Z ∞ 1
dx
xp = lim
x →∞
x1−p 1 − p
n
1
= lim
n→∞
n1−p− 1 1 − p
has a finite limit if and only if 1 − p < 0, it follows from the Integral Test that (4) converges if and only if p > 1.2
Proof:
If p = 1 or p ≤ 0, the series diverges. If p > 0 and p 6= 1, set f (x ) = x−p and observe that f0(x ) = −px−p−1 <0 for all x ∈ [1, ∞). Hence, f is nonnegative and decreasing on [1, ∞). Since,
Z ∞ 1
dx
xp = lim
x →∞
x1−p 1 − p
n
1
= lim
n→∞
n1−p− 1 1 − p
has a finite limit if and only if 1 − p < 0, it follows from the Integral Test that (4) converges if and only if p > 1.2
Proof:
If p = 1 or p ≤ 0, the series diverges. If p > 0 and p 6= 1, set f (x ) = x−p and observe that f0(x ) = −px−p−1 <0 for all x ∈ [1, ∞). Hence, f is nonnegative and decreasing on [1, ∞).Since,
Z ∞ 1
dx
xp = lim
x →∞
x1−p 1 − p
n
1
= lim
n→∞
n1−p− 1 1 − p
has a finite limit if and only if 1 − p < 0, it follows from the Integral Test that (4) converges if and only if p > 1.2
Proof:
If p = 1 or p ≤ 0, the series diverges. If p > 0 and p 6= 1, set f (x ) = x−p and observe that f0(x ) = −px−p−1 <0 for all x ∈ [1, ∞). Hence, f is nonnegative and decreasing on [1, ∞). Since,
Z ∞ 1
dx
xp = lim
x →∞
x1−p 1 − p
n
1
= lim
n→∞
n1−p− 1 1 − p
has a finite limit if and only if 1 − p < 0,it follows from the Integral Test that (4) converges if and only if p > 1.2
Proof:
If p = 1 or p ≤ 0, the series diverges. If p > 0 and p 6= 1, set f (x ) = x−p and observe that f0(x ) = −px−p−1 <0 for all x ∈ [1, ∞). Hence, f is nonnegative and decreasing on [1, ∞).Since,
Z ∞ 1
dx
xp = lim
x →∞
x1−p 1 − p
n
1
= lim
n→∞
n1−p− 1 1 − p
has a finite limit if and only if 1 − p < 0, it follows from the Integral Test that (4) converges if and only if p > 1.2
Proof:
If p = 1 or p ≤ 0, the series diverges. If p > 0 and p 6= 1, set f (x ) = x−p and observe that f0(x ) = −px−p−1 <0 for all x ∈ [1, ∞). Hence, f is nonnegative and decreasing on [1, ∞). Since,
Z ∞ 1
dx
xp = lim
x →∞
x1−p 1 − p
n
1
= lim
n→∞
n1−p− 1 1 − p
has a finite limit if and only if 1 − p < 0,it follows from the Integral Test that (4) converges if and only if p > 1.2
Proof:
If p = 1 or p ≤ 0, the series diverges. If p > 0 and p 6= 1, set f (x ) = x−p and observe that f0(x ) = −px−p−1 <0 for all x ∈ [1, ∞). Hence, f is nonnegative and decreasing on [1, ∞). Since,
Z ∞ 1
dx
xp = lim
x →∞
x1−p 1 − p
n
1
= lim
n→∞
n1−p− 1 1 − p
has a finite limit if and only if 1 − p < 0, it follows from the Integral Test that (4) converges if and only if p > 1.2
Theorem (Comparison Test)
Suppose that 0 ≤ ak ≤ bk for large k.
(i)If
∞
P
k =1
bk < ∞, then
∞
P
k =1
ak < ∞.
(ii)If
∞
P
k =1
ak = ∞, then
∞
P
k =1
bk = ∞.
Theorem (Comparison Test)
Suppose that 0 ≤ ak ≤ bk for large k.
(i)If
∞
P
k =1
bk < ∞, then
∞
P
k =1
ak < ∞.
(ii)If
∞
P
k =1
ak = ∞, then
∞
P
k =1
bk = ∞.
Theorem (Comparison Test)
Suppose that 0 ≤ ak ≤ bk for large k.
(i)If
∞
P
k =1
bk < ∞, then
∞
P
k =1
ak < ∞.
(ii)If
∞
P
k =1
ak = ∞, then
∞
P
k =1
bk = ∞.
Theorem (Comparison Test)
Suppose that 0 ≤ ak ≤ bk for large k.
(i)If
∞
P
k =1
bk < ∞, then
∞
P
k =1
ak < ∞.
(ii)If
∞
P
k =1
ak = ∞, then
∞
P
k =1
bk = ∞.
Theorem (Comparison Test)
Suppose that 0 ≤ ak ≤ bk for large k.
(i)If
∞
P
k =1
bk < ∞, then
∞
P
k =1
ak < ∞.
(ii)If
∞
P
k =1
ak = ∞, then
∞
P
k =1
bk = ∞.
