Problem: An infinite sequence $a_1$, $a_2$, $a_3$, \dots\ of real numbers satisfies
\[ a_{2n-1} + a_{2n} > a_{2n+1} + a_{2n+2} \qquad
  \mbox{and} \qquad a_{2n} + a_{2n+1} < a_{2n+2} + a_{2n+3} \]
for every positive integer $n$.
Prove that there exists a real number $C$ such
that $a_{n} a_{n+1} < C$ for every positive integer $n$.

Solution: It suffices to solve the problem for sufficiently large $n$.
Let $d_n = (-1)^{n-1}(a_{n+2}-a_n)$.
The assertion simply says that $d_1,d_2,\dotsc$ is strictly increasing.

We consider the following cases.
\begin{itemize}
  \ii Suppose that $d_k > 0$ for some $k$. Then,
  \[ a_{2n+1} = a_1 + (d_1+d_3+\dotsb+d_{2n-1}) \]
  clearly diverges to $+\infty$, and
  \[ a_{2n} = a_2 - (d_2+d_4+\dotsb+d_{2n-2}) \]
  clearly diverges to $-\infty$, so taking $C=0$ works.

  \ii Now assume that $d_k \le 0$ for all $k$.
  This implies that $(a_{2n+1})_{n\ge 0}$ is weakly decreasing
  and $(a_{2n})_{n\ge 1}$ is weakly increasing.
  Adding the two expressions above together, we see that
  \[ d_1 + a_1 + a_2 < a_{2n} + a_{2n+1} < a_1 + a_2. \]
  Since $(a_{2n+1})_{n\ge 0}$ is weakly decreasing,
  it either diverges to $-\infty$ or has a finite limit.
  Similarly, $(a_{2n})_{n\ge 1}$ either diverges to $+\infty$ or has a finite limit.
  \begin{itemize}
    \ii If $(a_{2n+1})_{n\ge 0}$ diverges to $-\infty$,
    then the first inequality above implies
    that $(a_{2n})_{n\ge 1}$ must diverge to $+\infty$,
    in which case $C=0$ works.

    \ii If $(a_{2n})_{n\ge 1}$ diverges to $+\infty$,
    then the second inequality implies that $(a_{2n+1})_{n\ge 0}$
    must diverge to $-\infty$, so again $C=0$ works.

    \ii Finally assume that both $\lim_{n\to\infty} a_{2n+1}$ and $\lim_{n\to\infty} a_{2n}$ exist,
    so the limit $L = \lim_{n\to\infty} a_{2n}a_{2n+1}$ also exists.
    The result follows taking $C > L$.
  \end{itemize}
\end{itemize}