# Zeno's paradox and the limit of an increasing function with negative second derivative

Consider the classic example where the infinite sum of consecutive halves has a finite limit (which I think is related to Zeno's paradoxes, and particularly the Dichotomy paradox). Here, the infinite sum of smaller and smaller yet positive quantities has a finite limit.

Now, think of $f(x)= a\sqrt{x}$, with $a>0$. It is always increasing, but the rate at which it increases is falling (second derivative is negative). And yet, the limit of this function when $x\rightarrow \infty$ is $\infty$.

I struggle to reconcile these facts. How a function which "growth rate" is positive yet falling can reach infinity? Even more, we can make $a$ "very small" (e.g. 0.000001), and the limit of this function still diverges. Conversely, I can "see" how $x^2$ reaches infinity. Just look at the graph and you can see it "exploding".

Is there any intuitive way to comprehend this?

• $\sqrt{x}$ is still increasing, no matter whether its rate of increase is decreasing whilst $2^{-n}$ is decreasing. – Zain Patel Jul 21 '17 at 11:24
• Are you aware of a proof that the series $1+1/2+1/3+\dotsb$ diverges? – Chappers Jul 21 '17 at 11:26
• Some infinite sums of smaller and smaller positive quantities have finite limits ($\sum 2^{-n}$), others don't ($\sum 1/n$). Some increasing functions with negative second derivatives remain finite ($1-2^{-x}$), others don't ($\sqrt x$). – Rahul Jul 21 '17 at 11:37
• @Chappers Nope. I can see it now here. So apparently, in a divergent series the rate of growth can slow down, but cannot slow down too fast. A clarification or proof of this would suffice to answer my question. Rahul's comment clearly shows two examples. – luchonacho Jul 21 '17 at 15:54
• Any series that decays at least as slowly as $1/n$ diverges. Any series that decays faster than $1/n^{1+\epsilon}$ for some $\epsilon>0$ converges. Unfortunately there are series that decay faster than $1/n$ that also diverge, for example $1/(n\log n)$, $1/(n\log n\log\log n)$, and so on... These facts are most easily seen via the integral test. – Rahul Jul 23 '17 at 1:59

Maybe look at it like this:

• For $\sqrt{x}$ to grow from $0$ to $1$, you need to move $x$ from $0$ to $1$.
• For $\sqrt{x}$ to grow from $1$ to $2$, you need to move $x$ from $1$ to $4$, a length of $3$
• For $\sqrt{x}$ to grow from $2$ to $3$, you need to move $x$ from $4$ to $9$, a length of $5$
• For $\sqrt{x}$ to grow from $3$ to $4$, you need to move $x$ from $9$ to $16$, a length of $7$
• For $\sqrt{x}$ to grow from $n$ to $n+1$, you need to move $x$ from $n^2$ to $(n+1)^2$, a length of $(n+1)^2-n^2=2n+1$

So, the reason why the "rate" of growth is slowing down is because for each next step up I want to take, I have to make longer and longer steps to the right. Indeed, the length of those steps becomes arbitrarily large as you go high enough.

However, the reason why the function itself still has a limit of $\infty$ is that no matter how high up I am, there is always a finite number of steps to the right I need to take to reach one step higher. I can't stop at any finite number, because no matter how high the number $M$ I stop at is, I can make $2M+1$ steps to the right and I will hit $M+1$.

So, I can't stop on one million, because I can make $2,000,002$ steps to the right and I will be at $1,000,001$ for example.

• Ok, but given @Rahul's comment, why that rule does not apply to $1-2^{-x}$, which is also increasing with negative second derivative? – luchonacho Jul 21 '17 at 15:49