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8.3 day 2
Tests for Convergence
Greg Kelly, Hanford High School, Richland, Washington
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View Full Document Review:
1
P
dx
x
∞
∫
0
P
1
P
x
dx
∞

∫
1
lim
b
P
b
x
dx

→∞
∫
1
1
1
lim
1
b
P
b
x
P
 +
→∞
 +
1
1
1
1
1
P
P
b
b
P
P
 +
 +

 +
 +
If
then
gets bigger
and bigger as
, therefore
the integral diverges
.
1
P
≤
1
P
b
 +
b
→ ∞
If
then
b
has a negative
exponent and
,
therefore the integral converges
.
1
P
1
0
P
b
 +
→
(P is a constant.)
→
1
x
e dx
∞

∫
1
lim
b
x
b
e dx

→∞
∫
1
lim
b
x
b
e

→∞

( 29
1
b
b
e
e


→∞

 
1
b
e
e
→∞

+
0
1
e
=
Converges
→
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View Full Document Does
converge?
2
1
x
e
dx
∞

∫
Compare:
to
for positive values of
x
.
2
1
x
e
1
x
e
For
2
2
x
1
1
1,
e
x
x
x
x
e
e
e
∴
<
→
2
1
x
e
1
x
e
For
2
2
x
1
1
1,
e
x
x
x
x
e
e
e
∴
<
Since
is always below
, we say that it is
“bounded above” by
.
2
1
x
e
1
x
e
1
x
e
Since
converges to a finite number,
must also converge!
1
x
e
2
1
x
e
→
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View Full Document Direct Comparison Test:
Let
f
and
g
be continuous on
with
for all
, then:
[
29
,
a
∞
( 29 ( 29
0
f x
g x
≤
≤
x
a
≥
2
( 29
a
f x dx
∞
∫
diverges if
diverges.
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This note was uploaded on 03/10/2008 for the course MATH 214 taught by Professor Riggs during the Fall '05 term at Cal Poly Pomona.
 Fall '05
 Riggs
 Differential Calculus

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