--- id: 5900f3ad1000cf542c50fec0 challengeType: 5 title: 'Problem 65: Convergents of e' forumTopicId: 302177 --- ## Description
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, $\sqrt{2} = [1; (2)]$ 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.
## Instructions
## Tests
```yml tests: - text: convergentsOfE() should return a number. testString: assert(typeof convergentsOfE() === 'number'); - text: convergentsOfE() should return 272. testString: assert.strictEqual(convergentsOfE(), 272); ```
## Challenge Seed
```js function convergentsOfE() { return true; } convergentsOfE(); ```
## Solution
```js // solution required ```