## Convergents of e

### Problem 65

Published on Friday, 12th March 2004, 06:00 pm; Solved by 13567

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

2 = 1 + | 1
| 2 + | 1
|  | 2 + | 1
|  |  | 2 + | 1
|  |  |  | 2 + ...

The infinite continued fraction can be written, 2 = [1;(2)], (2) indicates that 2 repeats ad infinitum. In a similar way, 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 2.

1 + | 1 | = 3/2
| 2 |

1 + | 1 | = 7/5
| 2 + | 1
|  | 2 |

1 + | 1 | = 17/12
| 2 + | 1 |
|  | 2 + | 1 |
|  |  | 2 |

1 + | 1 | = 41/29
| 2 + | 1
|  | 2 + | 1 |
|  |  | 2 + | 1 |
|  |  |  | 2 |

Hence the sequence of the first ten convergents for 2 are:

1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169, 577/408, 1393/985, 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, 8/3, 11/4, 19/7, 87/32, 106/39, 193/71, 1264/465, 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.