HOMEWORK #1 – SOLUTIONS
Page 623
8)
2n + 1
∞
2n + 1
; lim an = ⇒ lim
= lim
n→∞
n→∞
n
∞ n→∞ n
2n + 1
2n + 1− 2n
−2 <ε ⇒
<ε ⇒
n
n
an =
2+
1
1
1
1
<ε ⇒ <ε ⇒n>
n
n
ε
10)
4
4
; lim
=0
n + 1 n→∞ n + 1
4
4
4
−0 <ε ⇒
<ε ⇒
<ε ⇒
n +1
n +1
n +1
16
16
< ε2 ⇒ n > 2 −1
n +1
ε
an =
1
5n − 1
5n − 1
n3 = 5
an = 3 ; lim 3
= lim
2n + 1 n→∞ 2n + 1 n→∞ 2 + 1 2
n3
3
12)
22)
3
5−
1
n =2
3n
3n
1 3n 1 ⎛ 3 ⎞
n
n
an = n
; e + 1 < 2e ⇒ n
> ⋅ = ⋅⎜ ⎟
e +1
e + 1 2 en 2 ⎝ e ⎠
n
1 ⎛ 3⎞
lim ⋅ ⎜ ⎟ = ∞
n→∞ 2 ⎝ e ⎠
n
⎛ 3⎞
⎜⎝ ⎟⎠ > 1
e
3n
∴lim n
= ∞ ⇒ diverges
n→∞ e + 1
26)
an = n 2 + n − n
lim
n→∞
(
⎛ n2 + n + n ⎞
n + n − n ⋅⎜
⎟=
2
⎝ n + n + n⎠
2
)
⎛
⎞
n
lim ⎜
⎟
n→∞ ⎝
n2 + n + n ⎠
30)
⎛
⎞
⎜
⎟ 1
1
= lim ⎜
⎟=
n→∞
1
⎜ 1+ + 1 ⎟ 2
⎜⎝
⎟⎠
n
cos nπ
; − 1 ≤ cos nπ ≤ 1 ⇒
n2
−1 cos nπ 1
≤
≤ 2
n2
n2
n
cos nπ
⎛ −1 ⎞
⎛ 1⎞
lim ⎜ 2 ⎟ = 0 = lim ⎜ 2 ⎟ ⇒ lim
=0
n→∞ ⎝ n ⎠
n→∞ ⎝ n ⎠
n→∞
n2
an =
by the Squeeze Theorem.
34)
( n + 1) − 1
n − 1 an+1 ( n + 1) + 1
n2 + n
an =
;
=
=
> 1 (n ≥ 2)
( n − 1) n 2 + n − 2
n + 1 an
( n + 1)
Therefore, the sequence is increasing.
36)
n! a
an = n ; n+1 =
5
an
n +1
> 1, n ≥ 5
5
( n + 1)!
5 n+1
n!
5n
=
n +1
≤ 1, n = 1, 2, 3, 4
5
Therefore, the sequence is neither increasing nor decreasing.
40)
an =
6n − 1
6n − 1 6n + 18
; an =
<
=6
n+3
n+3
n+3
Hence, the sequence is bounded.
48)
1 n
1
⋅ 2 = 2 n−1; x3 = ⋅ 2 n−1 = 2 n−2 ;
2
2
1
1
1
 xn = xn−1 = ⋅ 2 n−( n−2 ) = ⋅ 2 2 = 2
2
2
2
1
1
xn+1 = ⋅ xn = ⋅ 2 = 1
2
2
x1 = 2 n ⇒ x2 =