Thomas Calculus 13th [Solutions]

Section 10.7 Power Series 761
interval of convergence is 0 x 16; the series
and its sum is
1
1 1 2
x 2 2 x 2 4 x
2 2
n
0
x 2
2
n
is a convergent geometric series when 0 x 16
46.
n
n 1
un
1 (ln x) 1
x x e x e when
u
n
n n (ln x)
lim 1 lim 1 ln 1 1 ln 1 ;
1
x e or e we obtain the
series
when
1 n and
n 0
e
1
x e
n
( 1) n
0
which both diverge; the interval of convergence is
1
e x e;
(ln x) n 1
1 ln x
n 0
47.
n 1 n x 1
2
lim u 2 2
n 1
1 lim x 1 3 1 2
3 1 2
3 lim |1| 1 x
3
1 x 2 x
u
2
n
n n x 1
n
2 x 2; at x 2 we have
2 x 2; the series
1 1 3
1
x x 2 x
2
1 3
2
1
3 3
2
2 1
x
3
n 0
n
n
(1) n
0
which diverges; the interval of convergence is
is a convergent geometric series when 2 x 2 and its sum is
48.
2
n 1
u
x 1
n
n 1
2
2
lim 1 lim 1 1 2 3 3;
u
n 1
2
n
n
n n 2 x 1
x x when x 3 we have
a divergent series; the interval of convergence is 3 x 3; the series
2 1
x
2
n 0
1 1 2
1
x
2
1 2 x
2
1 3 x
geometric series when 3 x 3 and its sum is
2
2
2
n
n
is a convergent
n
1 ,
0
49. Writing 2 x as 2
1 [ ( x 1)]
series
for
we see that it can be written as the power
n n n
2[ ( x 1)] 2( 1) ( x 1) . Since this is a geometric series with ratio ( x 1) it will converge
n 0 n 0
( x 1) 1or 0 x 2.
50. (a)
(b)
5 5 / 3 5 x
x
f ( x) , which converges for 1 or x 3.
3 x 1 ( x / 3) 3 3
3
n
0
n
n
3 3 / 2 3 x
3 n
x
g( x) x , which converges for 1 or x 2.
x 2 1 ( x / 2) 2 2 n 1
2
2
n 0 n 0
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