Recursive sequence and limiting value

Recursively defined sequences appear throughout the HSC Sequences and Series topic, and proving their convergence requires more than just finding a limit numerically. In this problem we use a fixed-point equation to identify a candidate limit, then build a contraction inequality to give a rigorous, epsilon-free proof that the sequence indeed converges. You’ll walk away with a powerful two‑step method – solve the steady-state equation, then bound the error – that works for many rational recursions, including those linked to continued fractions.

Problem Statement

Let $(x_n)$ be defined by

$$x_1=16,\qquad x_{n+1}=2+\frac{1}{x_n}\quad (n\ge 1).$$

• Find exact values of $x_2,x_3,x_4$.
• Assuming $x_n\to L>0$, show that $L=1+\sqrt{2}$.
• Prove that

$$|x_{n+1}-L|=\frac{|x_n-L|}{Lx_n},$$

and deduce that for $n\ge 2$,

$$|x_{n+1}-L|<\frac14|x_n-L|.$$

• Hence show $x_n\to 1+\sqrt{2}$.

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Solutions

(i)
We start by computing the next few terms directly from the recurrence.

$$x_2=2+\frac1{16}=\frac{33}{16},\quad x_3=2+\frac{16}{33}=\frac{82}{33},\quad x_4=2+\frac{33}{82}=\frac{197}{82}.$$

For $n\ge 2$, we have $x_n>2$ because each term is $2+\frac{1}{x_{n-1}}$ with positive second part.

(ii)
If the sequence converges to a limit $L$, then as $n\to\infty$, both $x_n$ and $x_{n+1}$ approach $L$, so the recurrence becomes the fixed‑point equation for $L$.

Taking limits in the recurrence gives

$$L=2+\frac1L \implies L^2-2L-1=0.$$

So $L=1\pm\sqrt2$. Since terms are positive and $>2$ from $n\ge2$, we take

$$L=1+\sqrt2.$$

(iii)
To obtain a contraction inequality, we subtract the fixed‑point equation $L=2+\frac1L$ from the recurrence:

$$x_{n+1}-L=\left(2+\frac1{x_n}\right)-\left(2+\frac1L\right) =\frac{L-x_n}{Lx_n},$$

thus

$$|x_{n+1}-L|=\frac{|x_n-L|}{Lx_n}.$$

Now $L=1+\sqrt2>2$ and $x_n>2$ for $n\ge2$, so $Lx_n>4$. Hence

$$|x_{n+1}-L|<\frac14|x_n-L|,\quad n\ge2.$$

(iv)
The contraction factor $\frac14$ guarantees that the error shrinks by at least a factor of $\frac14$ each step from $n=2$ onward. Repeatedly applying the inequality gives geometric decay of the error:

$$|x_{n}-L|\le \left(\frac14\right)^{n-2}|x_2-L|\to 0.$$

Therefore $x_n\to L=1+\sqrt2$.

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Takeaways

• For recursive limits, first solve the fixed-point equation.
• A contraction inequality gives a rigorous convergence proof.
• The same method works for many rational recursions.
• The sequence $x_n$ is related to the continued fraction of $1+\sqrt{2}$. It can be expressed as

$$x_{n+1} = 2 + \frac{1}{x_n} = 2 + \frac{1}{2 + \frac{1}{x_{n-1}}} = \dots = 2 + \frac{1}{2 + \frac{1}{2 + \dots + \frac{1}{x_1}}}$$

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Further Readings

HSC Functions, HSC Induction, HSC Complex Numbers, HSC Last Resorts