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

Question

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?

Answer 1

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 $(n+1)^2-n^2=2n+1$

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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.