MAT 127: Calculus C, Fall 2009
Solutions to Midterm II
Problem 1 (15pts)
Determine whether each of the following sequences converges or diverges; justify your answer. You
do not have to determine the limit if the sequence converges.
(a; 5pts)
a
n
=
(

1)
n
n
4
3
n
4
+ 1
diverges
because
a
n
=
(

1)
n
n
4
3
n
4
+ 1
= (

1)
n
n
4
/n
4
3
n
4
/n
4
+ 1
/n
4
= (

1)
n
1
3 + 1
/n
4
;
as
n
→ ∞
the fraction approaches 1
/
3, while the entire expression keeps on jumping from near
1
/
3 (even
n
) to near

1
/
3 (odd
n
).
Grading:
correct answer 2pts; dividing by
n
4
, obtaining 1/3 as limit of the fraction, and some
comment on the sign 1pt each (1pt for
mce
); wrong answer with complete explanation containing
mce
in division by
n
4
or forgetting (

1)
n
max 2pts.
mce
=
mi
nor computa
tional error
(b; 5pts)
a
n
= 1 +
sin(
nπ/
2) ln
n
n
converges
because
(ln
n
)
/n
approaches 0, while

sin(
nπ/
2)
 ≤
1; so sin(
nπ/
2)(ln
n
)
/n
→
0 and
a
n
→
1.
Grading:
correct answer 2pts; correct statements regarding (ln
n
)
/n
and sin(
nπ/
2) 2pts and
1pt, respectively;
so
. . .
not required; wrong answer with complete explanation misidentifying
lim
n
→∞
(ln
n
)
/n
max 2pts.
(c; 5pts)
a
1
= 1
,
a
n
+1
=
a
n
+ 9
2
converges
because
•
a
n
≤
9 for all
n
: this is true for
n
=1; if this is true for
n
,
a
n
+1
= (
a
n
+9)
/
2
≤
(9+9)
/
2 = 9
and so this is true for
n
+1;
•
a
n
≤
a
n
+1
:
a
n
+1
= (
a
n
+9)
/
2
≥
(
a
n
+
a
n
)
/
2 =
a
n
.
So
{
a
n
}
converges by the Monotonic Sequence Theorem.
Grading:
correct answer 2pts;
bounded above
2pts;
increasing
1pt; last sentence not required;
wrong answer 0pts regardless of explanation.
Alternatively (bonus 2pts, whether or not in addition to a standard argument): the sequence
{
a
n
}
is obtained by starting with some
a
1
<
9, then taking
a
2
to be the midpoint of the segment [
a
1
,
9],
a
3
to be the midpoint of the segment [
a
2
,
9], and so on. This sequence of consecutive midpoints
approaches 9, since at each step the new distance to 9 is reduced to 1/2 of the old distance to 9.
Grading:
the substance must be identical to the above for full credit, but could be worded (with
a bit of care) in terms of the sequence
b
n
=9

a
n
; 2nd half of last sentence not required.
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Problem 2 (20pts)
Each of the following sequences converges (you do not need to check this); determine its limit (show
your work!).
(a; 6pts)
a
n
=
n
1 +
√
4
n
2
+ 1
a
n
=
n/n
1
/n
+
√
4
n
2
+ 1
/n
=
1
1
/n
+
radicalbig
4
n
2
/n
2
+ 1
/n
2
=
1
1
/n
+
radicalbig
4 + 1
/n
2
So
a
n
→
1
/
(0 +
√
4 + 0) = 1
/
2
Grading:
correct answer 3pts; dividing by
n
1pt; 2pts to the answer;
mce
1pt (from 6pts); taking
n
under
√
without squaring it resulting in limit of 0 2pts (from 6pts)
mce
=
mi
nor computa
tional error
(b; 7pts)
a
n
=
5
n
n
!
(reminder:
n
! = 1
·
2
·
. . .
·
n
)
1st Approach.
Since the limit exists and
a
n
+1
=
5
n
+ 1
a
n
,
ok
without
n
→∞
lim
n
→∞
a
n
=
lim
n
→∞
a
n
+1
=
lim
n
→∞
parenleftbigg
5
n
+ 1
a
n
parenrightbigg
=
parenleftbigg
lim
n
→∞
5
n
+ 1
parenrightbigg
·
(
lim
n
→∞
a
n
)
= 0
·
(
lim
n
→∞
a
n
)
= 0
2nd Approach.
a
n
≥
0 for all
n
and
a
5+
n
=
5
5
5!
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 Fall '07
 GuanYuShi
 Calculus, lim, wrong answer 0pts, correct answer 2pts

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