What is the limit of the sum $\sum_{i=1}^{n}\frac{1}{b_i}$ for the decreasing sequence of positive integers $b_n=\frac{x+2^n-1}{2^n}$?

Question

Let $x$ be a positive integer, let $b_k=\dfrac{x+2^k-1}{2^k}$, and let $n\in\mathbb{N}$ be as large as possible such that $b_n$ is a positive integer. Consider the sequence $(b_k)_{k=1}^n$. For example, taking $x=17$, we retrieve the sequence $(9,5,3,2)$. Now consider the sum $S(n)=\sum_{k=1}^{n}\frac{1}{b_k}$. How does this behave as $n\rightarrow\infty$?

Answer 1

If we factor $x-1=2^k y$ where $y$ is odd, then $b_n = 2^{k-n} y + 1$ for all $n$, and thus $b_k$ is the last integral element of the sequence. The sum in question is then $$ \sum_{i=1}^k \frac1{b_i} = \sum_{i=1}^k \frac1{2^{k-i} y + 1} < \sum_{i=1}^k \frac1{2^{k-i} y} = \frac1y\bigg( 2-\frac1{2^{k-1}} \bigg) < 2. $$ The actual best upper bound, when taking $x=2^k+1$ for larger and larger $k$, is approximately $1.2644997803484442092$.