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Convergents of e

Problem 65

Published on Friday, 12th March 2004, 06:00 pm; Solved by 15284

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.