Lecture 22Section 11.2 The Integral Test; Comparison Tests
Jiwen He
1
1.1
The Integral Test
The Integral Test
The Integral Test
Let ak = f (k), where f is continuous, decreasing and positive on [1, ∞), then
Z ∞
∞
X
ak converges iff
f (x)dx converges
1
k=1
n
X
Z
n
n−1
X
Since f decreases on [1, ∞)
ak . Since f is positive
ak <
f (x)dx <
1
k=2
k=1
Z
∞
∞
∞
X
X
on [1, ∞),
ak ≤
f (x)dx ≤
ak . Therefore, either both converge or
k=2
1
k=1
both diverge.
1.2
Applying the Integral Test
The p-Sereis
(The p-Sereis)
∞
X
1
1
1
= 1 + p + p + · · · converges
kp
2
3
iff
p > 1.
k=1
Proof.
Z ∞
∞
X
1
1
converges
iff
dx converges.
p
k
xp
1
k=1
Z ∞
We know that
1
dx converges iff p > 1.
p
x
1
∞
X 1
It follows that
converges iff p > 1.
kp
k=1
1
Example (p = 1): Harmonic Series
(The p-Sereis)
∞
X
1
1
1
= 1 + p + p + · · · converges
p
k
2
3
iff
p > 1.
k=1
Harmonic Series (p = 1)
When p = 1, this is the harmonic series
∞
X
1
1 1
= 1 + + + ··· ,
k
2 3
k=1
and the related improper integral is
Z ∞
∞
1
dx = ln x1 = lim ln x = ∞.
x→∞
x
1
Since this impproper integral diverges, so does the harmonic series.
Example (p = 2)
(The p-Sereis)
∞
X
1
k=1
kp
=1+
1
1
+ p + · · · converges
2p
3
iff
p > 1.
(p = 2)
When p = 2, this is
∞
X
1
1
1
= 1 + 2 + 2 + ··· ,
k2
2
3
k=1
and the related improper
Z ∞ integral is
∞
1
dx = −x−1 1 = 1 − lim x−1 = 1.
2
x→∞
x
1
By the integral test,∞
Z
∞
∞
X 1
X 1
1
=1+
≤1+
dx = 2
2
2
2
k
k
x
1
k=1
Example
Show that
∞
X
k=2
k=2
1
diverges.
k ln k
The related
Z ∞ improper integral
Z ∞ is
∞
1
1
dx =
du = ln uln 2 = lim ln x − ln ln 2 = ∞.
x→∞
x ln x
2
ln 2 u
Since this impproper integral diverges, so does the infinite series.
Show that
∞
X
k=2
1
converges.
k(ln k)2
The related
Z ∞ improper integral
Z ∞ is
∞
1
1
1
1
dx =
du = −u−1 ln 2 =
− lim x−1 =
.
2
2
x→∞
x(ln x)
ln 2
ln 2
2
ln 2 u
Since this impproper integral diverges, so does the infinite series.
2
The Integral Test for the nth Remainder
P∞
k=n+1
ak , n ≥ 1
Let ak = f (k), where ∞
f is continuous,
positive on [n, ∞), then
Z ∞ decreasing andX
∞
X
ak ≤
f (x)dx ≤ an +
ak
n
k=n+1
Z ∞
or equivalently
f (x)dx ≤
n+1
Let
P∞
Let
R∞
P∞
k=1
Z
k=n+1
∞
ak ≤
f (x)dx
n
k=n+1
ak be convergent. Its nth remainder is defined as
∞
∞
n
∞
X
X
X
X
Rn =
ak =
ak −
ak =
ak − sn
k=n+1
n
∞
X
k=1
k=1
k=1
R∞
k=1 ak be convergent, so do 1 f (x)dx. Then, ∀ > 0, ∃N s.t. for n > N ,
f (x)dx < , and
∞
∞
n
X
X
X
0 < Rn =
ak =
ak −
ak < k=n+1
Example
Use a partial sum to approximate
∞
X
k=1
k=1
k=1
1
with an error < 0.01
k2 + 1
∞
remainder
X
Estimating the nth
Rn =
Z ∞
n series
X
1
1
1 of the
−
<
dx
2+1
k2 + 1
k2 + 1
x
n
k=1
k=1
∞
π
= tan−1 xn = − tan−1 n.
