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Problem 65

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Convergents of $e$

The square root of $2$ can be written as an infinite continued fraction.

$\sqrt{2}=1+{\displaystyle \frac{1}{2+{\displaystyle \frac{1}{2+{\displaystyle \frac{1}{2+{\displaystyle \frac{1}{2+\dots }}}}}}}}$

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 $\sqrt{2}$.

$\begin{array}{rl}& 1+{\displaystyle \frac{1}{2}}={\displaystyle \frac{3}{2}}\\ & 1+{\displaystyle \frac{1}{2+{\displaystyle \frac{1}{2}}}}={\displaystyle \frac{7}{5}}\\ & 1+{\displaystyle \frac{1}{2+{\displaystyle \frac{1}{2+{\displaystyle \frac{1}{2}}}}}}={\displaystyle \frac{17}{12}}\\ & 1+{\displaystyle \frac{1}{2+{\displaystyle \frac{1}{2+{\displaystyle \frac{1}{2+{\displaystyle \frac{1}{2}}}}}}}}={\displaystyle \frac{41}{29}}\end{array}$

Hence the sequence of the first ten convergents for $\sqrt{2}$ are:

$1,{\displaystyle \frac{3}{2}},{\displaystyle \frac{7}{5}},{\displaystyle \frac{17}{12}},{\displaystyle \frac{41}{29}},{\displaystyle \frac{99}{70}},{\displaystyle \frac{239}{169}},{\displaystyle \frac{577}{408}},{\displaystyle \frac{1393}{985}},{\displaystyle \frac{3363}{2378}},\dots$

What is most surprising is that the important mathematical constant,

$e=[2;1,2,1,1,4,1,1,6,1,\dots ,1,2k,1,\dots ]$.

The first ten terms in the sequence of convergents for $e$ are:

$2,3,{\displaystyle \frac{8}{3}},{\displaystyle \frac{11}{4}},{\displaystyle \frac{19}{7}},{\displaystyle \frac{87}{32}},{\displaystyle \frac{106}{39}},{\displaystyle \frac{193}{71}},{\displaystyle \frac{1264}{465}},{\displaystyle \frac{1457}{536}},\dots$

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 $e$.

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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。

$\sqrt{2}=1+{\displaystyle \frac{1}{2+{\displaystyle \frac{1}{2+{\displaystyle \frac{1}{2+{\displaystyle \frac{1}{2+\dots }}}}}}}}$

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：

$\begin{array}{rl}& 1+{\displaystyle \frac{1}{2}}={\displaystyle \frac{3}{2}}\\ & 1+{\displaystyle \frac{1}{2+{\displaystyle \frac{1}{2}}}}={\displaystyle \frac{7}{5}}\\ & 1+{\displaystyle \frac{1}{2+{\displaystyle \frac{1}{2+{\displaystyle \frac{1}{2}}}}}}={\displaystyle \frac{17}{12}}\\ & 1+{\displaystyle \frac{1}{2+{\displaystyle \frac{1}{2+{\displaystyle \frac{1}{2+{\displaystyle \frac{1}{2}}}}}}}}={\displaystyle \frac{41}{29}}\end{array}$

因此$\sqrt{2}$的前十个逼近值为：

$1,{\displaystyle \frac{3}{2}},{\displaystyle \frac{7}{5}},{\displaystyle \frac{17}{12}},{\displaystyle \frac{41}{29}},{\displaystyle \frac{99}{70}},{\displaystyle \frac{239}{169}},{\displaystyle \frac{577}{408}},{\displaystyle \frac{1393}{985}},{\displaystyle \frac{3363}{2378}},\dots$

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

$e=[2;1,2,1,1,4,1,1,6,1,\dots ,1,2k,1,\dots ]$。

$e$的前十个逼近值为：

$2,3,{\displaystyle \frac{8}{3}},{\displaystyle \frac{11}{4}},{\displaystyle \frac{19}{7}},{\displaystyle \frac{87}{32}},{\displaystyle \frac{106}{39}},{\displaystyle \frac{193}{71}},{\displaystyle \frac{1264}{465}},{\displaystyle \frac{1457}{536}},\dots$

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。

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