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Given that

$\left\{ \begin{matrix} a_{n+1}=\frac{1+c_n}{1+\frac{1}{b_n}}\\ b_{n+1}=\frac{1+a_n}{1+\frac{1}{c_n}}\\ c_{n+1}=\frac{1+b_n}{1+\frac{1}{a_n}}\\ \end{matrix} \right. $
where $a_0b_0c_0=1$ and $a_0, b_0, c_0>0$
Find $\displaystyle\lim_{n\to\infty} a_n$, $\displaystyle\lim_{n\to\infty} b_n$ and $\displaystyle\lim_{n\to\infty} c_n$

My work:

  • $a_nb_nc_n=1$, it can be proved by induction
  • By checking the first 100 terms with excel, $a_n, b_n$ and $c_n$ probably converge to $1$.
  • The sequence is not monotonic.
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2 Answers 2

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Convergence

First I use the identity $a_nb_nc_n = 1$ and manipulate the first expression $$ a_{n+1} = \frac{1+c_n}{1+ \frac1{b_n}} = \frac{1+c_n}{1+a_n c_n} $$

From here it is clear to see that if $a_n = 1$, then $a_{n+1} = 1$ and hence by induction the sequence $a_n$ converges to $1$. From here on we assume $a_n \ne 1$.

The equation can be rearranged into

$$c_n \left ( a_{n+1} a_n - 1 \right) = 1 - a_{n+1} $$

From here, we use the facts that $c_n$ is positive (an easy induction shows this) and $a_n \ne 1$ to see that $ a_{n+1} a_n - 1 $ and $ a_{n+1} - 1$ have opposite signs. Note that the equation implies that $a_{n+1} \ne 1$

In the case where $a_n <1$, we first note that if $a_{n+1} <1$ then we have a contradiction. Noting that $a_{n+1} \ne 1$ also holds we then have $a_{n+1} > 1$. As this holds we can then conclude that $ a_n a_{n+1} -1 < 0$ and hence that $a_{n+1} < \frac 1{a_n}$. From here we conclude that in this case

$$ a_{n+1} \in \left ( 1, \frac 1 {a_n} \right ) $$

In the second case where $a_n > 1$, a similar line of reasoning gives that that

$$ a_{n+1} \in \left (\frac 1 {a_n} , 1 \right ) $$

Going again to the case where $a_n < 1$ we can iterate our result to find that

$$ a_{n+2} \in \left ( a_n, 1 \right ) $$

Similarly for $a_n > 1$ we have that

$$ a_{n+2} \in \left ( 1, a_n \right ) $$ This means that in this case where $a_0 > 1$, the sequence $a_0, a_2, a_4, a_6, \ldots$ is strictly decreasing and bounded below by 1, and hence must converge. Similarly in the case where $a_0 <1 $, the sequence $a_0, a_2, a_4, a_6, \ldots$ is strictly increasing and bounded above by 1, and hence must converge. We call this limit $A$. A similar analysis can be done with the sequence $a_1, a_3, a_5, a_7, \ldots$ We call the limit of this sequence $A'$. We can do the same analysis for $\{b_n\}$ and $\{c_n \}$ to get $B$, $B'$, $C$ and $C'$.

Values

At this point all we need to do is solve the system of equations (all of these can be found by taking the limit of all $c_n \left ( a_{n+1} a_n - 1 \right) = 1 - a_{n+1} $ type relations)

$$ C(A'A - 1) = 1 - A' $$ $$ C'(A'A - 1) = 1 - A $$ $$ A(B'B - 1) = 1 - B'$$ $$ A'(B'B - 1) = 1 - B$$ $$ B(C'C - 1) = 1 - C'$$ $$ B'(C'C - 1) = 1 - C$$ $$ABC = 1$$ $$A'B'C' = 1$$

Multiplying the first, third and fifth equations together, and applying the seventh we get that

$$ (A'A - 1)(B'B - 1)(C'C - 1) = (1- A')(1-B')( 1 - C') $$

Multiplying the second, fourth and sixth equations together and applying the seventh we get that

$$ (A'A - 1)(B'B - 1)(C'C - 1) = (1- A)(1-B)( 1 - C) $$

Hence we have

$$ (1- A')(1-B')( 1 - C') = (1- A)(1-B)( 1 - C) $$

Recalling our earlier analysis, one of $A$ and $A'$ is $\ge 1$ (It was a decreasing sequence that approached the limit from above) and one was $\le 1$ (This is the increasing sequence that approached that limit from below) The same holds for $B$ and $B'$ as well as $C$ and $C'$. Suppose now that none of $A$, $B$ and $C$ are $1$. Then the right hand side of the above equation is nonzero, and hence none of $A'$, $B'$ and $C'$ are $1$ either. But then both sides of the equation have opposite signs, and we have a contradiction. Hence one of $A$, $B$ and $C$ is $1$. Without any loss of generality we can suppose that it is $A$.

From here we can substitute $A= 1$ into the first equation to get

$$ C (A' - 1) = 1 - A' $$

Both sides have opposite signs unless $A' = 1$, hence $A' = 1$.

