Theorem: Continued Fraction Expansion of Euler's Number

The constant Euler's number $e$ has the continued fraction expansion:
{{begin-eqn}}
{{eqn | l = e
      | r = \sqbrk {2; 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, \ldots }
      | c =
}}
{{eqn | l = 
      | r = \sqbrk {1, 0, 1, 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, \ldots }
      | c =
}}
{{end-eqn}}

Proof:
From the recursive definition of continued fractions, we have:
{{begin-eqn}}
{{eqn | l = p_i
      | r = a_i p_{i - 1} + p_{i - 2}
      | c =
}}
{{eqn | l = q_i
      | r = a_i q_{i - 1} + q_{i - 2}
      | c =
}}
{{end-eqn}}
Let:
{{begin-eqn}}
{{eqn | l = \sqbrk {a_0, a_1, a_2, a_3, a_4, a_5, a_6, a_7, a_8, a_9, \ldots}
      | r = \sqbrk {1, 0, 1, 1, 2, 1, 1, 4, 1, 1, \ldots}
}}
{{end-eqn}}
In other words:
:$a_{3 n + 1} = 2 n$
and:
:$a_{3 n + 0} = a_{3 n + 2} = 1$
Then $p_i$ and $q_i$ are as follows:
:<nowiki>$\begin{array}{r|cccccccccc}
\ds i &  0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9  \\
\hline
p_i & 1 & 1 & 2 & 3 & 8 & 11 & 19 & 87 & 106 & 193 \\
q_i & 1 & 0 & 1 & 1 & 3 & 4 & 7 & 32 & 39 & 71 \\
\hline
\end{array}$</nowiki>
Furthermore, $p_i$ and $q_i$ satisfy the following $6$ recurrence relations:
{{begin-eqn}}
{{eqn | n = 1
      | l = p_{3 n + 0}
      | m = \paren {a_{3 n + 0} } p_{3 n - 1} + p_{3 n - 2}
      | mo= =
      | r = p_{3 n - 1} + p_{3 n - 2}
}}
{{eqn | n = 2
      | l = p_{3 n + 1}
      | m = \paren {a_{3 n + 1} } p_{3 n + 0} + p_{3 n - 1}
      | mo= =
      | r = 2 n p_{3 n + 0} + p_{3 n - 1}
}}
{{eqn | n = 3
      | l = p_{3 n + 2}
      | m = \paren {a_{3 n + 2} } p_{3 n + 1} + p_{3 n + 0}
      | mo= =
      | r = p_{3 n + 1} + p_{3 n + 0}
}}
{{eqn | n = 4
      | l = q_{3 n + 0}
      | m = \paren {a_{3 n + 0} } q_{3 n - 1} + q_{3 n - 2}
      | mo= =
      | r = q_{3 n - 1} + q_{3 n - 2},
}}
{{eqn | n = 5
      | l = q_{3 n + 1}
      | m = \paren {a_{3 n + 1} } q_{3 n + 0} + q_{3 n - 1}
      | mo= =
      | r = 2 n q_{3 n + 0} + q_{3 n - 1},
}}
{{eqn | n = 6
      | l = q_{3 n + 2}
      | m = \paren {a_{3 n + 2} } q_{3 n + 1} + q_{3 n + 0}
      | mo= =
      | r = q_{3 n + 1} + q_{3 n + 0},
}}
{{end-eqn}}
Our ultimate aim is to prove that:
:$\ds \lim_{n \mathop \to \infty} \frac {p_n} {q_n} = e$
In the pursuit of that aim, let us define the integrals:
{{begin-eqn}}
{{eqn | l = A_n
      | r = \int_0^1 \frac {x^n \paren {x - 1}^n} {n!} e^x \rd x
}}
{{eqn | l = B_n
      | r = \int_0^1 \frac {x^{n + 1} \paren {x - 1}^n} {n!} e^x \rd x
}}
{{eqn | l = C_n
      | r = \int_0^1 \frac {x^n \paren {x - 1}^{n + 1} } {n!} e^x \rd x
}}
{{end-eqn}}
=== Lemma ===
{{:Continued Fraction Expansion of Euler's Number/Proof 1/Lemma}}{{qed|lemma}}
We assert that $A_n$, $B_n$ and $C_n$ all converge to $0$ as $n \to \infty$:
{{begin-eqn}}
{{eqn | l = \lim_{n \mathop \to \infty} A_n
      | r = \frac {\frac {x^{n + 1} \paren {x - 1}^{n + 1} } {\paren {n + 1}!} e^x} {\frac {x^n \paren {x - 1}^n } {n!} e^x}
      | c = Radius of Convergence from Limit of Sequence: Real Case
}}
{{eqn | r = \frac {x \paren {x - 1} } {\paren {n + 1} }
}}
{{eqn | r = 0
}}
{{eqn | l = \lim_{n \mathop \to \infty} B_n
      | r = \frac {\frac {x^{n + 2} \paren {x - 1}^{n + 1} } {\paren {n + 1}!} e^x} {\frac {x^{n + 1} \paren {x - 1}^n} {n!} e^x}
      | c = Radius of Convergence from Limit of Sequence/Real Case
}}
{{eqn | r = \frac {x \paren {x - 1} } {\paren {n + 1} }
}}
{{eqn | r = 0
}}
{{eqn | l = \lim_{n \mathop \to \infty} C_n
      | r = \lim_{n \mathop \to \infty} B_n - \lim_{n \mathop \to \infty} A_n
}}
{{eqn | r = 0
}}
{{end-eqn}}
We now have:
{{begin-eqn}}
{{eqn | l = \lim_{n \mathop \to \infty} A_n
      | m = \lim_{n \mathop \to \infty} \paren {q_{3 n} e - p_{3 n} }
      | mo= =
      | r = 0
}}
{{eqn | l = \lim_{n \mathop \to \infty} B_n
      | m = \lim_{n \mathop \to \infty} \paren {p_{3 n + 1} - q_{3 n + 1} e}
      | mo= =
      | r = 0
}}
{{eqn | l = \lim_{n \mathop \to \infty} C_n
      | m = \lim_{n \mathop \to \infty} \paren {p_{3 n + 2} - q_{3 n + 2} e}
      | mo= =
      | r = 0
}}
{{end-eqn}}
from which we conclude:
{{begin-eqn}}
{{eqn | l = \lim_{n \mathop \to \infty} \paren {p_n - q_n e}
      | r = 0
}}
{{eqn | l = \lim_{n \mathop \to \infty} p_n 
      | r = q_n e
}}
{{eqn | l = \lim_{n \mathop \to \infty} \frac {p_n} {q_n}
      | r = e
}}
{{end-eqn}}
{{qed}}