Real Analysis Fall 2004 Take Home Test 1 SOLUTIONS 1. Use the ...

For any given ε > 0 there exists N 2 > 0 such that for n > N 2 , |x n − L| < 2ε
L . 2
Let N = max{N 1 , N 2 }. Then for n > N,
1
Thus lim
n→∞ x n
= L.
1
∣ − 1 x n L∣ = |x n − L|
|x n ||L|
≤ |x n − L|
L 2 / 2
< 2ε/L2
L 2 / 2
= ε.
1
3. Let x n+1 = , x 1 > 0. Prove that the sequence {x n } converges and then compute
3 + x n
the limit of the sequence.
1
< 1 3 + x n 3
= 3
10 for all n ≥ 1. Hence 3 10 < x n < 1 3
First Proof (My Preference). Since x 1 > 0, x n+1 =
x n+1 = 1
3 + x n
< 1
3 + 1 3
1
(b) If lim x n = L exists, then L = lim x n+1 = lim =
n→∞ n→∞ n→∞ 3 + x n
Thus
L = −3 + √ 13
.
2
(a) If x 1 > L, then x 2 =
1
3 + L
for all n ≥ 1. Moreover,
for all n > 1.
1
lim (3 + x n) = 1
3 + L
n→∞
= L. (1)
L 2 + 3L − 1 = 0. (2)
1
< 1
3 + x 1 3 + L = L from (1). More generally, if x 2n−1 > L,
1
then x 2n = < 1
3 + x 2n−1 3 + L = L. On the other hand, if x 1 < L, then x 2n−1 < L
and x 2n > L. Without loss of generality, assume x 1 > L. Define the sequence {s n }
as s n = |x n − L|. Then claim that s n+1 < s n for all n ≥ 1. If not, then s n+1
≥ 1.
s n
Assume n = 2k. Then
s 2k+1
s 2k
= x 2k+1 − L
L − x 2k
=
1
3+x 2k
− L
≥ 1
L − x 2k
1 − 3L − Lx 2k ≥ 3L − 3x 2k + Lx 2k − x 2 2k
(x 2 2k − 2Lx 2k + L 2 ) − L 2 − 6L + 3x 2k + 1 ≥ 0
(x 2k − L) 2 − 1 − 3L + 3x 2k + 1 ≥ 0 (using (2))
(x 2k − L)[x 2k − L + 3] ≥ 0.