Problem 65

Convergents of e

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 , ... , 1,2*k*,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 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*.

**
These problems are part of
Project Euler
and are licensed under
CC BY-NC-SA 2.0 UK
**