We can substitute this into the third and fourth equations to get

$$BB' -1 = 1 - B'$$ $$BB' -1 = 1 - B$$

Here we can immediately conclude that $B = B'$ and hence that $B^2 - 1 = 0$ and hence that $B = B' = 1$> We repeat this process to conclude that $C = C' = 1$

This gives us the expected values for the limits.

An Edge Case

If you have followed through this answer carefully you may have noticed that it does not cover the edge case where one of $a_0$, $b_0$ and $c_0$ is 1 and the other two aren't. Here we resolve this edge case. Suppose without generality that $a_0 = 0$ (and hence as discussed in the first section $a_n = 1$ for every $n$. We can than use the fact that $b_n c_n = 1$ to write the equation for $b_{n+1}$ as

$$b_{n+1} = \frac{ 2} {1 + b_n} = f(b_n)$$ where $f(x) = \frac 2{1+x}$.

Now note that $$b_0 \in (0, \infty)$$ $$b_1 \in f(0, \infty) = (0,2)$$ $$b_2 \in f(0,2) = \left (2, \frac 23 \right)$$ $$b_3 \in f \left (2, \frac 23 \right) = \left (\frac 23 , \frac 65 \right ) $$

The equation for $b_{n+1} can be rearranged to show that

$$|b_{n+1} - 1| = \frac{ |b_n - 1| }{1+b_n}$$

This shows that $|b_n - 1|$ is decreasing, and hence given our previous analysis for $n \ge 3$, $|b_n - 1| \le \frac 13$ and hence $1+ b_n \ge \frac 53 $. This means that for $n \ge 3$

$$|b_{n+1} - 1| \le \frac 35 |b_n - 1| $$

Hence, by induction $$|b_{n+3} - 1 | \le \left (\frac 35 \right ) ^ n |b_3 - 1| $$

This shows that the limit of $\{b_n\}$ is 1. Since $c_n = \frac 1 {b_n}$, the limit of $\{c_n \}$ is also 1.

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  • $\begingroup$ How $c_n$ is always positive ? Even $c_0$ can be negative. $\endgroup$ Jul 6, 2023 at 18:00
  • $\begingroup$ Ah oops. For some reason I thought that $a_0$, $b_0$, and $c_0$ are all positive. $\endgroup$
    – Ben Martin
    Jul 7, 2023 at 1:37
  • $\begingroup$ WIth negative numbers introduced into the mix there are significant problems with the sequence even being well defined. If at any point during the sequence any of the numbers becomes $-1$ the next step will have undefined numbers. Simply guaranteeing that none of $a_0$ ,$b_0$ and $c_0$ are not $-1$ doesn't do the trick either. setting them to $1$, $-3$, and $-\frac 13$, runs into the $-1$ issue on the second step. $\endgroup$
    – Ben Martin
    Jul 7, 2023 at 1:55
  • $\begingroup$ Furthermore, when negative numbers are included, if there is a limit it doesn't need to be 1. The case where $a_0 = 1$, $b_0 = -2$ and $c_0 = - \frac 12$ is an example of this. $\endgroup$
    – Ben Martin
    Jul 7, 2023 at 12:08
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    $\begingroup$ Sorry, it is my mistake. a, b, c should be positive. I have edited my question. $\endgroup$
    – eoj
    Jul 7, 2023 at 14:26
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The question is about the limiting behavior of the recursion $$ a_{n+1}=\frac{1+c_n}{1+\frac{1}{b_n}}, \quad b_{n+1}=\frac{1+a_n}{1+\frac{1}{c_n}}, \quad c_{n+1}=\frac{1+b_n}{1+\frac{1}{a_n}} $$ with initial values such that $\,a_0b_0c_0 = 1.\,$ The first issue is that there are several divisions in the recursion and, for simplicity, assume that none of them are divisions by zero. Notice the invariant relation $\, a_{n+1}b_{n+1}c_{n+1} = a_n b_n c_n.\,$ With the initial value assumption this implies $\,a_n b_n c_n=1\,$ for all $\,n\,.$

Define the quantities (assuming $\,a_0\ne 1$) $$ x_n := (-1/2)^n,\, q_1 = \frac{3(a_0-1)}{a_0+a_0b_0+1},\, q_2 = \frac{3(a_0-1)(a_0+a_0b_0-2)}{(a_0+a_0b_0+1)^2}. $$ Verify by simple algebra and induction that $$ a_n = 1+\frac{q_1^2 x_n}{q_1-q_2x_n},\, b_n = 1+\frac{(3q_2-2q_1^2)x_n}{q_1 +(q_1^2-q_2)x_n},\, c_n = \frac{q_1-q_2x_n}{q_1-(q_1^2-2q_2)x_n}. $$ Now $\,x_n\,$ converges to $0$ which implies that the sequences $\,a_n,b_n,c_n\,$ converge to $1$.

The special case when $\,a_0=1\,$ is interesting. Define the integer sequence $\,s_n := (1-(-2)^n)/3.\,$ Due to the invariant relation, $\,c_0=1/b_0.\,$ In general, $$ a_n = 1,\quad b_n = \frac{b_0-s_n(2+b_0)}{1-s_n(2+b_0)}, \quad c_n = 1/b_n $$ for all integer $\,n.\,$ Also, now $\,b_n\,$ converges to $1$ which implies $\,c_n\,$ does also.

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