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 $e$.