 ## Convergents of e ### Problem 65

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 $\sqrt{2}$.

$1 + \dfrac{1}{2} = \dfrac{3}{2}\\ 1 + \dfrac{1}{2 + \dfrac{1}{2}} = \dfrac{7}{5}\\ 1 + \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{2}}} = \dfrac{17}{12}\\ 1 + \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{2}}}} = \dfrac{41}{29}$

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 10th convergent is $1 + 4 + 5 + 7 = 17$.

Find the sum of digits in the numerator of the 100th convergent of the continued fraction for $e$.