Convergents of e

The square root of \( 2 \) can be written as an infinite continued fraction.

\( \sqrt{2} = 1 + \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{2 + ...}}}} \)

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 \( \sqrt{2} \).

\[ \begin{align} &1+ \dfrac{1}{2} = \dfrac{3}{2} \\ &1+ \dfrac{1}{2 + \dfrac{1}{2}} = \dfrac{7}{5}\\ &1+ \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{2}}} = \dfrac{17}{12}\\ &1+ \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{2}}}} = \dfrac{41}{29}\\ \end{align} \]

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

\( 1, \dfrac{3}{2}, \dfrac{7}{5}, \dfrac{17}{12}, \dfrac{41}{29}, \dfrac{99}{70}, \dfrac{239}{169}, \dfrac{577}{408}, \dfrac{1393}{985}, \dfrac{3363}{2378}, ... \)

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:

\( 2, 3, \dfrac{8}{3}, \dfrac{11}{4}, \dfrac{19}{7}, \dfrac{87}{32}, \dfrac{106}{39}, \dfrac{193}{71}, \dfrac{1264}{465}, \dfrac{1457}{536}, ... \)

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 \( e \).