Exercise 2.4.1

Question

\begin{enumerate}[label=(\alph*)] 
  \item Prove that the sequence defined by $x_{1}=3$ and
    $$
    x_{n+1}=\frac{1}{4-x_{n}}
    $$
    converges.
  \item Now that we know $\lim x_{n}$ exists, explain why $\lim x_{n+1}$ must also exist and equal the same value.
  \item Take the limit of each side of the recursive equation in part (a) to explicitly compute $\lim x_{n}$.
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] \item $x_2 = 1$ makes me conjecture $x_n$ is monotonic. For induction suppose $x_n > x_{n+1}$ then we have $$ 4 - x_n < 4 - x_{n+1} \implies \frac{1}{4 - x_n} > \frac{1}{4 - x_{n+1}} \implies x_{n+1} > x_{n+2} $$ Thus $x_n$ is decreasing, to show $x_n$ is bounded notice $x_n$ cannot be negative since $x_n < 3$ means $x_{n+1} = 1/(4-x_n) > 0$. therefore by the monotone convergence theorem $(x_n)$ converges. \item Clearly skipping a single term does not change what the series converges to. \item Since $x = \lim (x_n) = \lim (x_{n+1})$ we must have $$ x = \frac{1}{4 - x} \iff x^2 - 4x + 1 = 0 \iff (x - 2)^2 = 3 \iff x = 2 \pm \sqrt 3 $$ $2 + \sqrt 3 > 3$ is impossible since $x_n < 3$ thus $x = 2 - \sqrt 3$. \end{enumerate}

Exercise 2.4.1

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Exercise 2.4.1

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