Theorem: Continued Fraction Expansion of Euler's Number/Proof 1/Lemma

:For $n \in \Z , n \ge 0$:
{{begin-eqn}}
{{eqn | l = A_n
      | r = q_{3 n} e - p_{3 n}
}}
{{eqn | l = B_n
      | r = p_{3 n + 1} - q_{3 n + 1} e
}}
{{eqn | l = C_n
      | r = p_{3 n + 2} - q_{3 n + 2} e
}}
{{end-eqn}}

Proof:
To prove the assertion, we begin by demonstrating the relationships hold for the initial conditions at $n = 0$:
{{begin-eqn}}
{{eqn | l = A_0
      | r = \int_0^1 e^x \rd x
}}
{{eqn | r = \bigintlimits {e^x} {x \mathop = 0} {x \mathop = 1}
      | c = Primitive of Exponential Function
}}
{{eqn | r = e - 1
}}
{{eqn | r = q_0 e - p_0
}}
{{eqn | l = B_0
      | r = \int_0^1 x e^x \rd x
}}
{{eqn | r = \bigintlimits {x e^x - e^x} {x \mathop = 0} {x \mathop = 1}
      | c = Primitive of x by Exponential of a x
}}
{{eqn | r = 1
}}
{{eqn | r = p_1 - q_1 e
}}
{{eqn | l = C_0
      | r = \int_0^1 \paren {x - 1} e^x \rd x
}}
{{eqn | r = \int_0^1 x e^x \rd x - \int_0^1 e^x \rd x
}}
{{eqn | r = B_0 - A_0
}}
{{eqn | r = 1 - \paren {e - 1}
}}
{{eqn | r = 2 - e
}}
{{eqn | r = p_2 - q_2 e
}}
{{end-eqn}}
The final step needed to validate the assertion, we must demonstrate that the following three recurrence relations hold:
{{begin-eqn}}
{{eqn | n = 1
      | l = A_n
      | r = -B_{n - 1} - C_{n - 1}
}}
{{eqn | n = 2
      | l = B_n
      | r = -2 n A_n + C_{n - 1}
}}
{{eqn | n = 3
      | l = C_n
      | r = B_n - A_n
}}
{{end-eqn}}
To prove the '''first''' relation, we note that the derivative of the integrand of $A_n$ is equal to the sum of the integrand of $A_n$ with the integrand of $B_{n - 1}$ and the integrand of $C_{n - 1}$.
By integrating both sides of the equation, we verify the first recurrence relation:
{{begin-eqn}}
{{eqn | l = \map {\frac \d {\d x} } {\frac {x^n \paren {x - 1}^n } {n!} e^x}
      | r = \frac {x^n \paren {x - 1 }^n} {n!} e^x + \frac {x^n \paren {x - 1}^{n - 1} } {\paren {n - 1}!} e^x + \frac {x^{n - 1} \paren {x - 1}^n} {\paren {n - 1}!} e^x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \int_0^1 \map {\frac \d {\d x} } {\frac {x^n \paren {x - 1}^n} {n!} e^x}
      | r = \int_0^1 \frac {x^n \paren {x - 1}^n} {n!} e^x \rd x + \int_0^1 \frac {x^n \paren {x - 1}^{n - 1} } {\paren {n - 1}!} e^x \rd x + \int_0^1 \frac {x^{n - 1} \paren {x - 1}^n} {\paren {n - 1}!} e^x \rd x
      | c = Integrating both sides of the equation over the interval from $0$ to $1$
}}
{{eqn | ll= \leadsto
      | l = \intlimits {\frac {x^n \paren {x - 1}^n} {n!} e^x} {x \mathop = 0} {x \mathop = 1}
      | r = A_n + B_{n - 1} + C_{n - 1}
      | c = Fundamental Theorem of Calculus
}}
{{eqn | ll= \leadsto
      | l = 0
      | r = A_n + B_{n - 1} + C_{n - 1}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = A_n
      | r = -B_{n - 1} - C_{n - 1}
      | c = rearranging
}}
{{end-eqn}}
To prove the '''second''' relation, we note that the derivative of the integrand of $C_n$ is equal to the sum of the integrand of $B_n$ with two times n times the integrand of $A_{n}$ minus the integrand of $C_{n - 1}$.
