A Continued Fraction With Two Convergences?

First, we need to formalize what it means by

$7 - \dfrac{12}{7 - \dfrac{12}{7 - \dfrac{12}{\ldots}}}$

The common definition is this: define the sequence $a_1 = 7, a_{n+1} = 7 - \frac{12}{a_n}$ , then the expression above is defined to be $\displaystyle \lim_{n \to \infty} a_n$ .

If the sequence converges to a limit $L$ , then it satisfies $L = 7 - \frac{12}{L}$ . But there are two solutions, $L = 3$ and $L = 4$ . Which one is correct? Moreover, it's still possible that neither of them is correct, if the sequence doesn't converge.

This is a pretty common problem; I posted a solution to a similar problem . The steps are generally in this form:

• Find the candidates for the limit
• Prove that the sequence is monotonic (either $a_{n+1} \ge a_n$ for all $n$ , or $a_{n+1} \le a_n$ for all $n$ )
• Prove that it's bounded
• Conclude by the monotone convergence theorem that it converges
• Hope that you're lucky enough that all but one of the limits are ruled out by the bounds
• By process of elimination, the sequence must converge to the surviving limit

For this problem, we will follow the same steps:

• Find the candidates for the limit: this was done above, giving $L = 3, 4$
• Prove that it's bounded below by 4 ( $a_n > 4$ for all $n$ )
• Note that this rules out 3 (it "doesn't come close" to 3, so the limit cannot be 3)
• Prove that the sequence is monotonically decreasing ( $a_{n+1} < a_n$ for all $n$ )
• By monotone convergence theorem, this sequence converges
• The only remaining option is if it converges to 4, so the limit is 4

Step 1 : Find the candidates

This was done above.

Step 2 : Sequence is bounded below by 4

We proceed by induction. Note that $a_1 = 7 > 4$ , so this serves as the base case. For the inductive step, note that since $a_n > 4$ , we have $\frac{12}{a_n} < 3$ , and so $a_{n+1} = 7 - \frac{12}{a_n} > 7 - 3 = 4$ .

Step 3 : This rules out 3 as a limit

We use the epsilon-delta definition of limit for this. A sequence $\{a_n\}$ converges to a limit $L$ precisely when for every $\epsilon > 0$ , there exists $N > 0$ such that for all $n > N$ , we have $|a_n - L| < \epsilon$ .

Suppose, for the sake of contradiction, that $L = 3$ . Since the sequence is bounded below by 4, we have $a_n > 4$ for all $n$ . This means $a_n - 3 > 1$ . Since $a_n - 3 > 1 > 0$ , $a_n - 3$ is positive, so $|a_n - 3| = a_n - 3$ . Now consider $\epsilon = 1$ . We have $|a_n - L| = |a_n - 3| = a_n - 3 > 1$ for all $n$ , contradiction with the requirement that there exists some $N$ such that $|a_n - L| < 1$ for all $n > N$ . Thus the assumption that $L = 3$ is incorrect; 3 cannot be the limit.

Step 4 : The sequence is monotonically decreasing

This needs the use of Step 2. Since $a_n > 4$ , we know that $(a_n - 4)(a_n - 3) > 0 \cdot 1 > 0$ , so it is positive. We can rearrange that into $a_n^2 - 7a_n + 12 > 0$ or $a_n^2 > 7a_n - 12$ . Since $a_n > 4$ , $a_n$ is positive, so dividing by $a_n$ doesn't flip the inequality sign: $a_n > 7 - \frac{12}{a_n} = a_{n+1}$ . This completes the proof.

Step 5 : The sequence converges

We apply the monotone convergence theorem. The sequence is monotonically decreasing, and it is bounded below, so it converges.

Step 6 : The sequence converges to 4

We know, from Step 1, that if the sequence converges, then it converges to either 3 or 4. From Step 5, we know that the sequence does converge. From Step 3, the limit cannot be 3. The only remaining option is that the limit is 4; that is, the value of the expression above is $\boxed{4}$ .

I literally don't know how I solved it I just came up with the answer :)

To save myself some typing, I'll tell you that I'm using the formalism of the following article. Generalized Continued Fraction

In this problem, $a_n = -12$ and $b_n = 7$ for all $n$ . Then $A_n = 7A_{n-1} - 12A_{n-2}$ and $B_n=7B_{n-1}-12B_{n-2}$ . The seed values for these relations are as follows.

$A_0=b_0=7,A_1=b_0b_1+a_1=37,B_0=1,B_1=b_1=7$

We wish to solve the recurrence relations for $A_n$ and $B_n$ . The standard trick for solving linear recurrence relations such as these is to solve the characteristic polynomial equation for the relation.

The characteristic equation for $A_n$ is $x^2 = 7x-12$ , and its solutions are $x=3$ or $x=4$ . So the solution of the recurrence relation is of the form $A_n=c_13^n+c_24^n$ . Using the seed values, we obtain $c_1=-9,c_2=16$ , so the solution is $A_n=-9\cdot 3^n+16\cdot 4^n$ .

$B_n$ has the same characteristic equation, so $B_n=d_13^n+d_24^n$ . Using the seed values, we obtain $d_1=-3,d_2=4$ , so the solution is $B_n=-3\cdot 3^n+4\cdot 4^n$ .

Now define a sequence $x_n$ as follows.

$x_n=\frac{A_n}{B_n}=\frac{-9\cdot 3^n+16\cdot 4^n}{-3\cdot 3^n+4\cdot 4^n}$

Dividing the numerator and denominator by $4^n$ yields the following.

$x_n=\frac{-9\left(\frac{3}{4}\right)^n+16}{-3\left(\frac{3}{4}\right)^n+4}$

Finally, take the limit of the sequence.

$\lim_{n\to\infty}x_n=\lim_{n\to\infty}\frac{-9\left(\frac{3}{4}\right)^n+16}{-3\left(\frac{3}{4}\right)^n+4}=\frac{-9\cdot 0+16}{-3\cdot 0 +4}=4$