## Convergents of e

### Problem 65

Published on Friday, 12th March 2004, 06:00 pm; Solved by 20156; Difficulty rating: 15%The square root of 2 can be written as an infinite continued fraction.

√2 = 1 + | 1 |
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2 + | 1 |
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2 + | 1 |
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2 + | 1 |
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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 |
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2 |

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

2 + | 1 |
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2 + | 1 |
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2 |

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

2 + | 1 |
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2 + | 1 |
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2 + | 1 |
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2 |

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

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:

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*.