You'll find in the wikipedia Logarithm article a section called Feynman's algorithm.
The paragraph makes claims about uniqueness that, when interpreted mathematically, is incorrect; see this math.stackexchange post,
Feynman's Algorithm for computing a logarithm of a number in [1,2]
But there can be little doubt that the algorithm work:
We define for $n \ge 1$
$$\tag 1 H_n = \frac{1}{2^n}$$
Let $x$ be a real number between $1$ and $2$ and set $P_0 = 1$
Let $P_n$ be recursively defined. If $P_n * (1 + H_{n+1}) \gt x$ set
$\tag 2 P_{n+1} = P_n$
Else set
$\tag 3 P_{n+1} = P_n * (1 + H_{n+1})$
Claim; The monotone increasing sequence $P_n$ converges to $x$.
Provide a proof that this is true.
My Work
We know that the bounded sequence converges, and by construction it looks like it is doing everything it can to 'get to $x$', but this is not the usual binary subdivision of intervals that is easy to work with.
When an update occurs at $n+1$ we get $P_{n+1} = P_n + H_{n+1} * P_n$. So we 'add in' the new term $H_{n+1}$ and then $H_{n+1}$ applied to other terms 'shifts them down' into lower order terms. It might be a matter of solving equations along with each step of the algorithm, but it isn't easy to see how we are guaranteed to build $x$ that has the form,
$\tag 4 x = 1 + \sum_{k=1}^\infty b_k H_k \; \text{ with } b_k \in \{0,1\}$
Also, I searched for a proof of this but couldn't find one. So any comments with links would be much appreciated.
Best Answer
Let $y$ be the limit of the $P_n$ and suppose for a contradiction that $y<x$. There is a minimal $m$ such that $y(1+H_m)<x$.
Then $1+H_m$ must be one of the terms in the product defining $y$, for the sequence of $P_n$ is increasing and so $P_{m-1}(1+H_m) \leq y(1+H_m) < x$. Similarly every $1+H_{m+k}$ for $k\geq 0$ is a factor in $y$, and $$ y = (1+a_1 2^{-1})(1+a_2 2^{-2})\cdots (1+a_{m-1}2^{-(m-1)}) (1+2^{-m})(1+2^{-m-1})(1+2^{-m-2})\cdots$$ where each $a_i$ is 0 or 1. Now $a_{m-1}$ might not be zero, but there is a biggest $r$ such that $a_{r-1}=0$ (at least one must be zero since $\prod_{k=1}^\infty (1+2^{-k}) >2 \geq x$) and we have $$y = P_{r-2} (1+2^{-r})(1+2^{-r-1})\cdots$$ Note that $$ 1+2^{-r+1} < (1+2^{-r})(1+2^{-r-1})\cdots$$ since the rhs is greater than $1+ 2^{-r}+2^{-r-1}+2^{-r-2}+\cdots = 1+2^{-r+1}$. That means $P_{r-2} (1+H_{r-1}) < y < x$, contradicting $a_{r-1}=0$