For a monotone decreasing function $f:\mathbb{R} \rightarrow (0,\infty)$ Show that $a_n \to \infty$ 
Let $f:\mathbb{R} \rightarrow (0,\infty)$ be a monotonic decreasing function. Suppose $a_1=1$ and $a_{n+1}=a_{n} + f(a_n)$. Show that $a_n \to \infty$

I've got that $a_n$ is a monotonic increasing sequence as well since the range of $f$ is the positive real numbers. How do I proceed from there?
 A: As you observed, $\{a_n\}$ is an increasing sequence, so it is enough to show that $\{a_n\}$ is not bounded above.
Suppose that there is some $M>0$ such that $a_n\leq M$ for all $n$, and let $c=f(M)>0$. Then since $f$ is decreasing, we have $a_2=a_1+f(a_1)\geq a_1+c$, $a_3=a_2+f(a_2)\geq a_2+c\geq a_1+2c$, and in general
$$ a_{n+1}=a_n+f(a_n)\geq a_n+c\geq\dots\geq a_1+nc $$
Since $c>0$, $a_1+nc$ will be larger than $M$ for sufficiently large $n$, which contradicts the assumption that $a_n\leq M$ for all $n$. Therefore $a_n\to\infty$ as $n\to\infty$.
A: First, you have
$$a_{n+1} - a_n = f(a_n) > 0$$
so the sequence is increasing. Therefore, there are two possibilities : either the sequence converges, or it tends to $+\infty$.
Let's suppose that $(a_n)$ converges to a certain limit $L \in \mathbb{R}$. Then there exists $N \in \mathbb{N}$ such that for all $n \geq N$, one has $a_n \leq L+1$. Because $f$ is decreasing, you get $f(a_n) \geq f(L+1)$, i.e. $a_{n+1}-a_n \geq f(L+1) $. Now let $n$ tend to $+\infty$ in this inequality : you get $0 \geq f(L+1)$ which is absurd.
So $(a_n)_{n \in \mathbb{N}}$ must tend to $+\infty$.
