Quiz 6 for MATH 105 SECTION 205
March 27, 2015
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Given Name
Student Number
1. (1 point) Is the series
∞
X
k=1
√
3
k4 + 1
√
convergent or divergent?
k5 + 9
convergent
1.
Answer.
We get
√
√
3
3
1
k4 + 1
k4
1
1
For the term √
, it behaves like √ = 5 4 = 7 as k → ∞, so let’s compare it with 7 .
k5 + 9
k5
k2−3
k6
k6
√
3 4
√ k +1
k5 +9
1
7
k6
Since
∞
X
1
7
6
k=1 k
√
∞
3
X k4 + 1
√
k=1
k5 + 9
p
√
3
6
(k 4 + 1)2 · k 7
k4 + 1 7
· k6 = p
→ 1,
= √
6
k5 + 9
(k 5 + 9)3
is a p-series with p = 7/6, then
∞
X
1
7
k=1
as k → ∞.
converges. By the limit comparison test, then the series
k6
converges.
∞
X
(k!)2
convergent or divergent?
2. (1 point) Is the series
(2k)!
k=1
2.
Answer.
Let ak =
convergent
(k!)2
, then
(2k)!
ak+1
ak
=
[(k+1)!]2
[2(k+1)]!
(k!)2
(2k)!
=
(k + 1)!
k!
2
·
(2k)!
(2k + 2)!
1
(2k + 2)(2k + 1)
k2
1
→ , as k → ∞.
2
4k + 6k + 2
4
= k2 ·
=
By the ratio test, then the series
∞
X
k=1
√
3
k4 + 1
√
is convergent.
k5 + 9
∞ h
π n
π n i
X
3. (2 points) Evaluate the series
cos
− 2 · − tan
.
3
6
n=2
3.
√
1
3
− +
2
3
Answer.
Since cos
∞ h
X
π 3
cos
n=2
π √3
1
= and tan
=
, then
2
6
3
π n
3
− 2 · − tan
π n i
6
=
=
=
=
=
=
=
∞ X
cos
−2
∞ X
− tan
π n
n=2
6
√ !n
∞
X
3
1 n
−2
−−
2
3
n=2
n=2
√ 2
1 2
− 33
2
√
−2·
1 − 12
1 + 33
1
1
√
−2·
2
3+ 3
√
3− 3
1
−2·
2
√6
1
3
−1+
2
√ 3
1
3
− +
.
2
3
∞
X
k=2
Let f (x) =
3
n=2
∞ X
4. (a) (3 points) Use the integral test to show that the series
Answer.
π n
1
is convergent.
+ 1)
k 2 (k
1
, then f (x) is positive and decreasing. By the partial fractional decom+ 1)
x2 (x
position, we have
1
A
B
C
= + 2+
.
+ 1)
x
x
x+1
x2 (x
Both sides of the above identity multiplied by x2 (x + 1), then
1 = Ax(x + 1) + B(x + 1) + Cx2
= Ax2 + Ax + Bx + B + Cx2
= (A + C)x2 + (A + B)x + B.
Then we get
A + C = 0,
A + B = 0,
and B = 1.
So we get
B = 1,
A = −1,
and C = 1.
So we get
1
1
1
1
=− + 2 +
.
x2 (x + 1)
x x
x+1
Then
Z
Z
Z
Z
1
1
1
dx +
|dx +
dx
f (x) dx = −
2
x
x
x+1
1
= − ln |x| − + ln |x + 1| + C
x
x + 1
1
+ C.
= − + ln x
x Hence we have
Z
∞
b
Z
f (x) dx
b
1
1
= lim − + ln 1 + b→∞
x
x 1
1
1 = lim
+ ln 1 + + 1 − ln 2
b→∞
b
b
= 0 + ln 1 + 1 − ln 2
f (x) dx =
1
lim
b→∞
1
= 1 − ln 2.
So by the integral test, we have the series
∞
X
k=2
1
is convergent.
+ 1)
k 2 (k
(b) (1 point) Use the comparison test to show that the series
∞
X
k=2
Answer.
Notice that k 2 (k + 1) = k 3 + k 2 ≥ k 3 , then
0<
Since
∞
X
k=1
1
is convergent.
k 2 (k + 1)
1
1
≤ 3,
k 2 (k + 1)
k
for all k ≥ 1.
∞
∞
X
X
1
1
is
a
p-series
with
p
=
3,
then
converges. By the comparison test, then the series
k3
k3
k=1
k=1
1
converges.
2
k (k + 1)
(c) (2 points) Evaluate the series
∞
X
k=2
1
.
+ 1)
k 2 (k
∞
X
1
π2
Hint:
=
and the result of the partial fraction decomposition in Problem 4 (a).
k2
6
k=1
Answer.
By the computation in part (a), we have
1
1
1
1
=− + 2 +
.
+ 1)
k k
k+1
k 2 (k
Notice that
∞
X
π2
1
=
,
k2
6
and
k=1
∞ X
1
k=1
1
−
= 1.
k k+1
Then we have
∞
X
k=2
1
2
k (k + 1)
∞ X
1
1
1
=
+
−
k2 k + 1 k
k=2
∞
∞ X
X
1
1
1
=
−
−
k2
k k+1
=
=
k=2
π2
k=2
1
−1−
6
2
π2 3
− .
6
2
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