Simple Infinite Continued Fraction is Uniquely Determined by Limit

Theorem
Let $(a_n)_{n\geq 0}$ and $(b_n)_{n\geq 0}$ be simple infinite continued fractions in $\R$.

Let $(a_n)_{n\geq 0}$ and $(b_n)_{n\geq 0}$ have the same limit.

Then they are equal.

Proof 1
Follows immediately from Continued Fraction Expansion of Limit of Simple Infinite Continued Fraction equals Expansion Itself.

Proof 2
Note that by Simple Infinite Continued Fraction Converges, they do indeed have a limit.

The result will be achieved by the Second Principle of Mathematical Induction.

Suppose $\left[{a_0, a_1, a_2, \ldots}\right] = \left[{b_0, b_1, b_2, \ldots}\right]$ have the same value.

First we note that if $\left[{a_0, a_1, a_2, \ldots}\right] = \left[{b_0, b_1, b_2, \ldots}\right]$ then $a_0 = b_0$ since both are equal to the integer part of the common value.

This is our basis for the induction.

Now suppose that for some $k \ge 1$, we have:
 * $a_0 = b_0, a_1 = b_1, \ldots, a_k = b_k$.

Then all need to do is show that $a_{k+1} = b_{k+1}$.

Now:
 * $\left[{a_0, a_1, a_2, \ldots}\right] = \left[{a_0, a_1, \ldots, a_k, \left[{a_{k+1}, a_{k+2}, \ldots}\right]}\right]$

and similarly
 * $\left[{b_0, b_1, b_2, \ldots}\right] = \left[{b_0, b_1, \ldots, b_k, \left[{b_{k+1}, b_{k+2}, \ldots}\right]}\right]$.

As these have the same value and have the same first $k$ partial quotients, it follows that:
 * $\left[{a_{k+1}, a_{k+2}, \ldots,}\right] = \left[{b_{k+1}, b_{k+2}, \ldots}\right]$.

But now $a_{k+1} = b_{k+1}$ as each is equal to the integer part of the value of this simple infinite continued fraction.

Hence the result.

Also see

 * Irrational Number is Limit of Unique Simple Infinite Continued Fraction
 * Simple Finite Continued Fraction is Almost Determined by Value