(a) Consider the sequence of ratios:

0. Since s_{n}^{→} 0 as n → ∞ , 1 /s_{n}^{=}

n !

k_{n} → ∞

as n → ∞ by Theorem 4.2.13. If k < 0, then k^{n} is alternately negative and

positive, and so the limit does not exist (odd numbered terms → -∞ , even

numbered terms → ∞ ).

18. Given that ( s_{n}^{)} is a convergent sequence. Suppose s_{n}^{→} s . Let ( t_{n}^{)} be

the sequence ( t_{n}^{) = (} b, b, b, b, . . . ). Then t_{n}^{→} b . Since s_{n}^{≤} b for all

n , it follows that s ≤ b by Theorem 4.2.4. Now let ( u_{n}^{)} be the sequence

( u_{n}^{) = (} a, a, a, a, . . . ). Then u_{n}^{→} a . Since a ≤ s_{n}^{for all} n , it follows that

a ≤ s , again by Theorem 4.2.4.

2

Section 4.3

3. (a) s_{1}^{= 1 and} s_{n} +1^{=}^{1}

5^{(} s_{n}^{+ 7)} for all n >

1.

Note: s_{2}^{=}

8

5

, s_{3}^{=}

43

25

, s_{4}^{=}

218

125

. . . .

( s_{n} ) appears to be increasing: Proof by induction –

1 ∈ S : s_{2}^{=}^{1}

5^{(} s_{1}^{+ 7) =}^{1}

5^{(8) =}^{8}

5^{>} 1 = s_{1} . Therefore 1 ∈ S .

Assume that k ∈ S . That is, assume s_{k} +1^{≥} s_{k}^{.}

Prove that k + 1 ∈ S . That is, prove that s_{k} +2^{≥} s_{k} +1^{.}

s_{k} +2^{=}^{1}

5^{(} s_{k} +1^{+ 7)} ≥^{1}

5^{(} s_{k}^{+ 7) =} s_{k} +1 . Therefore, k + 1 ∈ S .

Thus, S = N — the sequence is increasing.

( s_{n} ) is bounded: Claim s_{n}^{<} 2 for all n ; by induction. Let S be the

set of positive integers for which s_{n}^{<}

2.

1 ∈ S : s_{1}^{= 1} <

2. Thus 1 ∈ S .

Assume that k ∈ S . That is, assume s_{k}^{<}

2.

Prove that k + 1 ∈ S . That is, prove s_{k} +1^{<}

2.

s_{k} +1^{=}^{1}

5^{(} s_{k}^{+ 7)} ≤^{1}

5^{(2 + 7) =}^{9}

5^{<}

2. Therefore, k + 1 ∈ S and S = N .

Now, since ( s_{n}^{)} is increasing and bounded above, s_{n}^{→} s ≤

2.

For large n , s_{n}^{≈} s and s_{n} +1^{≈} s . Therefore, for large n

s =

1

5

( s + 7) ⇒ 5 s = s + 7 ⇒ s =

7

4

.

(c) s_{1}^{= 2 and} s_{n} +1^{=}^{1}

4^{(2} s_{n}^{+ 7)} for all n >

1.

( s_{n} ) is increasing: By induction –

1 ∈ S : s_{2}^{=}^{1}

4^{(2} s_{1}^{+ 7) =}^{1}

4^{(11) =}^{11}

4^{>} 2 = s_{1} . Therefore 1 ∈ S .

Assume that k ∈ S . That is, assume s_{k} +1^{≥} s_{k}^{.}

3

Prove that k + 1 ∈ S . That is, prove that s_{k} +2^{≥} s_{k} +1^{.}

#### You've reached the end of your free preview.

Want to read all 7 pages?

- Fall '08
- Staff
- Math, lim, Natural number, Sn