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Let f(k) = (k-1)/2^k

Let g(n) = S(f(k)) for k=0..n

For n > 0,

g(n) = f(n) + g(n-1)

For n = 0,

g(n) = f(n)

The above defines g(n) as a recursive function, which is trivial to implement in C.

double g(int n) {

assert(n >= 0);

if (n == 0)

return f(n);

else

return f(n) + g(n - 1);

}

double f(double k) {

return (k-1)/pow(2,k);

}