Assignment 1 (MATH 214 B1)
1. Determine whether the given sequence converges or not. If it converges, find its limit.
n−1
n2 + 1
3n2 − 5
(c) cn =
2 + 3n − 4n2
n2 − 2
n+3
(−1)n
(d) dn = √
n
(a) an =
(b) bn =
Solution. (a) We have
lim an = lim
n→∞
n→∞
n(1 − 1/n)
1 (1 − 1/n)
n−1
= lim 2
= lim
= 0.
2
2
n→∞
n→∞
n +1
n (1 + 1/n )
n (1 + 1/n2 )
(b) We have
n2 (1 − 2/n2 )
1 − 2/n2
n2 − 2
= lim
= lim n
= ∞.
n→∞ n(1 + 3/n)
n→∞
n→∞ n + 3
1 + 3/n
lim bn = lim
n→∞
(c) We have
n2 (3 − n52 )
3n2 − 5
=
lim
= lim
n→∞ 2 + 3n − 4n2
n→∞ n2 ( 22 + 3 − 4)
n→∞
n
n
lim cn = lim
n→∞
3 − n52
3
=− .
2
3
4
n2 + n − 4
(d) Observe that
1
(−1)n
1
−√ ≤ √
≤√ .
n
n
n
Since
1
1
lim − √ = lim √ = 0,
n→∞
n n→∞ n
by the Sandwich Theorem for Sequences we obtain
(−1)n
√
= 0.
n→∞
n
lim dn = lim
n→∞
2. Determine whether the given sequence converges or not. If it converges, find its limit.
( 2 )n
( √10 )n
(a) an = −
(b) bn =
3
3
n
n
2 −1
3 −1
(d) dn = n
(c) cn = n
2 +1
2 +3
Solution. (a) We have r = −2/3. Since |r| < 1, we obtain
( 2 )n
lim an = lim −
= 0.
n→∞
n→∞
3
1
(b) Since
√
10/3 > 1, the sequence diverges.
(c) We have
2n − 1
2n (1 − 1/2n )
1 − 1/2n
=
lim
=
lim
= 1.
n→∞ 2n + 1
n→∞ 2n (1 + 1/2n )
n→∞ 1 + 1/2n
lim
(d) We have
( 3 )n 1 − 1/3n
3n − 1
3n (1 − 1/3n )
=
lim
=
lim
= ∞.
n→∞ 2n + 3
n→∞ 2n (1 + 3/2n )
n→∞ 2
1 + 3/2n
lim
3. Determine whether the given geometric series converges or diverges. If it converges,
find its sum.
∑∞
(a) n=1 1/2n
∑∞
(c) n=0 (−3)n+1 4−n
(b)
(d)
∑∞
n=2 (−3)
∑∞
n=1
n
/5n−1
32n /23n
Solution. (a) We have a = 1/2 and r = 1/2. Since |r| < 1, the series converges and
its sum is
a
1/2
=
= 1.
1−r
1 − 1/2
s=
(b) We have a = a2 = (−3)2 /5 = 9/5 and r = −3/5. Since −1 < r < 1, the series
converges and its sum is
s=
a
9/5
9
=
= .
−3
1−r
8
1− 5
(c) We have a = a0 = −3 and r = −3/4. Hence, the series converges and its sum is
s=
a
−3
12
=
=− .
−3
1−r
7
1− 4
(d) We have
an = 9n /8n = (9/8)n .
It follows that r = 9/8. Since r > 1, the series diverges.
4. Determine whether the given series converges or diverges. Give reasons for your
answers. If a series converges, find its sum.
∞ (
∞
)
∑
∑
1
1
(a)
−
(b)
2−1/n
2
2
n
(n
+
1)
n=1
n=1
2
(c)
∞
∑
1−n
100n
n=1
(d)
∞
∑
2n−1 − 1
5n−1
n=1
Solution. (a) Let an := 1/n2 − 1/(n + 1)2 for n = 1, 2, . . .. We have
(
(1
)
1) (1
1)
1
1
sn = a1 +a2 +· · ·+an = 1− 2 + 2 − 2 +· · ·+ 2 −
= 1−
.
2
2
2
3
n
(n + 1)
(n + 1)2
It follows that limn→∞ sn = 1. Therefore, the series converges and its sum is 1.
(b) Since limn→∞ 2−1/n = 1 ̸= 0, by the nth term test for divergence, the series
diverges.
(c) Since
1
1−n
=−
̸= 0,
n→∞ 100n
100
lim
by the nth term test for divergence, the series diverges.
(d) We have
∞
∞
∞
∑
2n−1 − 1 ∑ 2n−1 ∑ 1
=
−
.
n−1
n−1
n−1
5
5
5
n=1
n=1
n=1
The two geometric series on the right of the above equation converge:
∞
∑
1
5
2n−1
=
=
5n−1
1 − 2/5
3
n=1
and
∞
∑
n=1
1
5n−1
=
1
5
= .
1 − 1/5
4
Therefore, the given series converges and
∞
∑
2n−1 − 1
5 5
5
= − =
.
n−1
5
3
4
12
n=1
5. Use the the Integral Test to determine whether the given series converges or diverges.
∞
∞
∑
∑
1
1
(a)
(b)
2
n
2n + 1
n=1
n=1
∞
∞
∑ 1
∑ 1
√
(c)
(d)
n
n ln2 n
n=1
n=2
Solution. (a) Let f (x) := 1/x2 . Then f is continuous, positive, and decreasing on
[1, ∞). We have
∫
n
lim
n→∞
1
[ 1 ]n
(
1
1)
dx = lim −
= lim 1 −
= 1.
n→∞
x2
x 1 n→∞
n
By the integral test, the series
∑∞
1
n=1 n2
converges.
3
(b) Let g(x) :=
1
2x+1 .
Then g is continuous, positive, and decreasing on [1, ∞). We
∫
have
[1
]n
1
dx = lim
ln(2x + 1) = ∞.
n→∞ 1 2x + 1
n→∞ 2
1
∑∞
1
By the integral test, the series n=1 2n+1
diverges.
√
(c) Let h(x) := 1/ x. Then h is continuous, positive, and decreasing on [1, ∞). We
n
lim
∫
have
[ √ ]n
[ √
]
dx
√ = lim 2 x 1 = lim 2 n − 2 = ∞.
n→∞ 1
n→∞
x n→∞
∑∞ 1
By the integral test, the series n=2 √n diverges.
n
lim
(d) Let u(x) := 1/(x ln2 x). Then u is continuous, positive, and decreasing on [2, ∞).
We have
∫
lim
n→∞
2
n
[ 1 ]n
( 1
1
1 )
1
=
lim
−
=
.
dx
=
lim
−
2
n→∞
ln x 2 n→∞ ln 2 ln n
ln 2
x ln x
By the integral test, the series
∑∞
1
n=1 n ln2 n
4
converges.