Math 111-002
Assignment # 9 - Answers
1. Determine whether the sequence converges or diverges. If it converges,
find the limit.
3 + 5n2
4+n
Answer. We have, for n ≥ 4,
(a) an =
an =
3 + 5n2
5n2
≥
−−−→ ∞
4+n
4 + 4 n→∞
So limn→∞ an = ∞.
(b) an = 8.3 + (0.715)n
Answer. Since 0 < 0.715 < 1, we get limn→∞ (0.715)n = 0. Then
lim 8.3 + (0.715)n = 8.3 + lim (0.715)n = 8.3 + 0 = 8.3
n→∞
n→∞
7n
100 + 8n
Answer. We have, since 0 < 7/8 < 1,
n
7n
7
7n
≤ n =
0≤
−−−→ 0.
n
n→∞
100 + 8
8
8
(c) an =
By the squeeze property, limn→∞
7n
= 0.
100 + 8n
7n
(d) an =
Answer. We can repeat the argument from above:
n + 8n
since 0 < 7/8 < 1,
n
7n
7n
7
0≤
≤ n =
−−−→ 0.
n
n→∞
n+8
8
8
By the squeeze property, limn→∞
1
7n
= 0.
n + 8n
nπ
(e) an = cos
n+2
Answer. First, note that
n
1
= lim
= 1.
n→∞ n + 2
n→∞ 1 + 2/n
lim
Then, because the cosine is continuous,
nπ
nπ
= cos lim
= cos π = −1.
lim cos
n→∞ n + 2
n→∞
n+2
(f) an = ln(n + 2) − ln n
Answer. Here,
ln(n + 2) − ln n = ln
n+2
n
2
= ln 1 +
.
n
Then, using the continuity of ln as 1,
2
2
= ln lim 1 +
= ln 1 = 0.
lim ln(n+2)−ln n = lim ln 1 +
n→∞
n→∞
n→∞
n
n
(g) an = n1/n
1
Answer. We have n1/n = e n
lim
x→∞
ln n
. Using L’Hôpital,
ln x
1/x
= lim
= 0.
x→∞
x
1
The, by the continuity of the exponential,
1
lim n1/n = lim e n
n→∞
n→∞
ln n
1
= elimn→∞ n
ln n
= e0 = 1.
(h) an = cos nπ
Answer. We have cos nπ = (−1)n . So the values of the sequence
constantly alternate between 1 and −1, and so the limit cannot
exist.
(ln n)2
(i) an =
n
2
Answer. Again we use L’Hôpital (twice). We have
(ln x)2
= lim
lim
x→∞
x→∞
x
2
x
ln x
2
= lim ln x = 0.
x→∞ x
1
Then, as the limit of the function is zero at infinity, we get
(ln n)2
(ln x)2
= lim
= 0.
n→∞
x→∞
n
x
lim
(j) an = arctan(ln n)
Answer. We have limn→∞ ln n = ∞. So
lim arctan(ln n) = lim arctan m =
n→∞
m→∞
π
.
2
n
3
(k) an = 1 +
n
Answer. Here
n
3
3
1+
= exp n ln 1 +
.
n
n
Using L’Hôpital,
3
ln 1 +
3
3
x
lim n ln 1 +
= lim x ln 1 +
= lim
n→∞
x→∞
x→∞
n
x
1/x
3
3
− 2/ 1 +
x
x
= lim
2
x→∞
−1/x
3
= 3.
= lim x→∞
3
1+
x
By the continuity of the exponential,
3
3
lim exp n → ∞n ln 1 +
= exp lim n ln 1 +
= exp(3) = e3 .
n→∞
n→∞
n
n
3