
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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Transcribed Image Text:For each of the following series decide if it is Absolutely Convergent, Conditionally Convergent, or
Divergent. You must justify all answers.
∞
√In(n)
(a) (-1) V
n=4
n
3n
(b) (2n²±³)³n
3n2-1
n=1
(c)
Ĩ
n=1
e(n²)
1x5x9...x(4n-3)
Q2) AC, CC, D.
124
(-1) √In (n)
n
consider Elanl
Tim
n=1
Positive
Continuous
D
Σ
√In (n)
n=4 n
Integral Test
decreasing. f(x) = Vin(x) (In(x))
x
X
f'(x) = (( In (x))". x')- (x + (In(x))"})")
x²
f'(x) = (In (x)) = (x = (x)/2+)
x²
f'(x) = (In (x))" - 21m (1) vz
X2
21n (x)-1
214(x)/2
'f'(x)='
f'(x) = 2/n(x)-1
2x2 14 (x)2
2n+3 3n
3n²-1
Root test
I'm me lank
Tim 2n²+33
1700 1137²+
lim 2n²+3 1/2
x²
еспа)
(×5×9x(4-3)
h=1
Ratio Test
Tim
n->00
an+1
an
Tim
1-78
(n+1)²
\x5x9x (4(n+1)-3)
e(2)
1x5x9(4-3)
lim
n->x
eln t
e(n)
1+1)²
1x+x9(4n-3)
1x4x9
(4-3)(4(4+1)-3)
Iim
(n+1)2
1x519 (4n-3) (4×4)
(4+1)²
lim
1-700
Tim
1-720
612
(4n+1)
x²+2+1 (4+1)
(45-3)
11-200
Гом
(3n²-1
12+3/20
1-200 .3-46
=
1/n²
(를) : 음시
So by Root Test
this series is Ac.
Tim
n->∞
=
2n+1
e
(4n+1).
Diverges
Diverges
+∞ > 1
So by Ratio Test
This Series
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