= Solution
Let $(u_n)$ be increasing and bounded above, and let $L=\sup\{u_n:n\in\mathbb N\}$. For every $\varepsilon>0$, the definition of the <supremum> supplies $N$ with $L-\varepsilon<u_N\leq L$. Monotonicity gives $L-\varepsilon<u_n\leq L$ for every $n\geq N$, proving $u_n\to L$. This is the <monotone bounded sequence> theorem.
For the recurrence, $f(x_n)>0$ makes $(x_n)$ increasing. If it were bounded above, it would converge to some $L$. Since $x_n\leq L$ and $f$ is decreasing, $f(x_n)\geq f(L)>0$, whence
$$
x_n=1+\sum_{j=1}^{n-1}f(x_j)\geq1+(n-1)f(L)\longrightarrow\infty,
$$
a contradiction. Therefore \b[$x_n\to\infty$].
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