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: :$\begin{array}{r|cccccccccc} \displaystyle i &  0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9  \\ \hline p_i & 1 & 1 & 2 & 3 & 8 & 11 & 19 & 87 & 106 & 193 \\ q_i & 1 & 0 & 1 & 1 & 3 & 4 & 7 & 32 & 39 & 71 \\ \hline \end{array}$ Furthermore, $p_i$ and $q_i$ satisfy the following $6$ recurrence relations: {{begin-eqn}} {{eqn | n = 1       | l = p_{3 n + 0}       | r = \paren {a_{3 n + 0} } p_{3 n - 1} + p_{3 n - 2}       | rr= = p_{3 n - 1} + p_{3 n - 2} }} {{eqn | n = 2       | l = p_{3 n + 1}       | r = \paren {a_{3 n + 1} } p_{3 n + 0} + p_{3 n - 1}       | rr= = 2 n p_{3 n + 0} + p_{3 n - 1} }} {{eqn | n = 3       | l = p_{3 n + 2}       | r = \paren {a_{3 n + 2} } p_{3 n + 1} + p_{3 n + 0}       | rr= = p_{3 n + 1} + p_{3 n + 0} }} {{eqn | n = 4       | l = q_{3 n + 0}       | r = \paren {a_{3 n + 0} } q_{3 n - 1} + q_{3 n - 2}       | rr= = q_{3 n - 1} + q_{3 n - 2}, }} {{eqn | n = 5       | l = q_{3 n + 1}       | r = \paren {a_{3 n + 1} } q_{3 n + 0} + q_{3 n - 1}       | rr= = 2 n q_{3 n + 0} + q_{3 n - 1}, }} {{eqn | n = 6       | l = q_{3 n + 2}       | r = \paren {a_{3 n + 2} } q_{3 n + 1} + q_{3 n + 0}       | rr= = q_{3 n + 1} + q_{3 n + 0}, }} {{end-eqn}} Our ultimate aim is to prove that: :$\displaystyle \lim_{n \mathop \to \infty} \frac {p_n} {q_n} = e$ 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       | r = \lim_{n \mathop \to \infty} \paren {q_{3 n} e - p_{3 n} }       | rr = = 0 }} {{eqn | l = \lim_{n \mathop \to \infty} B_n       | r = \lim_{n \mathop \to \infty} \paren {p_{3 n + 1} - q_{3 n + 1} e}       | rr = = 0 }} {{eqn | l = \lim_{n \mathop \to \infty} C_n       | r = \lim_{n \mathop \to \infty} \paren {p_{3 n + 2} - q_{3 n + 2} e}       | rr = = 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}}