## Matrix & Series Methods ( Comparison Tests )

Matrix & Series Methods ( Comparison Tests )
Saeed Almheiri MTH306JohnsonS17
Assignment Comparison Tests due 05/22/2017 at 11:59pm PDT
1. (1 point) Determine whether the following series converges
or diverges.

n=1
7
6+4
n
Input C for convergence and D for divergence:

(incorrect)
2. (1 point) Determine whether the following series converges
or diverges:

n=1
n

n
6 +2
Input C for convergence and D for divergence:

(incorrect)
3. (1 point) Match each of the following with the correct
statement.
C stands for Convergent, D stands for Divergent.
1.

n=3
6
n
5 −49
2.

n=1
7+4
n
6+1
n
3.

n=1
ln(n)
5n
4.

n=1
6
n(n+2)
5.

n=1
1
6+
√2
n
6

(incorrect)
4. (1 point) Each of the following statements is an attempt
to show that a given series is convergent or divergent using the
Comparison Test (NOT the Limit Comparison Test.) For each
statement, enter C (for ”correct”) if the argument is valid, or
enter I (for ”incorrect”) if any part of the argument is flawed.
(Note: if the conclusion is true but the argument that led to it
was wrong, you must enter I.)
1. For all n > 1, n
6−n
3 <
1
n
2
, and the series ∑
1
n
2 converges,
so by the Comparison Test, the series ∑
n
6−n
3 converges.
2. For all n > 1, arctan(n)
n
3 <
π
2n
3
, and the series π
2 ∑
1
n
3 converges,
so by the Comparison Test, the series ∑
arctan(n)
n
3
converges.
3. For all n > 2, ln(n)
n
2 >
1
n
2
, and the series ∑
1
n
2 converges,
so by the Comparison Test, the series ∑
ln(n)
n
2 converges.
4. For all n > 1, ln(n)
n
2 <
1
n
1.5
, and the series ∑
1
n
1.5 converges,
so by the Comparison Test, the series ∑
ln(n)
n
2
converges.
5. For all n > 2, 1
n
2−1
<
1
n
2
, and the series ∑
1
n
2 converges,
so by the Comparison Test, the series ∑
1
n
2−1
converges.
6. For all n > 2, n
n
3−5
<
2
n
2
, and the series 2∑
1
n
2 converges,
so by the Comparison Test, the series ∑
n
n
3−5
converges.

(incorrect)
5. (1 point) The three series ∑An, ∑Bn, and ∑Cn have terms
An =
1
n
7
, Bn =
1
n
4
, Cn =
1
n
.
Use the Limit Comparison Test to compare the following series
to any of the above series. For each of the series below, you
must enter two letters. The first is the letter (A,B, or C) of the
series above that it can be legally compared to with the Limit
Comparison Test. The second is C if the given series converges,
or D if it diverges. So for instance, if you believe the series
converges and can be compared with series C above, you would
enter CC; or if you believe it diverges and can be compared with
series A, you would enter AD.
1
1.

n=1
4n
4 +n
7
1309n
11 +8n
4 +4
2.

n=1
4n
3 +n
2 −4n
8n
10 −4n
9 +7
3.

n=1
4n
2 +4n
6
4n
7 +8n
3 −4

(incorrect)
6. (1 point) The series

n=1
n
k
r
n
converges when 0 < r < 1 and
diverges when r > 1. This is true regardless of the value of the
constant k. When r = 1 the series is a p-series. It converges if
k < −1 and diverges otherwise. Each of the series below can be
compared to a series of the form

n=1
n
k
r
n
. For each series determine
the best value of r and decide whether the series converges.
A.

n=1
(3+n(7)
n
)
−5
r = converges or diverges (c or d)?
B.

n=1
n
π
5
2n
5
n +n
9
r = converges or diverges (c or d)?
C.

n=1
n
.5 +4
n
5 +7
r = converges or diverges (c or d)?
D.

n=1

4n
2 +5n+3
−8n
6
n+9 +5n+7

n
8
r = converges or diverges (c or d)?

(incorrect)
7. (1 point) Use the limit comparison test to determine
whether

n=6
an =

n=6
9n
3 −4n
2 +6
4+2n
4
converges or diverges.
(a) Choose a series

n=6
bn with terms of the form bn =
1
n
p
and
simplified fraction. For n ≥ 6,
lim
n→∞
an
bn
= lim
n→∞
(b) Evaluate the limit in the previous part. Enter ∞ as infinity
and −∞ as -infinity. If the limit does not exist, enter DNE.
lim
n→∞
an
bn
=
(c) By the limit comparison test, does the series converge, diverge,
or is the test inconclusive?
• Choose
• Converges
• Diverges
• Inconclusive