But
(
n
+ 1)!
n
!
=
n
+ 1
,
while
n
n
(
n
+ 1)
n
+1
=
1
n
+ 1
parenleftbigg
n
n
+ 1
parenrightbigg
n
.
Thus
a
n
+1
a
n
= 3
parenleftbigg
n
n
+ 1
parenrightbigg
n
→
3
e
>
1
as
n
→ ∞
, so series (C) diverges.
Consequently, of the given infinite series,
only
A
and
B
converge.
CalC7g23d
003
10.0 points
Find the value of
lim
x
→ ∞
x
3
3
x
.
1.
limit = 0
correct
2.
limit = 3
3.
none of the other answers
4.
limit =
1
3
5.
limit =
∞
6.
limit =
∞
Explanation:
Set
f
(
x
) =
x
3
,
g
(
x
) = 3
x
=
e
x
ln3
.
Then
f, g
are everywhere differentiable func
tions such that
lim
x
→ ∞
f
(
x
) =
lim
x
→ ∞
g
(
x
) =
∞
Thus L’Hospital’s Rule applies, in which case
lim
x
→ ∞
f
(
x
)
g
(
x
)
=
lim
x
→ ∞
f
′
(
x
)
g
′
(
x
)
.
Now
f
′
(
x
) = 3
x
2
,
g
′
(
x
) = (ln 3) 3
x
.
But then
lim
x
→ ∞
f
′
(
x
)
g
′
(
x
)
=
lim
x
→ ∞
3
x
2
(ln 3) 3
x
,
which, up to a constant, is the same limit we
started with except that
x
3
in the numerator
has become
x
2
.
Consequently, if we apply
L’Hospital’s rule sufficiently often, we finally
end up with
lim
x
→ ∞
f
(
x
)
g
(
x
)
=
lim
x
→ ∞
constant
3
x
.
Thus
lim
x
→ ∞
x
3
3
x
= 0
.
CalC12b50s
004
10.0 points
If the
n
th
partial sum
S
n
of an infinite series
∞
summationdisplay
n
=1
a
n
Version 001 – tester – srinivasan – (54690)
3
is given by
S
n
= 3

n
2
n
,
find
a
n
for
n >
1.
1.
a
n
=
n

2
2
n
correct
2.
a
n
=
n

2
2
n
−
1
3.
a
n
= 3
parenleftbigg
n

2
2
n
parenrightbigg
4.
a
n
=
3
n

2
2
n
5.
a
n
= 3
parenleftbigg
3
n

2
2
n
parenrightbigg
6.
a
n
= 3
parenleftbigg
n

2
2
n
−
1
parenrightbigg
Explanation:
By definition, the
n
th
partial sum of
∞
summationdisplay
n
=1
a
n
is given by
S
n
=
a
1
+
a
2
+
· · ·
+
a
n
.
In particular,
a
n
=
braceleftbigg
S
n

S
n
−
1
,
n >
1,
S
n
,
n
= 1.
Thus
a
n
=
S
n

S
n
−
1
=
n

1
2
n
−
1

n
2
n
=
2 (
n

1)
2
n

n
2
n
when
n >
1. Consequently,
a
n
=
n

2
2
n
for
n >
1.