Project Euler Problem 65 Statement
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.
Solution
The numerator, n, for the continued fraction follows a predictable pattern (2, 3, 8, 11, 19, 87, …) and we are exploiting that to solve this problem:
The multiplier, m, follows the infinite continued fraction [2; 1,2,1, 1,4,1, 1,6,1, … , 1,2k,1, …].
So, n_{6} = 4*19 + 11 = 87 and n_{10} = 1*1264 + 193 = 1457
Instead of the numerator, which grows to hundreds of digits rapidly, we are asked for the numerator’s digit sum because it’s a much smaller number to enter as an answer for Project Euler. It has absolutely nothing to do with solving the problem. For example, the numerator for the 30,000^{th} convergent is thousands of digits long, yet the digit sum is only 6 digits.
HackerRank version
HackerRank Project Euler 65 raises the limit to 30,000 from 100. This solution works for both.
Python Source Code
nx, ny = 1, 2
L = int(input("Numerator of Nth convergent for e, N="))
for i in range(2, L+1):
m = 2*i//3 if i%3==0 else 1
nx, ny = ny, m*ny + nx
print (sum(map(int, str(ny))),ny)
Last Word
 More reading: e Continued Fraction from Mathworld
 Continued fraction for e from Todd and Vishal's blog

Run of the first 9 iterations at end of loop  i  nx  ny  m   2  2  3  1   3  3  8  2   4  8  11  1   5  11  19  1   6  19  87  4   7  87  106  1   8  106  93  1   9  193  1264  6   10  1264  1457  1 