Convergents of e (65)

In Problem 65, we're told that Euler's number $\mathrm{e}$ can be represented as a continued fraction with coefficients

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

We're asked to find the sum of the digits in the numerator of the 100th convergent of this number.

Using the library functions for continued fractions, we only need to implement the sequence of coefficients to give us the 100 coefficients needed. Then we simplify the fraction, take the numerator and then take the digit sum.