The Jeep Problem and Nash's Friends The classical jeep problem is the following. A jeep can carry a maximum load of fuel of 1 gallon, and it travels $l$ miles with $l$ gallons of fuel. The jeep moves along a straight line, and is required to cross a desert $x$ miles wide in the most economic way, that is minimizing the required fuel. Let us say that $x > 0$ is the abscissa of the starting point, and $0$ that of the ending point. At $x$ there is a fuel station, where the jeep can load the fuel, while at any point $y < x$, there is a dumping station where the jeep can dump part of its fuel, in order to load it in a future trip.
The solution in this case was found by Fine, The Jeep Problem, Americ. Math. Monthly, Vol. 54 (1947), 24-31. The minimum required quantity of fuel $f(x)$ is the piecewise linear function having slope $2n+1$ on the interval $\left[ D_n, D_{n+1} \right]$, where $D_0=0$ and 
\begin{equation}
D_n = \sum_{k=1}^{n} \frac{1}{2k-1}
\end{equation}
for $n > 1$. From this result Fine easily derives the asymptotic formula for $f(x)$:
\begin{equation}
f(x) = \frac{1}{4} \exp(2x - \gamma) + \mathcal{O}(e^{-2x}),
\end{equation}
where $\gamma$ is Euler's constant.
Now let $N$ be a positive integer and let $g(x,N)$ be the minimum required fuel to cross the desert when the jeep can dump (and subsequently load) the fuel only at the points 
\begin{equation}
y_1 = \frac{x}{N}, y_2 = 2 \frac{x}{N}, \dots, y_{N-1}= \frac{N-1}{N} x.
\end{equation}
Suppose to choose for every $x \geq 0$ the positive integer $N(x)$ such that $x^2 / N(x) \rightarrow 0$ as $x \rightarrow \infty$.
Fine states at the end of his paper without proof that $[g(x,N(x)) - f(x)]/f(x)$ is not larger that $1 / 2$ for large $x$, meaning that
\begin{equation}
\limsup_{x \rightarrow \infty} \frac{g(x,N(x))-f(x)}{f(x)} \leq \frac{1}{2}.
\end{equation}
Does anyone know how to prove this result?
Thank you very much for your attention.
PS I discovered the Jeep Problem in the book "A Beautiful Mind" by Sylvia Nasar. In chapter 17, she tells a curious story in which Nash was challenged by one of his friends to give an upper bound for the minimum required quantity of fuel. Nasar writes in the book that "there is no optimal solution to the problem, as it turns out". She is not explicit about what version of the jeep problem Nash and his friends were discussing: it seems from her words that it was the version later analyzed by Rote and Zhang, Optimal Logistics for Expeditions: the Jeep Problem with Complete Refilling. In any case, also for this version there is an optimal solution, and this existence property is true for all the versions I know of the jeep problems (see the references in https://en.wikipedia.org/wiki/Jeep_problem and http://mathworld.wolfram.com/JeepProblem.html ), so I think Nasar simply made a wrong statement (maybe she simply meant that an optimal solution was not known at that time, or simply she quoted a wrong statement by someone).
PSS A very short and elegant proof that $f(x)$ is the minimum required quantity of fuel is found in Gale, The Jeep Once More or Jeeper by the Dozen, Americ. Math. Monthly, 77 (1970), 493-501.
 A: Finally, I found the answer to my question.
First of all, we can easily give a recursive solution to the problem as follows. 
Let us note that if $P$ is a feasible plan of trips which allows the jeep to arrive at $0$, then we can find another feasible plan $P'$ which arrives at $0$ such that $P'$ is made up a number of trips among $x=y_N$ and $y_{N-1}$, followed by a number of trips among $y_{N-1}$ and $y_{N-2}$, and so on, up to a number of trips among $y_1$ and $y_0=0$.
Indeed, assume that the jeep makes $m$ round trips starting at $x$, followed by a last one-way trip $A_{m+1}$ from $x$ to $y_{N-1}$. The i-th of these round trips is made up of the one-way trip $A_i$ from $x$ to $y_{N-1}$, followed by a round trip $B_i$ starting and ending at $y_{N-1}$, and by a one-way trip $C_i$ from $y_{N-1}$ to $x$. Let $g_i$ be the fuel of the jeep when it starts the trip $A_i$. Suppose we replace the sequence of trips $A_1$, $B_1$, $C_1$, ..., $A_m$, $B_m$, $C_m$, $A_{m+1}$ with the sequence $A_1$, $C_1$, ..., $A_m$, $C_m$, $A_{m+1}$ and deposit the quantity $g_i - 2(x - y_{N-1})$ at $y_{N- 1}$ in $A_i$, $i=1,\dots,m$, and $g_{m+1} - (x - y_{N-1})$ at $y_{N-1}$ in $A_{m+1}$. Then we are now in a position to make the trips $B_1$, $B_2$, ... , $B_m$. When we do this the final configuration is not altered. By induction, we can so get the desired $P'$. We call a plane like $P'$ a standard plan.
