The question above is Question A4 on the 2015 Putnam. I have written the solution I had come up with down below. I ended up coming up with the right answer, but my proof is long winded and hard to understand at some parts. I wanted to find out what score this would get on the actual Putnam, but I have not been able to find any resources that give a specific rubric for Putnam questions or give appropriate insight into how the Putnam is graded. Does anyone have such a resource and could direct me to it, or would be willing to wade through the absolute slugfest that is my proof to determine what score I would get? Many thanks in advance.
(Also sorry for the terrible formatting, I'm a newbie at latex ='))
$\textit{Solution:}$ L = $\frac{4}{7}$. First, consider comparing
$\Sigma_{n\in A}\frac{1}{2^{n}}$ vs $\Sigma_{n\in B}\frac{1}{2^{n}}$
where $A, B \subset \mathbb{Z}^{+}$. It suffices to compare which elements are in each staring from 1, and finding the first element that is only included in one of the sets. Assume WLOG that the first element that is only included in one of the sets is $m$, and WLOG assume B is the set that includes it. Since for all $n < m$, $A$ and $B$ both include or exclude the element, to compare the two sums it is sufficient to compare
$\Sigma_{n\in A, n \geq m}\frac{1}{2^{n}}$ vs $\Sigma_{n\in B, n \geq m}\frac{1}{2^{n}}$
which can be rewritten as
$\Sigma_{n\in A, n \geq m + 1}\frac{1}{2^{n}}$ vs $\frac{1}{2^{m}} + \Sigma_{n\in B, n \geq m + 1}\frac{1}{2^{n}}$
Even if B includes no elements greater than $m + 1$ and A includes all of them, then we can compare
$\Sigma_{n = m+1}^{\infty}\frac{1}{2^{n}}$ vs $\frac{1}{2^{m}}$
$\frac{1}{2^{m}}*\Sigma_{n = 0}^{\infty}\frac{1}{2^{n}}$ vs $\frac{1}{2^{m}}$
$\frac{1}{2^{m}}*\frac{1}{1-\frac{1}{2}}$ vs $\frac{1}{2^{m}}$
$\frac{1}{2^{m}}$ vs $\frac{1}{2^{m}}$
meaning even in the worst case the sum with B is still at least as big as the sum with A. This means that
$\Sigma_{n\in A}\frac{1}{2^{n}} \leq \Sigma_{n\in B}\frac{1}{2^{n}}$.
Now, we show that $0 \leq x < 1 \implies \Sigma_{n\in S_{x}}\frac{1}{2^{n}} \geq \Sigma_{n\in C}\frac{1}{2^{n}} = \frac{4}{7}$, where $C = \{1, 4, 7, ...\}$ to show that 4/7 is a lower bound.
Note that since for all $x$ in the given range, $\lfloor 1*x \rfloor = \lfloor x \rfloor = 0$, all sums must include the term corresponding to n = 1. So in order for $S_{x}$ to produce a lower sum than C, it can't include 2 or 3. Values of $x$ in $\frac{i}{n} \leq x < \frac{i + 1}{n}$ for odd $i$ don't include the term in the sum corresponding to $n$, since $\lfloor n*x \rfloor = i$, so the only values in $0 \leq x < 1$ that don't include 2 are $\frac{1}{2} \leq x < 1$, and the only values in that range that don't include 3 are $\frac{1}{2} \leq x < \frac{2}{3}$. If $S_{x}$ produces a lower sum then $\exists n. n \in S_{x}$ and $n + j \notin S_{x}$ if $j = 1, 2, or 3$, or in other words there must be a sequence of values for which the parity of $\lfloor n*x\rfloor$ doesn't change for 3 consecutive values of n. Because $0 \leq x < 1$, the quantity $\lfloor n*x\rfloor$ can only change by at most 1 when n increases by 1, so in order for the parity of $\lfloor n*x\rfloor$ to not change for 3 consecutive terms, $2x < 1$, meaning $x < \frac{1}{2}$. So in order for the sum produced by $S_{x}$ to be lower than that produced by $C$, $x < \frac{1}{2}$ and $\frac{1}{2} \leq x < \frac{2}{3}$, which is not possible, meaning the sum produced by $C$ is a lower bound.
The value of the sum produced by $C$ can be computed as follows:
$\Sigma_{n\in C}\frac{1}{2^{n}} = \Sigma_{n=0}^{\infty}\frac{1}{2^{3n+1}} = \frac{1}{2}*\Sigma_{n=0}^{\infty}\frac{1}{8} = \frac{1}{2}*\frac{1}{1-\frac{1}{8}} = \frac{4}{7}$
meaning $\frac{4}{7}$ is a lower bound.
Now let $\epsilon > 0$. To show that 4/7 is the greatest real number that is a lower bound, we will show that
$\exists x. \Sigma_{n\in S_{x}}\frac{1}{2^{n}} < 4/7 + \epsilon$.
We first show if $\frac{2m+1}{3m+2} \leq x < \frac{2}{3}$ for positive integer $m$, then $\forall n < 3m + 2. \lfloor n*x\rfloor$ is odd iff $n $mod$ 3 \not= 1$.
This is because $\forall n < 3m + 2. \frac{2}{3}*n - \frac{2m+1}{3m+2}*n < 1/3$, which is true because if $n = 3m+2$, then $\frac{2}{3}*n = 2m + \frac{4}{3}$ and $\frac{2m+1}{3m+2}*n = 2m + 1$, so $\frac{2}{3}*n - \frac{2m+1}{3m+2}*n = 1/3$, and $(\frac{2}{3} - \frac{2m+1}{3m+2})*n$ is a linear function of $n$ with a positive coefficient, so it grows with bigger $n$. Since $\frac{2m+1}{3m+2} \leq x < \frac{2}{3} \implies \frac{2m+1}{3m+2}*n \leq n*x < \frac{2}{3}*n \implies \forall n < 3m+2. \frac{2}{3}n - \frac{1}{3} < n*x < \frac{2}{3}*n$. Therefore, if $n = 3i + j$ for some positive integer $i$ and $j \in {0, 1, 2}$, then $2i + \frac{2j - 1}{3} < n*x < 2i + \frac{2j}{3}$, and $\lfloor n*x\rfloor = 2*i + j -1$, which proves the above statement.
Therefore, if $\frac{2m+1}{3m+2} \leq x < \frac{2}{3}$
$\Sigma_{n\in S_{x}}\frac{1}{2^{n}} = \Sigma_{i = 1}^{m}\frac{1}{2^{3*i+1}} + \Sigma_{n\in S_{x}, n\geq 3m+2}\frac{1}{2^{n}} < \Sigma_{i = 1}^{\infty}\frac{1}{2^{3*i+1}} + \Sigma_{i = 3m+2}^{\infty}\frac{1}{2^{i}} = \frac{4}{7} + \frac{1}{2^{3m+1}}$
Choosing $m > \frac{-log_{2}(\epsilon)-1}{3}$, we get
$\Sigma_{n\in S_{x}}\frac{1}{2^{n}} < \frac{4}{7} + \frac{1}{2^{3m+1}} < \frac{4}{7} + \frac{1}{2^{-log_{2}(\epsilon)}} = \frac{4}{7} + \epsilon$, which finishes the proof.