Thomas Calculus 13th [Solutions]

720 Chapter 10 Infinite Sequences and Series
96. (a)
2 n 1 n 1
L 4 4 4 4
1 3, L2 3 , L
3 3 3 , , L 3 lim lim 3
3 n
L
3 n
n n
3
(b) Using the fact that the area of an equilateral triangle of side length s is
3
4
2
s , we see that
3
A 1 4 ,
2
3
2
3 3 3 3 3 3
A 1 1
2 A1 3 4 3 4 12 , A3 A2
3(4) 4 2
3 4 12 27
,
2 2
2 3 3 3
A 1 1
4 A3 3(4) 4 , A
3 5 A4
3(4) 4
, ,
4
3 3
n k 1 n
3 2 3 3 3
1
n
k
3
3
1 k
k
k
A
1 4
n 3(4) 3 3(4) 3 3 .
4 4 2 1
3 4 9 4
k
9
k 2 k 2 k 2
n
3
1
3 3 36 3 3 3
lim A lim 3 3 4 k
1
3 8 8
n A
3 3 3 3 1
4 k 1 4 1
4
n n
9
4 20 4 5 4 5 5
k 2
9
10.3 THE INTEGRAL TEST
1. f ( x ) 1 is positive, continuous, and decreasing for 1;
2
x
1 b
1 1 b
dx lim
1 x b 1 x b
x 1
x dx lim
2 2
1
lim 1 1 1 1
b
1
2
b
x
dx converges 1
2
n 1 n
converges
f x 1 is positive, continuous, and decreasing for 1;
x
2. ( )
0.2
1
b
1
5 0.8
b
dx dx lim x
1 x b 1 x b
4 1
x lim
0.2 0.2
lim
b
5 0.8 5 1
4 4 1 x
1
n 1 n
b dx diverges
0.2
0.2
diverges
3. f ( x ) 1 is positive, continuous, and decreasing for 1;
2
x 4
1 b 1 dx
1 x 4 b 1 x 4
x dx lim
2 2
1
b
1 1 1 1 1 1 1 1
lim tan x lim tan b tan tan 1 1
b
1 b
1
n
2
1 n 4
2 2 2 2 2 2 4 2 2 1
2
x 4
converges
dx converges
4. f ( x ) 1 is positive, continuous, and decreasing for x 1; 1
b
1
b
dx lim dx lim ln x 4
x 4
1 x 4
b 1 x 4
b
1
lim ln | b 4| ln 5
1 dx diverges 1
b
1 x 4
4
n 1 n
diverges
5.
2x
2x
2
f ( x)
e is positive, continuous, and decreasing for x 1;
lim
b x
e dx e dx
1 b 1
1 2x
b
1 1 1
2x
2n
lim e lim e dx converges e converges
2 2b
2 2
b
1 b 2e 2e 2e
1
n 1
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