# 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 = \[2; 1,2,1, 1,4,1, 1,6,1 , ... ,
1,2k,1, ...\]. 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.