Question

Show that the sequence defined by \({a_1} = 2,{a_{n + 1}} = \frac{1}{{3 - {a_n}}}\) sequences \(0 < {a_n} \le 2\) and is decreasing . Deduce that the sequence is covering and find its limit.

Short answer

We will prove that the given sequences is converging and \(0 < {a_n} \le 2\) by using the induction.

Step-by-step solution

Given data:-

\(\begin{aligned}{}{a_1} &= 2\\{a_{n + 1}} &= \frac{1}{{3 - {a_n}}}\end{aligned}\)

Using induction to prove \(0 < {a_n} \le 2\).

Taking,

\(\begin{aligned}{}{a_{k + 1}} &= \frac{1}{{3 - {a_k}}}\\{a_k} &= \frac{1}{{3 - {a_{k - 1}}}}\\{a_{k + 1}} - {a_k} &= \frac{{{a_k} - {a_k} - 1}}{{\left( {3 - {a_k}} \right)\left( {3 - {a_{k - 1}}} \right)}}\\|{a_{k + 1}} - {a_k}| &= \frac{{|{a_k} - {a_{k + 1}}|}}{{|\left( {3 - {a_k}} \right)\left( {3 - {a_{k - 1}}} \right)|}}\\\end{aligned}\)

andfor \(0 < {a_k} < 2\) we have

\(|{a_{k + 1}} - {a_k}| \le |{a_k} - {a_{k - 1}}|\)

Then the sequences converges for

\(0 < {a_k} < 2\)

Solving further to evaluate the\(\mathop {\lim }\limits_{n \to \infty } {a_n}\).

\(\begin{aligned}{}{a_\infty } &= \frac{1}{{3 - {a_\infty }}}\\3{a_\infty } - {a_\infty }^2 - 1 &= 0\\{a_\infty } &= \frac{1}{2}(3 \pm \sqrt 5 )\end{aligned}\)

Hence, given sequence has \(0 < {a_k} < 2\) and\({a_\infty } = \frac{1}{2}(3 \pm \sqrt 5 )\).