Continued Fraction Expansion of Euler's Number

Theorem

The constant Euler's number $e$ has the continued fraction expansion:

 $\displaystyle e$ $=$ $\displaystyle \sqbrk {2; 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, \ldots }$ $\displaystyle$ $=$ $\displaystyle \sqbrk {1, 0, 1, 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, \ldots }$

Convergents

The convergents of the continued fraction expansion to Euler's number $e$ are:

$2, 3, \dfrac 8 3, \dfrac {11} 4, \dfrac {19} 7, \dfrac {87} {32}, \dfrac {106} {39}, \dfrac {193} {71}, \dfrac {1264} {465}, \dfrac {1457} {536}, \dfrac {2721} {1001}, \ldots$

These best rational approximations are accurate to $0, 0, 1, 1, 2, 3, 3, 4, 5, 5, \ldots$ decimals.

The fraction $\dfrac {878} {323}$ is exceptionally easy to remember:

$\dfrac {878} {323} = 2 \cdotp 71826 \, 625 \ldots$

although this does not occur in the above continued fraction expansion.

Proof 1

From the recursive definition of continued fractions, we have:

 $\displaystyle p_i$ $=$ $\displaystyle a_i p_{i - 1} + p_{i - 2}$ $\displaystyle q_i$ $=$ $\displaystyle a_i q_{i - 1} + q_{i - 2}$

Let:

 $\displaystyle \sqbrk {a_0, a_1, a_2, a_3, a_4, a_5, a_6, a_7, a_8, a_9, \ldots }$ $=$ $\displaystyle \sqbrk {1, 0, 1, 1, 2, 1, 1, 4, 1, 1, \ldots }$

In other words, $a_{3n+1} = 2n$ and $a_{3n+0} = a_{3n+2} = 1$

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:

 $\displaystyle (1) \ \$ $\displaystyle p_{3n+0}$ $=$ $\displaystyle \paren{ a_{3n+0} } p_{3n - 1} + p_{3n - 2},$ $\displaystyle = p_{3n - 1} + p_{3n - 2},$ $\displaystyle (2) \ \$ $\displaystyle p_{3n+1}$ $=$ $\displaystyle \paren{ a_{3n+1} } p_{3n + 0} + p_{3n - 1},$ $\displaystyle = 2n p_{3n + 0} + p_{3n - 1},$ $\displaystyle (3) \ \$ $\displaystyle p_{3n+2}$ $=$ $\displaystyle \paren{ a_{3n+2} } p_{3n + 1} + p_{3n + 0},$ $\displaystyle = p_{3n + 1} + p_{3n + 0},$ $\displaystyle (4) \ \$ $\displaystyle q_{3n+0}$ $=$ $\displaystyle \paren{ a_{3n+0} } q_{3n - 1} + q_{3n - 2},$ $\displaystyle = q_{3n - 1} + q_{3n - 2},$ $\displaystyle (5) \ \$ $\displaystyle q_{3n+1}$ $=$ $\displaystyle \paren{ a_{3n+1} } q_{3n + 0} + q_{3n - 1},$ $\displaystyle = 2n q_{3n + 0} + q_{3n - 1},$ $\displaystyle (6) \ \$ $\displaystyle q_{3n+2}$ $=$ $\displaystyle \paren{ a_{3n+2} } q_{3n + 1} + q_{3n + 0},$ $\displaystyle = q_{3n + 1} + q_{3n + 0},$

Our ultimate aim is to prove that:

 $\displaystyle \lim_{n \mathop \to \infty} \frac {p_n} {q_n}$ $=$ $\displaystyle e$

In the pursuit of that aim, let us define the integrals:

 $\displaystyle A_n$ $=$ $\displaystyle \int_0^1 \frac {x^n \paren {x - 1 }^n } {n!} e^{x} \rd x$ $\displaystyle B_n$ $=$ $\displaystyle \int_0^1 \frac {x^{n+1} \paren {x - 1 }^n } {n!} e^{x} \rd x$ $\displaystyle C_n$ $=$ $\displaystyle \int_0^1 \frac {x^n \paren {x - 1 }^{n+1} } {n!} e^{x} \rd x$

Lemma

For $n \in \Z , n \ge 0$:
 $\displaystyle A_n$ $=$ $\displaystyle q_{3 n} e - p_{3 n}$ $\displaystyle B_n$ $=$ $\displaystyle p_{3 n + 1} - q_{3 n + 1} e$ $\displaystyle C_n$ $=$ $\displaystyle p_{3 n + 2} - q_{3 n + 2} e$

We assert that $A_n$, $B_n$ and $C_n$ all converge to $0$ as $n \mathop \to \infty$:

 $\displaystyle \lim_{n \mathop \to \infty} A_n$ $=$ $\displaystyle \frac {\frac {x^{n+1 } \paren {x - 1 }^{n+1 } } {\paren {n+1 }!} e^{x} } {\frac {x^n \paren {x - 1 }^n } {n!} e^{x} }$ Radius of Convergence from Limit of Sequence/Real Case $\displaystyle \lim_{n \mathop \to \infty} A_n$ $=$ $\displaystyle \frac {x \paren {x-1 } } {\paren {n+1 } }$ $\displaystyle \lim_{n \mathop \to \infty} A_n$ $=$ $\displaystyle 0$ $\displaystyle \lim_{n \mathop \to \infty} B_n$ $=$ $\displaystyle \frac {\frac {x^{n+2 } \paren {x - 1 }^{n+1 } } {\paren {n+1 }!} e^{x} } {\frac {x^{n+1 } \paren {x - 1 }^n } {n!} e^{x} }$ Radius of Convergence from Limit of Sequence/Real Case $\displaystyle \lim_{n \mathop \to \infty} B_n$ $=$ $\displaystyle \frac {x \paren {x-1 } } {\paren {n+1 } }$ $\displaystyle \lim_{n \mathop \to \infty} B_n$ $=$ $\displaystyle 0$ $\displaystyle \lim_{n \mathop \to \infty} C_n$ $=$ $\displaystyle \lim_{n \mathop \to \infty} B_n - \lim_{n \mathop \to \infty} A_n$ $\displaystyle$ $=$ $\displaystyle 0$

We now have:

 $\displaystyle \lim_{n \mathop \to \infty} A_n$ $=$ $\displaystyle \lim_{n \mathop \to \infty} \paren {q_{3n}e - p_{3n} }$ $\displaystyle = 0$ $\displaystyle \lim_{n \mathop \to \infty} B_n$ $=$ $\displaystyle \lim_{n \mathop \to \infty} \paren {p_{3n+1} - q_{3n+1}e }$ $\displaystyle = 0$ $\displaystyle \lim_{n \mathop \to \infty} C_n$ $=$ $\displaystyle \lim_{n \mathop \to \infty} \paren {p_{3n+2} - q_{3n+2}e }$ $\displaystyle = 0$

From which, we conclude:

 $\displaystyle \lim_{n \mathop \to \infty} \paren {p_{n} - q_{n}e }$ $=$ $\displaystyle 0$ $\displaystyle \lim_{n \mathop \to \infty} p_{n}$ $=$ $\displaystyle q_{n}e$ $\displaystyle \lim_{n \mathop \to \infty} \frac {p_{n} } {q_{n} }$ $=$ $\displaystyle e$

$\blacksquare$

Historical Note

Leonhard Euler analyzed the Ricatti equation to prove that the number e has the continued fraction shown here.

Later, while proving the transcendence of e, Charles Hermite also proved this continued fraction.