Calculus II (MAC2312-02)
Test 4 (2015/08/06)
Name (PRINT):
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Page
Points Score
2
16
3
25
4
24
5
18
6
17
Total:
100
Page 1 of 6
Calculus II (MAC2312-02)
Page 2 of 6
1. Circle True of False.
(a) (2 points) True FALSE If we reorder the terms of a convergent, infinite series, the resulting series will have the same sum as the original series.
Solution: Riemann showed we may produce any sum by rearranging the terms of a
convergent, infinite series. On the other hand, if the series is absolutely convergent, any
rearrangement of its terms results in a series with the same sum.
(b) (2 points) TRUE
False If a series is absolutely convergent, then it is also convergent.
Solution: Absolute convergence is a stronger notion of convergence, which does not rely
on cancellations, since taking the absolute value removes sign changes. Convergence is
weaker, which may rely on cancellations. The stronger notion implies the weaker notion,
but not conversely.
2. In each part, determine whether the given sequence converges. If it does, find the limit.
(a) (6 points) an = ln(4n2 + 1) − ln(n2 + 1)
Solution: We have
an = ln(
4n2 + 1
).
n2 + 1
So,
lim an = lim ln(
n→∞
(b) (6 points) bn =
√
n
n→∞
4n2 + 1
4 + 1/n2
)
=
lim
ln(
) = ln(4).
n→∞
n2 + 1
1 + 1/n2
35+3n
Solution:
lim bn = lim
n→∞
n→∞
√
n
1
35+3n = lim (35+3n ) n = lim 35/n+3 = 30+3 = 27.
n→∞
n→∞
Calculus II (MAC2312-02)
Page 3 of 6
3. (15 points) Find the sum of the series
∞
X
n=1
2
if it converges.
n(n + 2)
Hint: sn may be expressed as a telescoping sum.
Solution: Let sN :=
P∞
2
n=1 n(n+2) .
Using partial fractions, we have
N
X
2
n(n + 2)
n=1
N X
1
1
=
−
n n+2
n=1
sN =
add and subtract
1
n+1
to get “unit telescoping pieces”
N X
1
1
1
1
=
−
+
−
n
n
+
1
n
+
1
n+2
n=1
X
N N X
1
1
1
1
=
−
+
−
n
n
+
1
n
+
1
n+2
n=1
n=1
1
1
1
1
=
−
+
−
.
1 N +1
2 N +2
So, limN →∞ sN = 1 + 1/2 = 3/2, which is the sum of the series.
4. (10 points) Find the sum of the series
∞
X
(0.4)n−1 − (0.3)n if it converges.
n=1
Solution: A geometric series
term a P
and common ratio r (where |r| < 1) converges
Pwith firstn−1
∞
n
to a/(1−r). Note that both ∞
(0.4)
and
n=1 (0.3) are geometric series with common
n=1
ratios, 0.4 and 0.3, strictly less than 1. So, they both converge. Their first terms, which in
this case correspond to n = 1 for both, are 0.41−1 = 1 and 0.31 = 0.3, respectively. This
gives
∞
X
n=1
(0.4)
n−1
− (0.3)
n
∞
∞
X
X
n−1
=
(0.4)
−
(0.3)n
n=1
n=1
1
0.3
=
−
1 − 0.4 1 − 0.3
= 1/0.6 − 0.3/0.7 = 5/3 − 3/7 = 26/21.
Calculus II (MAC2312-02)
Page 4 of 6
5. In each part, determine whether the given series converges or diverges. Show your complete
work.
∞
X
1
√
(a) (8 points)
n+1
n=1
Solution: This is almost a p series, but not quite: We don’t have np in the denominator
but (n + 1)p . If we change the index from n to j := n + 1, then j runs from 1 + 1 = 2
to ∞ and we have
∞
∞
X
X
1
1
√
√ ,
=
j
n+1
n=1
j=2
which makes it clear that we have a p series with p = 1/2 < 1, meaning the series is
divergent.
