57075/80/85
September 16
1. True or False (4 points for each):
(1) If
a
n
≥
0
,
a
n
+1
≤
a
n
for all
n
∈
N
and
lim
n
→∞
a
n
= 0
, then
∞
X
n
=1
a
n
is convergent.
(2) If
lim
n
→∞
(
a
n
+1

a
n
) = 0
, then the sequence
a
n
is convergent.
(3) Suppose the series
∞
X
n
=0
a
n
is convergent and the series
∞
X
n
=0
b
n
is divergent. Then the series
∞
X
n
=0
(
a
n
+
b
n
)
is also divergent.
(4) By L'Hospital's rule, we have
lim
x
→
0
sec
x
1

cos
x
= lim
x
→
0
sec
x
tan
x
sin
x
= lim
x
→
0
1
cos
2
x
= 1
.
(5) Although two series
∞
X
n
=0
a
n
and
∞
X
n
=0
b
n
are both divergent, the series
∞
X
n
=0
(
a
n
+
b
n
)
could
be convergent.
Answer (Mark T for True and F for False):
(1) F. For example,
a
n
= 1
/n
.
(2) F. For example,
a
n
=
n
X
k
=1
1
k
. Then
a
n
+1

a
n
= 1
/
(
n
+ 1)
.
(3) T. Proof: By contradiction. Let
c
n
=
a
n
+
b
n
and suppose
∞
X
n
=1
c
n
is convergent. Then by
the Theorem 8(on page 729 in the textbook),
∞
X
n
=1
c
n

a
n
is also convergent. Since
c
n

a
n
=
b
n
,
it implies
∞
X
n
=1
b
n
is convergent. This contradicts to
∞
X
n
=1
b
n
is divergent.
(4) F.
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 Fall '07
 Sadler
 Exponential Function, lim, Natural logarithm, lim lim bx

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