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

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

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 $\sqrt{2}$.

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

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

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

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:

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

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 $e$.