P
P∞ 1
1
1
√1 √1
√1
√
Or we may note that √n+1
> √n+n
= √12 √1n . Since ∞
n=1 2 n =
n=1 n is a p
2
series
p = 1/2 < 1, it diverges and by comparison test so does the larger series
P∞ with
√1 .
n=1 n+1
Or we could use the limit comparison test. Since
√1
n+1
lim
n→∞ √1
n
√
= lim √
n→∞
n
1
= lim p
= 1 > 0,
n→∞
n+1
1 + 1/n
P
P∞
√1
and we know that ∞
n=1 n is divergent, we conclude that
n=1
Or we could use the integral test. Since
Z
∞
1
meaning that
(b) (8 points)
1
is also divergent.
t
(1 + x)1/2 (1 + x)
dx = lim
= ∞,
t→∞
1/2 0
1
P∞
√1
√1
dx
is
divergent,
and
hence
so
is
n=1 n+1 .
1+x
1
√
dx := lim
t→∞
1+x
R∞
√1
n+1
Z
t
−1/2
∞
X
(−1)n
n=1
n2n
1
Solution: This is an alternating series with bn :=
. Since limn→∞ bn = 0, bn > 0,
n2n
1
1
and bn+1 =
< n = bn (i.e., bn is a decreasing sequence), by the alternating
n+1
(n + 1)2
n2
series test, this series converges.
∞
X
sin 3n
(c) (8 points)
1 + 2n
n=1
Solution: The terms of this series irregularly change sign due to the sin 3n factor. Since
the comparison test is applicable to series with positive terms, we look at the absolute
value of the terms and see that
sin 3n | sin 3n|
1
1
1 + 2n = 1 + 2n ≤ 1 + 2n ≤ 2n .
P
1
SinceP ∞
(geometric series with
ratio 1/2 < 1), by comparison
n=1 2n converges
P common
∞ sin 3n sin 3n
test, n=1 1+2n also converges and hence ∞
converges
(absolutely).
n=1 1+2n
Calculus II (MAC2312-02)
(d) (8 points)
Page 5 of 6
n
∞ X
5n + 2
n=1
8n + 5
Solution: Using the ratio test we have
s
5n + 2 n 5 + 2/n
5n + 2
n = lim
= 5/8 < 1.
lim
lim
8n + 5 = n→∞
n→∞
n→∞
8n + 5
8 + 5/n
So, the series converges (absolutely).
(e) (10 points)
∞
X
n=1
1
n + n cos2 3n
Solution: Note that 0 ≤ cos2 3n ≤ 1. So, 0 ≤ n cos2 3n ≤ n and n ≤ n+n cos2 3n ≤ 2n.
That is,
1
1
1
≤
≤ .
2
2n
n + n cos 3n
n
Since
∞
∞
X
1
1X1
=
2n
2 n=1 n
n=1
diverges (harmonic series), so does
∞
X
n=1
by comparison test.
1
,
n + n cos2 3n
Calculus II (MAC2312-02)
Page 6 of 6
6. (10 points) Find a power series representation for the function given by f (x) =
find the interval of convergence of the series.
x
and
+1
3x2
Hint: Recall the power series for 1/(1 − x).
Solution: Recall that
∞
X
1
=
xn
1 − x n=0
for |x| < 1. So,
∞
∞
∞
X
X
X
x
1
2 n
n n 2n
=x
=x
(−3x ) = x
(−1) 3 x =
(−1)n 3n x2n+1
3x2 + 1
1 − (−3x2 )
n=0
n=0
n=0
√
√
for | − 3x2 | < 1, i.e., |x| < 1/ √3, which
means
the
radius
of
convergence
is
1/
3 and the
√
interval of convergence is (−1/ 3, 1/ 3).
7. Circle True or False.
(a) (2 points) True
FALSE The element (differential) of area in polar coordinates is r2 dθ.
1 2
r dθ. This may
2
1
be remembered by noting that the area of a circle of radius r is πr2 = r2 2π.
2
Solution: The element (differential) of area in polar coordinates is
(b) (2 points) TRUE
rdθ.
False The element (differential) of arc length in polar coordinates is
Solution: This may be remembered by noting that the circumference of a circle of
radius r is 2πr = r 2π.
(c) (3 points) TRUE False
p The element (differential) of area for the surface of revolution
about the x axis is 2πy 1 + (y 0 )2 dx.
Solution: This is 2πyds, where ds =
p
p
p
dx2 + dy 2 = 1 + (dy/dx)2 dx = 1 + (y 0 )2 dx.