Math 2283 Groupwork Answers
November 17, 2016
TA: Mike Loper
Determine whether the series converges or diverges. As always support your answer with work. If
the series is geometric, determine what the series converges to.
(1)
∞
X
3n
√
n5 + 1
n=1
0≤ √
3n
3n
3
≤ √ = 3/2
5
5
n
n +1
n
∞
∞
X
X
3
3n
√
converges
as
it
is
p-series
with
p
=
3/2
>
1.
So
converges
3/2
5+1
n
n
n=1
n=1
by Direct Comparison Test.
We know
(2)
∞
X
3n−1 + 1
n=1
4n+1
∞
X
3n−1
n=1
4n+1
=
∞ n
X
1
3
1
n=1
=
1
12
3
∞ X
n=0
∞ X
3
4
4
4
n+1
n
3
4
n=0
∞
1 X 3 n
=
16
4
n=0
1
1
=
16 1 − 3/4
1
=
4
1
=
12
3
4
This is because the series is geometric with constant term 1/16 and initial ration 3/4 < 1.
And
∞
∞ X
1
1X 1 n
=
4n+1
4
4
n=1
n=1
1
1
=
16 1 − 1/4
1
=
12
This is because the series is geometric with first term 1/16 and ratio 1/4 < 1.
P
P∞
P∞ 3n−1 P∞
3n−1
1
1
Now since ∞
converges
and
converges,
then
n=1 n+1
n=1 n+1
n=1 n+1 +
n=1 n+1
4
4
4
4
converges and is equal to
1
1
1
+ =
12 4
3
1
Math 2283 Groupwork Answers
(3)
November 17, 2016
TA: Mike Loper
∞
X
R 1
en
dx = arctan(x) + C]
[Hint:
Integral
test.
Remember
x2 +1
e2n + 1
n=1
ex
Let f (x) = 2x
. Then it is obvious that f is nonnegative and continuous on [1, ∞).
e +1
Also, using the quotient rule we see that
f 0 (x) =
ex − e3x
ex (e2x + 1) − 2e2x ex
=
< 0 for x > 1.
(e2x + 1)2
(e2x + 1)2
So we may use the integral test. Using the u-substitution u = ex and du = ex dx we get
Z u=∞
Z ∞
du
ex
dx =
2x
2
e +1
u +1
u=e
1
= lim [arctan(u)]u=b
u=e
b→∞
=
Z
∞
Since
1
(4)
∞
X
n=1
pi
− arctan(e)
2
∞
X
ex
en
dx
converges,
then
converges by Integral Test.
e2x + 1
e2n + 1
n=1
en
3n
∞ ∞
X
X
e n
en
e/3
=
=
n
3
3
1 − e/3
n=1
n=1
since the series is geometric with initial term e/3 and ratio e/3 < 1.
Determine whether or not the sequences converge or diverge. If converge, what is the limit?
(5) an = ln(3n + 2) − ln(2n + 3)
lim an = lim(ln(3n + 2) − ln(2n + 3))
3n + 2
= lim ln
2n + 3
3n + 2
= ln lim
because ln(x) is continuous
2n + 3
3n + 2 1/n
= ln lim
2n + 3 1/n
3 + 2/n
= ln lim
2 + 3/n
3
= ln
2
(6) bn =
ln(n)
n
ln(n)
n
ln(x)
= lim
x→∞ x
1/x
L’H
= lim
x→∞ 1
= 0
lim bn = lim
2
Math 2283 Groupwork Answers
(7) cn =
November 17, 2016
TA: Mike Loper
(−1)n + n
(−1)n − n
(−1)n + n
(−1)n − n
(−1)n + n 1/n
(−1)n − n 1/n
(−1)n /n + 1
(−1)n /n − 1
0+1
0−1
−1
lim cn = lim
=
=
=
=
(8) Prove that if {an } converges and {bn } diverges, then {an + bn } must diverge.
Proof. Suppose for the sake of contradiction that {an + bn } converges. Then as {an } converges, {−an } must converge. But since the sum of two convergent sequences converge this
means that that
{bn } = {an + bn − an }
must converge as well. This contradicts the initial assumption so it must be true that
{an + bn } diverges.
3