4
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
6
$\begingroup$

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 $n\to\infty$.

$\endgroup$
  • $\begingroup$ No problem, happy to help. $\endgroup$ – carmichael561 Aug 25 '16 at 4:30
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.