Problem 65
Convergents of $e$
The square root of $2$ can be written as an infinite continued fraction.
$\sqrt{2} = 1 + \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{2 + \ldots}}}}$
The infinite continued fraction can be written, $\sqrt{2} = [1;(2)]$, $(2)$ indicates that $2$ repeats ad infinitum. In a similar way, $\sqrt{23} = [4;(1,3,1,8)]$.
It turns out that the sequence of partial values of continued fractions for square roots provide the best rational approximations. Let us consider the convergents for $\sqrt{2}$.
$\begin{aligned}
& 1 + \dfrac{1}{2} = \dfrac{3}{2} \\
& 1 + \dfrac{1}{2 + \dfrac{1}{2}} = \dfrac{7}{5}\\
& 1 + \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{2}}} = \dfrac{17}{12}\\
& 1 + \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{2}}}} = \dfrac{41}{29}
\end{aligned}$
Hence the sequence of the first ten convergents for $\sqrt{2}$ are:
$1, \dfrac{3}{2}, \dfrac{7}{5}, \dfrac{17}{12}, \dfrac{41}{29}, \dfrac{99}{70}, \dfrac{239}{169}, \dfrac{577}{408}, \dfrac{1393}{985}, \dfrac{3363}{2378}, \ldots$
What is most surprising is that the important mathematical constant,
$e = [2; 1, 2, 1, 1, 4, 1, 1, 6, 1, … , 1, 2k, 1, …]$.
The first ten terms in the sequence of convergents for $e$ are:
$2, 3, \dfrac{8}{3}, \dfrac{11}{4}, \dfrac{19}{7}, \dfrac{87}{32}, \dfrac{106}{39}, \dfrac{193}{71}, \dfrac{1264}{465}, \dfrac{1457}{536}, \ldots$
The sum of digits in the numerator of the $10$th convergent is $1+4+5+7=17$.
Find the sum of digits in the numerator of the $100$th convergent of the continued fraction for $e$.
e的有理逼近
$2$的算术平方根可以写成无限连分数的形式。
$\sqrt{2} = 1 + \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{2 + \ldots}}}}$
这个无限连分数可以简记为$\sqrt{2} = [1;(2)]$，其中$(2)$表示$2$无限重复。同样的，我们可以记$\sqrt{23} = [4;(1,3,1,8)]$。
可以证明，截取算术平方根连分数表示的一部分所组成的序列，给出了一系列最佳有理逼近值。让我们来考虑$\sqrt{2}$的逼近值：
$\begin{aligned}
& 1 + \dfrac{1}{2} = \dfrac{3}{2} \\
& 1 + \dfrac{1}{2 + \dfrac{1}{2}} = \dfrac{7}{5}\\
& 1 + \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{2}}} = \dfrac{17}{12}\\
& 1 + \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{2}}}} = \dfrac{41}{29}
\end{aligned}$
因此$\sqrt{2}$的前十个逼近值为：
$1, \dfrac{3}{2}, \dfrac{7}{5}, \dfrac{17}{12}, \dfrac{41}{29}, \dfrac{99}{70}, \dfrac{239}{169}, \dfrac{577}{408}, \dfrac{1393}{985}, \dfrac{3363}{2378}, \ldots$
最令人惊讶的莫过于重要的数学常数$e$有如下连分数表示
$e = [2; 1, 2, 1, 1, 4, 1, 1, 6, 1, … , 1, 2k, 1, …]$。
$e$的前十个逼近值为：
$2, 3, \dfrac{8}{3}, \dfrac{11}{4}, \dfrac{19}{7}, \dfrac{87}{32}, \dfrac{106}{39}, \dfrac{193}{71}, \dfrac{1264}{465}, \dfrac{1457}{536}, \ldots$
第$10$个逼近值的分子各位数字之和为$1+4+5+7=17$。
求$e$的第$100$个逼近值的分子各位数字之和。