Now, if $S$ is a standard plan which realizes the optimal solution, then clearly the plan $S_n$, $n=1,\dots,N-1$ obtained from $S$ deleting the trips starting at $x_m$, for $m > n$, realizes the optimal solution $g(y_n, n)$. So, if $k_n$ is the number of trips in $S$ between $y_{n-1}$ and $y_n$, we have
\begin{equation}
g(y_n,n) - g(y_{n-1},n-1) = (2 k_n - 1) \Delta,
\end{equation}
where $\Delta=x/N$. Moreover, since in the first $k_n - 1$ round trips the maximum fuel that can be deposited at $y_n$ is $1-2 \Delta$, while in the $k_n$-th trip is $1- \Delta$, we must have
\begin{equation}
(k_n - 1)(1 - 2 \Delta) + 1 - \Delta \geq g(y_{n-1},n-1),
\end{equation}
or
\begin{equation}
k_n (1 - 2 \Delta) + 1 + \Delta \geq g(y_{n-1},n-1).
\end{equation}
Now note that, since we want to minimize the consumed fuel, clearly $k_n$ is determined as the least positive integer satisfying the above inequality.
This solves the problem recursively. 
Now, note that if $k_n \geq 2$, then we must have 
\begin{equation}
(k_n - 1)(1 - 2 \Delta) +  \Delta < g(y_{n-1},n-1),
\end{equation}
from which we get
\begin{equation}
k_n < \frac{g(y_{n-1},n-1) + 1 - 3 \Delta}{1-2 \Delta},
\end{equation}
so that
\begin{equation}
\frac{g(y_n,n) - g(y_{n-1},n-1)}{\Delta} = 2k_n - 1 < \frac{2g(y_{n-1},n-1) + 1 - 4 \Delta}{1-2 \Delta} < \frac{2g(y_{n-1},n-1) + 1}{1-2 \Delta}.
\end{equation}
Since we have 
\begin{equation}
g(\Delta \left \lfloor {1/ \Delta} \right \rfloor, \left \lfloor {1/ \Delta} \right \rfloor) = 1,
\end{equation}
we compare $g(y_n,n)$ with the function
\begin{equation}
h(y)= c(N) \exp \left( \frac{2y}{1- 2 \Delta} \right) - \frac{1}{2},
\end{equation}
where
\begin{equation}
c(N) = \frac{3}{2 \exp \left( \frac{ 2 \Delta \left \lfloor {1/ \Delta} \right \rfloor}{1 - 2 \Delta} \right) }.
\end{equation}
The function $h$ is a convex, satisfies the differential equation
\begin{equation}
h'=\frac{2 h + 1}{1-2 \Delta},
\end{equation}
and the initial condition $h(\Delta \left \lfloor {1/ \Delta} \right \rfloor)=1$.
From the above inequality for $g(y_n,n)$ we then get by induction that
\begin{equation}
h(y_n) \geq g(y_n,n),
\end{equation}
for all $n \geq \left \lfloor {1/ \Delta} \right \rfloor$, so that in particular
\begin{equation}
g(x,N(x)) \leq c(N(x)) \exp \left( \frac{2x}{1- 2 \Delta} \right) - \frac{1}{2}.
\end{equation}
Now note that since $x^2 / N(x) \rightarrow 0$ as $x \rightarrow \infty$, we have $\Delta \rightarrow 0$ and $\Delta \left \lfloor {1/ \Delta} \right \rfloor \rightarrow 1$. Moreover we have 
\begin{equation}
\lim_{x \rightarrow \infty} \frac{\exp \left( \frac{2x}{1 - 2 \Delta} \right)}{e^{2x}} = 
\lim_{x \rightarrow \infty} \exp \left(\frac{4x \Delta}{1 - 2 \Delta} \right) = 1.
\end{equation}
We finally have so 
\begin{equation}
\limsup_{x \rightarrow \infty} \frac{g(x,N(x))}{f(x)} \leq 
\limsup_{x \rightarrow \infty} \frac{c(N(x)) \exp \left( \frac{2x}{1 - 2 \Delta} \right)}{\frac{1}{4} \exp \left( 2x - \gamma \right) + \mathcal{O}(e^{-2x})} = \frac{6}{e^{2 - \gamma}} < \frac{3}{2}.
\end{equation}
QED
