Determine the maximum real number $C$ such that
$$
\sum_{i=1}^{n} \frac{x_{i}}{x_{i-1}} \geq n+C,
$$
for all positive integers $n$ and all sequences of positive real numbers $x_{0}, x_{1}, \ldots, x_{n}$ such that $x_{0}=1$ and $x_{n}=2$.
We will show that $C=\log (2)$ (where log refers to the natural logarithm).
By the AM-GM inequality we can compute:
$$
\frac{\frac{x_{1}}{x_{0}}+\frac{x_{2}}{x_{1}}+\frac{x_{3}}{x_{2}}+\cdots+\frac{x_{n}}{x_{n-1}}}{n} \geq \sqrt[n]{\frac{x_{1}}{x_{0}} \times \frac{x_{2}}{x_{1}} \times \frac{x_{3}}{x_{2}} \times \cdots \times \frac{x_{n}}{x_{n-1}}}=\sqrt[n]{2},
$$
with equality if and only if there is some constant $r=\frac{x_{i}}{x_{i-1}}$ for all $i$. So equality can be achiveved by $x_{i}=2^{i / n}(r=\sqrt[n]{2})$ for all $i=0,1,2,3, \ldots, n$. Define
$$
S_{n}=n(\sqrt[n]{2}-1)
$$
and thus it suffices to:
- show $S_{n}$ is a decreasing sequence, and
- determine $C=\lim _{n \rightarrow \infty} S_{n}$.

Letting $t=\frac{1}{n}$, we see
$$
\begin{aligned}
n(\sqrt[n]{2}-1) & =\frac{1}{t}\left(2^{t}-1\right) \\
& =\frac{1}{t}\left(e^{t \log 2}-1\right) \\
& =\log 2+\frac{(\log 2)^{2} t}{2}+\frac{(\log 2)^{3} t^{2}}{6}+\frac{(\log 2)^{4} t^{3}}{24}+\cdots
\end{aligned}
$$
using the power series formula for the exponential function.
As this is a sum of decreasing functions of $t$, it is clear that we have a decreasing function of $t$ and its limit is $\log 2$. This completes the proof, and shows that $C=\log 2$.

Solution 2
Let $f(x)=\frac{1}{x}$. First let us consider the case when $1=x_{0}0$ we need to construct a sequence such that
$$
\sum_{i=1}^{n} \frac{x_{i}}{x_{i-1}}<n+\log (2)+\epsilon
$$

Since the Riemann integral is the limit of its upper sums, we can find increasing sequences $1=x_{0}<x_{1}<x_{2}<\ldots<x_{n}=2$ that satisfy the above inequality.