Homework 5
9.14. Let
s
n
=
a
n
n
p
, so lim

s
n
+1
s
n

is
lim

a
n
p
(
n
+ 1)
p

= lim

a

(
n
n
+ 1
)
p

=

a

lim

(
n
n
+ 1
)
p

=

a

lim((1

1
n
+ 1
)
p
) =

a

(lim(1

1
n
+ 1
))
p
=

a

(lim(1)

lim(
1
n
+ 1
))
p
=

a

where we used the scalar multiplication theorem for limits once, we
used the multiplication theorem for limits
p
times, and we used the
addition theorem for limits once. If

a

<
1 then by 9.12(a) we have
that lim
s
n
= 0.
If

a

>
1, then
a >
1 or
a <

1.
If
a >
1 then
all of the terms of
s
n
are positive, and so we have by 9.12(b) that
+
∞
= lim

s
n

= lim
s
n
.
If
a <

1 then

s
n

diverges to infinity by
9.12(b), but the even terms of
s
n
are positive while the odd terms of
s
n
are negative. Therefore, the limit does not exist in this case.
9.18. This problem introduces the
geometric series
.
(a) Observe that
(1

a
)(1+
a
+
a
2
+
· · ·
+
a
n
) = (1+
a
+
a
2
+
· · ·
+
a
n
)

(
a
+
a
2
+
· · ·
+
a
n
+1
)
= 1

a
n
+1
so if
a
= 1 then we can divide the left and right sides of this equation
by 1

a
to obtain
1 +
a
+
a
2
+
· · ·
+
a
n
=
1

a
n
+1
1

a
as desired.
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 Spring '09
 Multiplication, Scalar, Limits, lim, Limit of a sequence, np np np

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