By integrating both sides of the equation, we verify the second recurrence relation:
{{begin-eqn}}
{{eqn | l = \map {\frac \d {\d x} } {\frac {x^n \paren {x - 1}^{n + 1} } {n!} e^x}
      | r = \frac {x^{n + 1} \paren {x - 1}^n} {n!} e^x + 2 n \frac {x^n \paren {x - 1}^n} {n!} e^x - \frac {x^{n - 1} \paren {x - 1}^n} {\paren {n - 1}!} e^x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \int_0^1 \map {\frac \d {\d x} } {\frac {x^n \paren {x - 1}^{n + 1} } {n!} e^x}
      | r = \int_0^1 \frac {x^{n + 1} \paren {x - 1}^n} {n!} e^x \rd x + 2 n \int_0^1 \frac {x^n \paren {x - 1}^n} {n!} e^x \rd x - \int_0^1 \frac {x^{n - 1} \paren {x - 1}^n} {\paren {n - 1}!} e^x \rd x
      | c = Integrating both sides of the equation over the interval from $0$ to $1$
}}
{{eqn | ll= \leadsto
      | l = \intlimits {\frac {x^n \paren {x - 1}^{n + 1} } {n!} e^x} {x \mathop = 0} {x \mathop = 1}
      | r = B_n + 2 n A_{n} - C_{n - 1}
      | c = Fundamental Theorem of Calculus
}}
{{eqn | ll= \leadsto
      | l = 0
      | r = B_n + 2 n A_n - C_{n - 1}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = B_n
      | r = -2 n A_n + C_{n - 1}
      | c = rearranging
}}
{{end-eqn}}
To prove the '''third''' relation, we have:
{{begin-eqn}}
{{eqn | l = C_n
      | r = \int_0^1 \frac {x^n \paren {x - 1}^{n + 1} } {n!} e^x \rd x
      | c = 
}}
{{eqn | r = \int_0^1 \frac {x^n \paren {x - 1 }^n} {n!} e^x \paren {x - 1} \rd x
      | c = factoring out $\paren {x - 1}$
}}
{{eqn | r = \int_0^1 \frac {x^n \paren {x - 1}^n} {n!} e^x \paren x \rd x - \int_0^1 \frac {x^n \paren {x - 1}^n} {n!} e^x \paren 1 \rd x
      | c = separate integrals
}}
{{eqn | r = \int_0^1 \frac {x^{n + 1} \paren {x - 1}^n} {n!} e^x \rd x - \int_0^1 \frac {x^n \paren {x - 1}^n} {n!} e^x \rd x
      | c =
}}
{{eqn | r = B_n - A_n
      | c = 
}}
{{end-eqn}}
From the '''first''' relation, combined with the initial condition at $n = 0$ being satisfied, we have:
{{begin-eqn}}
{{eqn | l = A_n
      | r = -B_{n - 1} - C_{n - 1}
}}
{{eqn | r = -\paren {p_{3 n - 2} - q_{3 n -2 } e} - \paren {p_{3 n - 1} - q_{3 n - 1} e}     
}}
{{eqn | r = q_{3 n} e - p_{3 n}
}}
{{end-eqn}}
From the '''second''' relation, combined with the initial condition at $n = 0$ being satisfied, we have:
{{begin-eqn}}
{{eqn | l = B_n
      | r = -2 n A_n + C_{n - 1}
}}
{{eqn | r = -2 n \paren {q_{3 n} e - p_{3 n} } + \paren {p_{3 n - 1} - q_{3 n - 1} e}     
}}
{{eqn | r = p_{3 n + 1} - q_{3 n + 1} e
}}
{{end-eqn}}
From the '''third''' relation, combined with the initial condition at $n = 0$ being satisfied, we have:
{{begin-eqn}}
{{eqn | l = C_n
      | r = B_n - A_n
}}
{{eqn | r = \paren {p_{3 n + 1} - q_{3 n + 1} e} - \paren {q_{3 n} e - p_{3 n} }     
}}
{{eqn | r = p_{3 n + 2} - q_{3 n + 2 } e
}}
{{end-eqn}}
{{qed}}