# Concluding divergence based on first and second derivatives

I've just started a calc III course and we're currently reviewing calc II. While doing some problems about infinite sequences, I had the thought that if $\{a_n\} = f(n)$ for $n = 1, 2, 3, ...$ and if, for all values of $x$ greater than some $c$, $f$ is continuous with $f '(x) > 0$ and $f ''(x) > 0$ (or $f '(x) < 0$ and $f ''(x) < 0$), $\{a_n\}$ must be divergent. I saw my prof at office hours today and he agreed with the intuition. He said that if I could find or come up with a proof for this, I could use it on our quiz tomorrow to prove that sequences diverge. I haven't found any sort of proof and am having trouble figuring it out on my own (or maybe my intuition was wrong), so would anyone be able to confirm/deny this and help come up with a proof?

Nicholas

• Do you know the mean value theorem? May 2 '17 at 21:25

You can prove that $f$ is not bounded, since then it will follow that $f$ is increasing and unbounded and therefore diverge to $\infty$ as $x \to \infty$. There is a sneek peek at the proof's outline bellow, but please try by yourself first!
Assume it is bounded. Since $f^\prime > 0$ the function $f$ is increasing and therefore the limit $$\underset{x \to \infty} \lim f(x) = L$$ exists and finite ($f$ is increasing and bounded). Convince yourself that since $f$ has a finite limit at $x \to \infty$ it holds that $\underset{x \to \infty} \lim f^\prime(x) = 0$. But since $f^{\prime\prime} > 0$ we know that $f^\prime > 0$ is increasing to the limit $0$, which is a contradiction. I'm sure you will be able to cover the other details and cases by yourself :)