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
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
Calculus: Early Transcendentals
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Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
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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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