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?

  • $\begingroup$ Do I have to prove that ${a_n}$ is not a cauchy sequence and hence it does not converge? How do i do that? $\endgroup$ – 112358 Aug 25 '16 at 4:16

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 $n\to\infty$.

  • $\begingroup$ No problem, happy to help. $\endgroup$ – carmichael561 Aug 25 '16 at 4:30
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    $\begingroup$ @112358 Welcome to Math.SE! If you find that this post fully answers your question, then I encourage you to upvote it (to show your appreciation to the answerer and to indicate to others that this is a good answer when they stumble on this question in the future) and perhaps to "accept" it (which you can do by clicking the checkmark beneath the answer --- this indicates that your question is fully solved). You don't have to, of course, but these are usually good things. $\endgroup$ – davidlowryduda Aug 25 '16 at 4:46

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