Thomas Calculus 13th [Solutions]

708 Chapter 10 Infinite Sequences and Series
103. (a) If a 2n 1, then
2
2n 2n 1 and
2 2
4 4 1 2 1 2
b a n n n n n n
2 2 2
2 2 2 2 ,
2
2
c a
2n 2n
1
2 2
2 2 2 2
2
2 4 3 2
a b n n n n n n n n
4 3 2 2
2
2
n n n n n n c
4 8 8 4 1 2 2 1 .
2 1 2 2 4 4 1 4 8 4
(b)
a
2
2
a
2
2
2
2n
2n
2
2n
2n
1
lim lim 1 or
a
a
a
a
2
2
lim lim sin lim sin 1
a
2
2
a
2
104. (a)
1 (2 n) ln 2n
1 0
n n
2n
n
2
n
2n
lim 2 n lim exp lim exp lim exp e 1; n ! 2 n ,
Stirlings approximation
2
2n
n 1 (2 n)
!
n
2
n
e
e
n n for large values of n
n
e
n
(b) n n n!
40 15.76852702 14.71517765
50 19.48325423 18.39397206
60 23.19189561 22.07276647
n
e
105. (a)
n
1
n
c c 1
c
ln 1
n n n cn n cn
(b) For all
lim lim lim 0
c
n
0, there exists an
(ln ) c
N e such that
1 1 1 1
c c c
n n n n
0 lim 0
(ln ) c
ln c
n e ln n ln n ln
1
c
106. Let { a n } and { b n } be sequences both converging to L. Define { c n } by c2n b n and c2n
1 a n , where
n 1, 2, 3, . For all 0 there exists N 1 such that when n N 1 then an
L and there exists N2
such that when n N 2 then bn
L . If n 1 2max{ N1, N 2},
then cn
L , so { c n } converges to L.
107.
108.
n 1 n
1 1 0
n n
n
n n
n
e
lim lim exp ln lim exp 1
1 n
1 0
n
n
lim x lim exp ln x e 1, because x remains fixed while n gets large
n
109. Assume the hypotheses of the theorem and let be a positive number. For all there exists an N 1 such that
when n N 1 then an L an L L a n,
and there exists an N 2 such that when n N2
then cn L cn L cn
L . If n max{ N1, N 2},
then L an bn cn
L
b L lim b L.
n
n
n
110. Let 0. We have f continuous at L there exists so that x L f ( x) f ( L ) . Also,
an
L there exists N so that for n N , an
L . Thus for n N , f ( an
) f ( L)
f ( a ) f ( L).
n
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