A Zeno's paradox revisited?

Relevant wiki: Harmonic Series

Let us note that, after the first step, before the circle's circumference is doubled, we have completed 1/100 of the required distance. After the radius is doubled, we note that this percentage does NOT change! Indeed, the new distance is then 200 units, but the already completed distance has been scaled accordingly as well and is of 2 units! Therefore, their ratio stays invariant! Of course, for any subsequent steps, this quantity will remain invariant. In other words, after the first step, we have completed 1/100th of the distance, irrespective of how the circle is scaled!

Now, the second step contributes to half what the first step was worth, that is: 1/200, or (1/2)/100. Again, by the argument above, this is scaling invariant. And so, after the second step, we will have completed: 1/100 + (1/2)/100 of the task.

It is now easy to see where this is going and in general, after the nth step, we will have done: $\frac{1}{100} +\frac{1/2}{100} + \frac{1/3}{100} + \cdots + \frac{1/n}{100} \equiv \frac{H_n}{100}$ where $H_n$ is the familiar harmonic series: $H_n = \sum_1^n \frac{1}{k}$

The question is then, what is the smallest n such that $\frac{H_n}{100} \geq 1$ or $H_n \geq 100$ ?

One quick way to go about this is using the fact that $H_n \approx \log(n)$ and therefore $n \approx e^{100}$ .

A more formal approach is to note the inequality : $\log(n+1) < H_n \leq 1+\log(n)$ Picking $H_n \approx 100$ , we find that $n \in (e^{99} , e^{100}-1]$ and again we see that $n \approx e^{100}$ is the best approximation.

Before beginning the explanation, I want to point out that the question wasn't stated very clearly, although it can be easily understood nonetheless. Let's first consider what it means to scale the circle. In order to scale it, we must have a point to scale it about (not necessarily the center). If we were to scale it at any point other than our current position, we would end up inside of the circle since the surface would be scaled away from where we were (unless we were scaled it with). Thus, we can reasonably conclude that we are scaling the circle about our current position. Now we can begin the problem.

Let's start off observing what happens each time we take a step. After the first step before the scaling, we are 1 unit in, 99 units to go. After scaling, both our distance traveled and distance left are scaled by a factor of $\frac{2}{1}$ . Thus, we have 198 units left to go by that point. Since we only care about the distance left to go, we only need to pay attention to that. The main problem when solving this is the factor by which we scale the circle by each time. Since we're adding 100 units to the circumference of the circle, the factor we multiply by is $\frac{C+100}{C}$ , which, as C gets larger, converges to 1. Thus, we now know the answer cannot be infinity since the factor converges.

Now let's setup an equation of what this would look like: After the first step and scaling, we have $\frac{2}{1}(100-1)$ ; After the second step and scaling, we have $\frac{3}{2}(\frac{2}{1}(100-1)-1)$ ; After the third step and scaling, we have $\frac{4}{3}(\frac{3}{2}(\frac{2}{1}(100-1)-1)-1)$ . Notice, when we expand this, the fractions being multiplied to $100-1=99$ cancel out to simply $n+1$ and the $-1$ s are multiplied by a partial sum of the harmonic series multiplied by $n+1)$ . In other words, the distance left to travel can be modeled by the formula $99(n+1)-(n+1)\sum_{k=2}^{n}\frac{1}{k}=(n+1)(99-\sum_{k=2}^{n}\frac{1}{k})$ We want to find when this is equal to zero, thus, we can divide $n+1$ on both sides, and we end up with $99-\sum_{k=2}^n \frac{1}{k}=0 \implies 100-H_n=0 \implies H_n=100$

Now, we want to find the nth harmonic series such that it is greater than or equal to 100. To do this, notice that the harmonic series can be modeled by the function $f(x)=\lfloor \frac{1}{x} \rfloor$ . Now, notice $\frac{1}{x+1}<f(x)\leq \frac{1}{x}$ . Thus, we know that the sum of the harmonic series is between the integrals of the two other functions, namely, $\int_{1}^{n}\frac{1}{x+1} dx < \int_{1}^{n} f(x) dx \leq \int_{1}^{n}\frac{1}{x} dx \implies \ln(n+1)-\ln(2)<H_n\leq \ln(n) \implies \ln(n+1)-\ln(2)<100\leq \ln(n) \implies n\approx \boxed{\lfloor e^{100} \rfloor}$

In a very similar fashion of other solution posted, I assumed that goal was to cover a 2π path but in our mission our steps are reduced following the sequence 1/2 ; 1/3 ; 1/4 and so on. Since our initial step is 2π/100 to achieve our goal we must perform N steps so

2π/100 ( 1+1/2+ 1/3+ …. 1/N) = 2π; simplifying ( 1+1/2+ 1/3+ …. 1/N) = 100

So N will be the term of the Harmonic Series which balance the above equation. It is well known since Euler for large number the Harmonic Series can be approximated by Ln (N) plus γ the Euler Constant. Since the answer is a multiple choice the closest is indeed e^100