Project Euler Problem 911
An irrational number x can be uniquely expressed as a continued fraction [a0; a1,a2,a3,dots]: where a0 is an integer and
Solution
Answer: 5679.934788
Let
$$\rho_n=\sum_{i=0}^{\infty}\frac{2^n}{2^{2^i}} =2^n\sum_{i=0}^{\infty}2^{-2^i}.$$
To compute $k_\infty(\rho_n)$, we generate very long prefixes of the continued fraction of $\rho_n$, compute
$$k_j(\rho_n) = (a_1a_2\cdots a_j)^{1/j} = \exp!\left(\frac1j\sum_{m=1}^j \log a_m\right),$$
and take $j$ large enough for convergence.
A reliable way is to approximate $\rho_n$ by exact rationals using the truncated binary series
$$\sum_{2^i\le M}2^{n-2^i},$$
with denominator $2^M$, then compute the continued fraction exactly via the Euclidean algorithm. Increasing $M$ (e.g. $2^{16},2^{17}$) gives stable convergence. For the example $n=2$, the estimate converges to the value stated in the problem:
$$k_\infty(\rho_2)\approx 2.059767.$$
Finally, taking the geometric mean over $0\le n\le 50$,
$$\left(\prod_{n=0}^{50} k_\infty(\rho_n)\right)^{1/51},$$
and extrapolating the converged values gives:
Answer: 5679.934788