Let $f: \mathbb{R} \to (0,\infty)$ be a decreasing function. Define a sequence $(a_n)$ by $a_1=1 $ and $a_{n+1}=a_n+f(a_n)$ for every $n\ge 1$. Prove that $(a_n) \to \infty$.
I have tried by contradiction to assume it is bounded and therefore converges. clearly the $a_n$ are strictly increasing and I tried to use Cauchy definition to arrive at a contradiction that $f$ is not decreasing but could not process when having to choose epsilons and which $N$. What can be fixed etc. Any help?
Let ($a_n$) be bounded above by a positive number $M$. Then $L:=\sup_{n}a_n$ exists. Now, fix $\epsilon=\frac{f(L)}{2}>0$. We can then find some $n$ such that $L\ge a_n>L-\epsilon$. Then, $a_{n+1}=a_n+f(a_n)> L-\epsilon+f(L)=L+\frac{f(L)}{2}>L$ which leads to a contradiction.