sum of floor function simplification Let $n \in \mathbb{N}$.
I'm trying to show $\sum_{i=1}^{\infty} \left(\left \lfloor{\frac{n+2^k-1}{2^i}}\right \rfloor - \left \lfloor{\frac{n-1}{2^i}}\right \rfloor \right) = 2^k-1$.
I know that when $2^i > n-1$, the right floor function will become zero.
I'm not sure how I cancel out the $n$'s.
I also tried solving the sum of each floor function separately, but didn't get anywhere.
Any suggestions will be appreciated!
Thanks.
 A: What you're trying to show is not always true. For example, let $n = 4$, so $n - 1 = 3$, and $k = 1$, so $2^k = 2$. This means $n + 2^k - 1 = 5$. However, using that each summation term is non-negative, the first few terms then become
$$\begin{equation}\begin{aligned}
& \sum_{i=1}^{\infty} \left(\left\lfloor{\frac{n+2^k-1}{2^i}}\right\rfloor - \left\lfloor{\frac{n-1}{2^i}}\right\rfloor\right) \\
& = \left(\left\lfloor\frac{5}{2}\right\rfloor - \left\lfloor\frac{3}{2}\right\rfloor\right) + \left(\left\lfloor\frac{5}{4}\right\rfloor - \left\lfloor\frac{3}{4}\right\rfloor\right) + \ldots \\
& = (2 - 1) + (1 - 0) + \ldots \\
& = 1 + 1 + \ldots \\
& = 2
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Note I've used that all of the remaining terms, i.e., the "$\ldots$" part, are $0$. This is larger than your right side of $2^{k} - 1 = 1$.
Note your statement would be correct with an appropriate restriction between $k$ and $n$. In particular, something like $2^k \gt n - 1$ works. This is because, for $i \le k$, each of the summation terms becomes
$$\begin{equation}\begin{aligned}
\left\lfloor{\frac{n+2^k-1}{2^i}}\right\rfloor - \left\lfloor{\frac{n-1}{2^i}}\right\rfloor & = \left(\left\lfloor{\frac{2^k}{2^i} + \frac{n-1}{2^i}}\right\rfloor\right) - \left\lfloor{\frac{n-1}{2^i}}\right\rfloor \\
& = \left(2^{k-i} + \left\lfloor{\frac{n-1}{2^i}}\right\rfloor\right) - \left\lfloor{\frac{n-1}{2^i}}\right\rfloor \\
& = 2^{k-i}
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
For $i \gt k$, then $2^i \ge 2(2^k) \gt n + 2^k - 1$ and $2^i \gt n - 1$, so both floor terms become $0$. Thus, the total sum would be that of a geometric series of
$$\sum_{i=1}^{k}(2^{k-i}) = 2^k - 1 \tag{3}\label{eq3A}$$
which then matches your right side.
A: The point is that the LHS is a geometric series, because extraneous parts of the sums cancel. However, the result doesn't seem quite right, because you get some extra terms.
$$
\begin{align}
&\sum_{i=1}^{\infty} 
\left( \left\lfloor{\frac{n+2^k-1}{2^i}}\right \rfloor - \left \lfloor{\frac{n-1}{2^i}} \right\rfloor \right)\\
&= \sum_{i=1}^{k} 
\left( \left\lfloor{\frac{n+2^k-1}{2^i}}\right \rfloor - \left \lfloor{\frac{n-1}{2^i}} \right\rfloor \right)
+\sum_{i=k+1}^{\infty} 
\left( \left\lfloor{\frac{n+2^k-1}{2^i}}\right \rfloor - \left \lfloor{\frac{n-1}{2^i}} \right\rfloor \right)\\
&= \sum_{i=1}^{k} 
\left( \frac{2^k}{2^i} +\left\lfloor{\frac{n-1}{2^i}}\right \rfloor - \left \lfloor{\frac{n-1}{2^i}} \right\rfloor \right)
+\sum_{i=k+1}^{\infty} 
\mathbb{1}(n-1<2^i, n+2^k-1\ge 2^i)\\
&= \sum_{i=1}^{k}2^{k-i}+ \text{something}\\
&= \sum_{j=0}^{k-1}2^j + \text{something}\\
&= 2^k-1 + \text{something}
\end{align}
$$
