Continued Fraction Expansion of Limit of Simple Infinite Continued Fraction equals Expansion Itself
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Theorem
Let $\sequence {a_n}_{n \mathop \ge 0}$ be a simple infinite continued fractions in $\R$.
Then $\sequence {a_n}_{n \mathop \ge 0}$ converges to an irrational number, whose continued fraction expansion is $\sequence {a_n}_{n \mathop \ge 0}$.
Proof
By Simple Infinite Continued Fraction Converges to Irrational Number, the value of $\sequence {a_n}_{n \mathop \ge 0}$ exists and is irrational.
Let $\sequence {b_n}_{n \mathop \ge 0}$ be its continued fraction expansion.
By Continued Fraction Expansion of Irrational Number Converges to Number Itself, $\sequence {a_n}_{n \mathop \ge 0}$ and $\sequence {b_n}_{n \mathop \ge 0}$ have the same value.
The result will be achieved by the Second Principle of Mathematical Induction.
First we note that if $\sqbrk {a_0, a_1, a_2, \ldots} = \sqbrk {b_0, b_1, b_2, \ldots}$ then $a_0 = b_0$ since both are equal to the integer part of the common value.
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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:
- $\sqbrk {a_0, a_1, a_2, \ldots} = \sqbrk {a_0, a_1, \ldots, a_k, \sqbrk {a_{k + 1}, a_{k + 2}, \ldots} }$
and similarly:
- $\sqbrk {b_0, b_1, b_2, \ldots} = \sqbrk {b_0, b_1, \ldots, b_k, \sqbrk {b_{k + 1}, b_{k + 2}, \ldots} }$
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As these have the same value and have the same first $k$ partial denominators, it follows that:
- $\sqbrk {a_{k + 1}, a_{k + 2}, \ldots} = \sqbrk {b_{k + 1}, b_{k + 2}, \ldots}$
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.
$\blacksquare$