2
− tan−1 n < 0.01, we need n ≥ tan( π2 − 0.01) ≈ 100.
n
X
1 1
1
1
1
sn =
= + +
+ ···
= 1.066 . . .
k2 + 1
2 5 10
10001
k=1
∞
X
Therefore
1
1.066 . . . <
< 1.076 . . .
2
k +1
For
π
2
k=1
1.3
Comparison Tests
Basic Series that Converge or Diverge
∞
∞
X
X
ak converges iff
ak converges, ∀j ≥ 1.
k=1
k=j
In determining whether a series converges, it does
P not matter where the summation begins. Thus, we will omit it and write
ak .
Basic Series that Converge
3
X
Geometric series:
xk ,
X 1
,
kp
p-series:
if |x| < 1
if p > 1
Basic Series that Diverge
Any series
X
for which lim ak 6= 0
ak
p-series:
k→∞
X 1
,
kp
if p ≤ 1
Basic Comparison Test
Basic Comparison Test
Suppose that 0 ≤ ak ≤X
bk for sufficiently large k.
X
If
bk converges, then so does
ak .
X
X
If
ak diverges, then so does
bk .
Comparison with p-Series
X
ak converges if
X
ak diverges if
0 ≤ ak ≤
0≤
c
,
kp
c
≤ ak ,
kp
p > 1.
p ≤ 1.
Comparison with Geometric Series
X
ak converges if
0 ≤ ak ≤ c xk ,
|x| < 1.
Examples
X
X 1
1
converges by comparison with
3
2k + 1
k3
X 1
1
1
< 3
and
converges.
2k 3 + 1
k
k3
X 1
X
k3
converges by comparison with
5
4
k + 4k + 7
k2
3
3
X
k
k
1
1
< 5 = 2
and
converges.
k 5 + 4k 4 + 7
k
k
k2
X
X 2
1
converges by comparison with
3
2
k −k
k3
X 2
1
1
2
< 3
= 3 , k ≥ 2, and
converges.
3
2
3
k −k
k − k /2
k
k3
4
Examples
X
X 1
1
diverges by comparison with
3k + 1
3(k + 1)
1
1
1X 1
>
and
diverges.
3k + 1
3(k + 1)
3
k+1
X
X 1
1
diverges by comparison with
ln(k + 6)
k+6
ln(k + 6)
→ 0 as k → ∞, ln(k + 6) < k + 6 for k large
k+6
X 1
1
1
>
for k large
and
diverges.
ln(k + 6)
k+6
k+6
Limit Comparison Test
Basic Comparison Test
ak
=L
Suppose that ak > 0 and bk > 0 for sufficiently large k, and that lim
k→∞ bk
for some L > 0. X
X
ak converges iff
bk converges.
ak
= L means that ak ≈ Lbk for large k
bk
ak
= 0, and
If lim
k→∞ bk
X
X
if
bk converges, then
ak converges.
lim
k→∞
If lim
k→∞
ak
= ∞, and
bk
X
if
ak converges, then
Example
X 1
k3 − 1
converges by comparison with
For large k,
k3
bk converges.
X 1
.
k3
1
1
differs little from 3 .
−1
k
k3
and
X
1
1
k3
÷ 3 = 3
→ 1 as k → ∞
−1 k
k −1
X 1
converges.
k3
5
Example
X 3k 2 + 2k + 1
k3
+1
For large k,
diverges by comparison with
X3
k
3k 2
3
3k 2 + 2k + 1
differs
little
from
= .
3
3
k +1
k
k
3k 3 + 2k 2 + k
3k 2 + 2k + 1 3
÷ =
→ 1 as k → ∞
3
k +1
k
3k 3 + 3
and
X3
diverges.
k
Example
√
X 5 k + 100
X 5
√
√ converges by comparison with
2k 2
2k 2 k − 9 k
√
√
5 k + 100
5 k
5
√
√ differs little from
√ = 2.
For large k,
2
2
2k
2k k − 9 k
2k k
√
√
5 k + 100
10k 2 k + 200k 2
5
√
√ ÷ 2 =
√
√ → 1 as k → ∞
2k 2 k − 9 k 2k
10k 2 k − 45 k
and
X 5
converges.
2k 2
Outline
Contents
1 Integral Test
1.1 Integral Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Application . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3 Comparison Tests . . . . . . . . . . . . . . . . . . . . . . . . . . .
6
1
1
1
3