(a) Since we only need to show that the sequence converges and not find the actual limit, we can use the Monotone Convergence theorem.
Therefore, we need to prove that the sequence is monotone and bounded. Let's begin by proving that the sequence is decreasing. After writing down a few terms, we may come to think that the sequence is decreasing.
x n = ( 3 , 1 , 1 3 , 3 11 , 11 41 , … ) . x_{n} = (3, 1, \frac{1}{3}, \frac{3}{11}, \frac{11}{41}, \dots). x n = ( 3 , 1 , 3 1 , 11 3 , 41 11 , … ) . To prove that it actually is, we use mathematical induction. The base case is obviously true since
1 = x 2 ≤ x 1 = 3. 1 = x_2 \leq x_1 = 3. 1 = x 2 ≤ x 1 = 3. We now prove the induction step by supposing that the statement is true for the pair x n + 1 , x n x_{n + 1}, x_{n} x n + 1 , x n
x n + 2 − x n + 1 = 1 4 − x n + 1 − 1 4 − x n = ( 4 − x n ) − ( 4 − x n + 1 ) ( 4 − x n + 1 ) ( 4 − x n ) = x n + 1 − x n ( 4 − x n + 1 ) ( 4 − x n ) > 0 \begin{aligned}
x_{n + 2} - x_{n + 1}
&= \frac{1}{4 - x_{n + 1}} - \frac{1}{4 - x_{n}} \\
&= \frac{(4 - x_{n}) - (4 - x_{n + 1})}{(4 - x_{n + 1}) (4 -
x_{n})} \\
&= \frac{x_{n + 1} - x_{n}}{(4 - x_{n + 1})(4 - x_{n})} > 0
\end{aligned} x n + 2 − x n + 1 = 4 − x n + 1 1 − 4 − x n 1 = ( 4 − x n + 1 ) ( 4 − x n ) ( 4 − x n ) − ( 4 − x n + 1 ) = ( 4 − x n + 1 ) ( 4 − x n ) x n + 1 − x n > 0 (b) Suppose x n x_{n} x n x x x ϵ > 0 \epsilon > 0 ϵ > 0
∃ N ∈ N , ∀ n ∈ N , n ≥ N ⟹ ∣ x n − x ∣ < ϵ . \exists N \in \mathbb{N}, \forall n \in \mathbb{N}, \quad n \geq N
\implies \abs{x_{n} - x} < \epsilon. ∃ N ∈ N , ∀ n ∈ N , n ≥ N ⟹ ∣ x n − x ∣ < ϵ . This also implies that for the given N N N
∀ n ∈ N , n ≥ N ⟹ ∣ x n + 1 − x ∣ < ϵ . \forall n \in \mathbb{N}, n \geq N
\implies \abs{x_{n + 1} - x} < \epsilon. ∀ n ∈ N , n ≥ N ⟹ ∣ x n + 1 − x ∣ < ϵ . since n + 1 > n n + 1 > n n + 1 > n
(c) Taking the limit on each side, we have
lim n → ∞ x n = lim n → ∞ x n + 1 = lim n → ∞ 1 4 − x n \begin{aligned}
\lim_{n \to \infty} x_{n}
&= \lim_{n \to \infty} x_{n + 1} \\
&= \lim_{n \to \infty} \frac{1}{4 - x_{n}}
\end{aligned} n → ∞ lim x n = n → ∞ lim x n + 1 = n → ∞ lim 4 − x n 1 By labeling lim n → ∞ x n \lim_{n \to \infty} x_{n} lim n → ∞ x n L L L
L = lim n → ∞ 1 lim n → ∞ 4 − x n = 1 4 − lim n → ∞ x n = 1 4 − L . \begin{aligned}
L &= \frac{\lim_{n \to \infty} 1}{\lim_{n \to \infty} 4 -
x_{n}} \\
&= \frac{1}{4 - \lim_{n \to \infty} x_{n}} \\
&= \frac{1}{4 - L}.
\end{aligned} L = lim n → ∞ 4 − x n lim n → ∞ 1 = 4 − lim n → ∞ x n 1 = 4 − L 1 . With some algebraic manipulations, we clearly obtain a quadratic equation:
L = 1 4 − L ⟹ ( 4 − L ) L = 1 ⟹ 4 L − L 2 = 1 ⟹ − L 2 + 4 L − 1 = 0. \begin{aligned}
L = \frac{1}{4 - L}
&\implies (4 - L)L = 1 \\
&\implies 4L - L^2 = 1 \\
&\implies -L^2 + 4L - 1 = 0.
\end{aligned} L = 4 − L 1 ⟹ ( 4 − L ) L = 1 ⟹ 4 L − L 2 = 1 ⟹ − L 2 + 4 L − 1 = 0. We can then use the quadratic formula
x = − b ± b 2 − 4 a c 2 a = − 4 ± 16 − 4 ( − 1 ) ( − 1 ) 2 ( − 1 ) = − 4 ± 12 − 2 = 2 ± 3 . \begin{aligned}
\bm{x} &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\
&= \frac{-4 \pm \sqrt{16 - 4(-1)(-1)}}{2(-1)} \\
&= \frac{-4 \pm \sqrt{12}}{-2} \\
&= 2 \pm \sqrt{3}.
\end{aligned} x = 2 a − b ± b 2 − 4 a c = 2 ( − 1 ) − 4 ± 16 − 4 ( − 1 ) ( − 1 ) = − 2 − 4 ± 12 = 2 ± 3 . Since x n ≤ 3 x_{n} \leq 3 x n ≤ 3 n n n 2 ± 3 > 3 2 \pm \sqrt{3} > 3 2 ± 3 > 3 2 − 3 2 - \sqrt{3} 2 − 3
