Calculus II
Practice Problems 1: Answers
Find the limits.
n
ln n 1. lim
n ∞
15
Answer. Here the general term is of the form f n , where f x x ln x function f and use l’Hôpital’s rule:
lim
x ∞
x
ln x 15
15 .
We can then switch to the
x
l H xlim∞ 15 lnx 114 1 x xlim∞ 15 lnx
14
l H xlim∞ 15 14x ln x 13 We can see where this is going: after 13 more steps, we get
lim
x ∞
x
ln x 15
x
1
lim
15! x ∞ ln x
l H
1
1
lim
15! x ∞ 1 x
∞
nk
∞ n!
2. lim
n
Answer. Well, this is just example 5 in section 9.1.
3. lim
n
n ∞
1
2
n
Answer.
lim
n
n ∞
1
n
2
lim 1 n ∞
1
n
2
1
2n 1 2
∞ n2 3n 1
4. lim
n
Answer.
5. lim
n ∞
2n 1 2
∞ n2 3n 1
lim
n
1 n
n!
4n2 4n 1
∞ n2 3n 1
lim
n
4
n
Answer. Both the numerator and denominator of the nth term have n factors; but each factor of the numerator
is larger than the corresponding factor of the denominator. Thus the nth term is greater than the quotient of
the last two factors:
n 1 n 1 n 1
1 n n
n 1 n 1
n!
n
n 1 n 2
1 1
Thus the sequence diverges to ∞.
6. limn
∞n
1 n
Answer. We can calculate this by l’Hôpital’s rule, replacing n by x. First we take logarithms: ln x 1
ln x x. Now
ln x l H
1 x
lim
lim
0
x ∞ x
x ∞ 1
x
Thus limn
∞n
1 n
exp limn
∞ ln x
x e0
1.
∞
7.
5n
∑ n 1
n 1 8
∞
∞
8.
n
5
∑ 8n 1 n 1
Answer.
1
8
∞
5n ∑ 8n n 1
∞
5n
1
8
∞
5n
∑ n
n 0 8
1
1 1
8 1
5
8
1
1
3
1
8
5n
∑ 8n 1
n 1
Answer.
∞
5n
∑ n
n 1 8 1 n∑
1 8n ∞
∞
9.
1
∑ 2k 2k 2
k 1
Answer. Suspecting a telescoping series, we find the partial fraction expansion:
so
n
1
∑ 2k 2k 2 k 1
1
2k 2k 2
1 1
2 2
1
4
1
4
1 1
2 2k
1
6
1
2k 2
1
2n
1 2n 2
1 1
2 2
1
2n 2
which converges to 1/4.
∞
10.
n
∑ n
n 1 2
Answer. This is just example 11 of section 9.2. However, it is interesting to note that we can actually sum
this series, using a clever trick due to Oresme (14th century. The formula for the sum of the geometric series
is classical and appears in Euclid).
∞ ∞
∞ ∞
∞
1
n
1 1
∑ 2n ∑ ∑ 2k ∑ 2n 1 ∑ 2n n 1 k n
n 1
n 0
n 1
using the results of examples 7 and 8.
2