Theorem (Comparison Test)
Suppose that 0 ≤ ak ≤ bk for large k.
(i)If
∞
P
k =1
bk < ∞, then
∞
P
k =1
ak < ∞.
(ii)If
∞
P
k =1
ak = ∞, then
∞
P
k =1
bk = ∞.
Example:
Determine whether the following series converges or diverges.
∞
X
k =1
3k k2+k
rlog k k
Example:
Determine whether the following series converges or diverges.
∞
X
k =1
3k k2+k
rlog k k
Theorem (Limit Comparison Test)
Suppose that ak ≥ 0 and bk >0 for large k and L := lim
n→∞
an
bn
exists as an extended real number.
(i)If 0 < L < ∞, then
∞
P
k =1
ak converges if and only if
∞
P
k =1
bk
converges.
(ii) If L = 0 and
∞
P
k =1
ak converges, then
∞
P
k =1
ak converges.
(iii) If L = 0 and
∞
P
k =1
bk diverges, then
∞
P
k =1
ak diverges.
Theorem (Limit Comparison Test)
Suppose that ak ≥ 0 and bk >0 for large k and L := lim
n→∞
an
bn
exists as an extended real number.
(i) If 0 < L < ∞, then
∞
P
k =1
ak converges if and only if
∞
P
k =1
bk
converges.
(ii) If L = 0 and
∞
P
k =1
ak converges, then
∞
P
k =1
ak converges.
(iii) If L = 0 and
∞
P
k =1
bk diverges, then
∞
P
k =1
ak diverges.
Theorem (Limit Comparison Test)
Suppose that ak ≥ 0 and bk >0 for large k and L := lim
n→∞
an
bn
exists as an extended real number.
(i)If 0 < L < ∞, then
∞
P
k =1
ak converges if and only if
∞
P
k =1
bk
converges.
(ii)If L = 0 and
∞
P
k =1
ak converges, then
∞
P
k =1
ak converges.
(iii) If L = 0 and
∞
P
k =1
bk diverges, then
∞
P
k =1
ak diverges.
Theorem (Limit Comparison Test)
Suppose that ak ≥ 0 and bk >0 for large k and L := lim
n→∞
an
bn
exists as an extended real number.
(i) If 0 < L < ∞, then
∞
P
k =1
ak converges if and only if
∞
P
k =1
bk
converges.
(ii) If L = 0 and
∞
P
k =1
ak converges, then
∞
P
k =1
ak converges.
(iii) If L = 0 and
∞
P
k =1
bk diverges, then
∞
P
k =1
ak diverges.
Theorem (Limit Comparison Test)
Suppose that ak ≥ 0 and bk >0 for large k and L := lim
n→∞
an
bn
exists as an extended real number.
(i) If 0 < L < ∞, then
∞
P
k =1
ak converges if and only if
∞
P
k =1
bk
converges.
(ii)If L = 0 and
∞
P
k =1
ak converges, then
∞
P
k =1
ak converges.
(iii)If L = 0 and
∞
P
k =1
bk diverges, then
∞
P
k =1
ak diverges.
Theorem (Limit Comparison Test)
Suppose that ak ≥ 0 and bk >0 for large k and L := lim
n→∞
an
bn
exists as an extended real number.
(i) If 0 < L < ∞, then
∞
P
k =1
ak converges if and only if
∞
P
k =1
bk
converges.
(ii) If L = 0 and
∞
P
k =1
ak converges, then
∞
P
k =1
ak converges.
(iii) If L = 0 and
∞
P
k =1
bk diverges, then
∞
P
k =1
ak diverges.
Theorem (Limit Comparison Test)
Suppose that ak ≥ 0 and bk >0 for large k and L := lim
n→∞
an
bn
exists as an extended real number.
(i) If 0 < L < ∞, then
∞
P
k =1
ak converges if and only if
∞
P
k =1
bk
converges.
(ii) If L = 0 and
∞
P
k =1
ak converges, then
∞
P
k =1
ak converges.
(iii)If L = 0 and
∞
P
k =1
bk diverges, then
∞
P
k =1
ak diverges.
Theorem (Limit Comparison Test)
Suppose that ak ≥ 0 and bk >0 for large k and L := lim
n→∞
an
bn
exists as an extended real number.
(i) If 0 < L < ∞, then
∞
P
k =1
ak converges if and only if
∞
P
k =1
bk
converges.
(ii) If L = 0 and
∞
P
k =1
ak converges, then
∞
P
k =1
ak converges.
(iii) If L = 0 and
∞
P
k =1
bk diverges, then
∞
P
k =1
ak diverges.
Example:
(1)
∞
P
k =1
√k
k k
(2)
∞
P
k =1
√ 1 k 3k −1
Example:
(1)
∞
P
k =1
√k
k k
(2)
∞
P
k =1
√ 1 k 3k −1
Example:
(1)
∞
P
k =1
√k
k k
(2)
∞
P
k =1
√ 1 k 3k −1