Fibonacci, inverted

First, note that we don't necessarily need a program to find the answer, but it will be very tedious (and possibly difficult) to do by hand. The solution is to first define a fibonacci method to find the $n$ th Fibonacci number. The method I use here is the (horribly inefficient) recursive method.

public static int fibonacci(int n)
{
    if (n == 0)
        return 0;
    else if (n == 1)
        return 1;
    else
        return (fibonacci(n-1) + fibonacci(n-2));
}

This gives us the $n$ th Fibonacci number, $F_n.$ Next we need our main method to compute the sum, so here's a hypothetical method:

public static void main(String[] args)
{
    double sum;
    for (int n = 1; n < 20; n++)
    {
        sum += 1.0 / (double)(fibonacci(n));
    }
    System.out.println(sum);
}

The main problem is that there's no guarantee this will accurately give the first $5$ digits of $S,$ since it is a finite sum whereas $S$ is a sum of infinitely many terms. To make sure we have the correct approximation, we need to change the main method.

public static void main(String[] args)
{
    double sum;
    for (int i = 2; i < 40; i++)
    {
        for (int n = 1; n < i; n++)
        {
            sum += 1.0 / (double)(fibonacci(n));
        }
    }
    System.out.println(sum);
}

This prints out $38$ approximations of $S.$ We can see that the values of the first $5$ digits do not change at all after the $23^{rd}$ approximation. Therefore, the value of $\lfloor 10000 \times S \rfloor$ is unchanged after the $23^{rd}$ approximation. So our first 5 digits are $3.3598.$ Multiplying by $10000$ gives $\lfloor 10000 \times S\rfloor = 10000 \times 3.3598 = \boxed{33598}$

The number $3.35988566624 ...$ is known as the Reciprocal Fibonacci Constant. One of my favorite properties of the constant is that it contains the early sequence of digits, $666$ , which is related to Fibonacci numbers in that $-2\sin (666^\circ) = \frac{1+\sqrt{5}}{2} = \phi,$ the golden ratio.

Python:

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from fractions import Fraction as frac
from math import floor
memo = {1:1,2:1}
def fib(n):
    if n in memo:
        return memo[n]
    memo[n] = fib(n-1)+fib(n-2)
    return memo[n]
memo2 = {1:frac(1,fib(1))}
def s(n):
    if n in memo2:
        return memo2[n]
    memo2[n] = s(n-1)+frac(1,fib(n))
    return memo2[n]
infinity = 200
for i in xrange(1, infinity):
    fib(i), s(i)
print float(s(infinity))
print floor(10000*s(infinity))