Solutions to Practice Problems, Math 220
1 Evaluate the following limits:
2n
1
lim 1 +
and
n→∞
n
Solution
n
1
lim 1 −
.
n→∞
n
2n n 2
1
1
1+
=
1+
n
n
and so
2n
1
lim 1 +
= e2 .
n→∞
n
Similarly,
n n −1
1
n
1−
=
n
n−1
and so
lim
n→∞
1
1−
n
n
= e−1 .
2 Use the definition of limit to evaluate
n
a: limn→∞ 2 + n1
b: limn→∞ 3nn
c: limn→∞
sin(n)+cos(n)
3
n
Solution
n
a: We claim that the sequence diverges. If limn→∞ 2 + n1 = x
for some real number x, then since
n
n
√
1
1
n
=2 1+
< 2n · e
2+
n
2n
and
2+
1
n+1
n+1
> 2n+1 = 2n · 2,
we have
n+1 n
√
1
1
2+
− 2+
> (2 − e) · 2n > 0.7.
n+1
n
If we take = 0.35, then there exists N such that n > N implies
n
1
2+
− x < 0.35.
n
1
2
On the other hand,
n+1 n
n
n
1
1
1
1
2+
≤ 2+
− 2+
−x + x− 2+
n+1
n
n
n
and this is less than 0.35 + 0.35 = 0.7, a contradiction.
b: We have limn→∞ 3nn = 0. To see this, note that 3nn < 2−n for
all n ∈ N (you can show this by induction). If we thus take
N = − log()/ log(2), then n > N implies that
n
n
− 0 = n < 2−n < 2−N = ,
n
3
3
as desired.
n
sin(n)+cos(n)
c: limn→∞
= 0. Here, we use the inequalities
3
−2 < sin(n) + cos(n) < 2,
which lead to
−(2/3) <
n
sin(n) + cos(n)
3
n
< (2/3)n .
Taking N = − log()/ log(3/2), and assuming n > N implies
that
n
sin(n) + cos(n)
− 0 (2/3)n < (2/3)N = .
3
3 For each of the following sequences, determine whether or not the
sequence converges or diverges. Prove your answer.
a: a1 = 1, an+1 = an + a1n
√
b: b1 = 3, bn+1 = 1 + bn − 1
Solution
a: If lim an = a, then we would have a = a + a1 and so 0 = 1/a, a
contradiction. It follows that an diverges.
b: We claim that bn ≤ 3. To see this, use induction. Since b1 = 3,
let use suppose that bk ≤ 3. Then
p
√
bk+1 = 1 + bk − 1 ≤ 1 + 2 < 3.
Similarly, we have bn > 2. To see this, again use induction.
Next we claim
that bn is decreasing. Indeed,
if bn+1 /bn > 1
√
√
then 1 + bn − 1 > bn or bn − 1 < bn − 1, contradicting
bn > 2. It follows, via the Monotone Convergence
Theorem
√
that bn converges, say to b. Since we have 1 + b − 1 = b and
b ≥ 2, we conclude that lim bn = 2.
P
P
4 Suppose that
an converges. Either prove that
bn converges or
give a counterexample, where
a: bn = an /n
3
√
b: bn = an /n, an ≥ 0
c: bn = an sin(n)
d: bn = n1/n an
Solution Upon reflection, you can skip this question, which involves
rather more of the chapter on series than we’ve covered.
√
5 If xn = 3 n, show that (xn ) satisfies lim |xn+1 − xn | = 0, but that
it is not a Cauchy sequence.
Solution Since
√
√
3
n+1− 3n =
1
1
< (
(
(
(n +
+ (n(n + 1)) 1/3) + n 2/3)
3n 2/3)
√
√
√
3
3
it follows that lim n + 1 − n = 0. On the other hand, since 3 n is
unbounded, it fails to converge and hence is not a Cauchy sequence.
1)( 2/3)
6 Prove that the set S of odd integers is countable.
Solution As a subset of Z, S is countable.