<i><font size="4">"The tail of a monstrously huge rabbit is tied to a pole in the ground by an infinitely stretchy elastic cord. A flea sits on the pole watching the rabbit (hungrily). The rabbit sees the flea, leaps into the air and lands one kilometer from the pole (with its tail still attached to the pole by the elastic cord). The flea gives chase and leaps into the air landing on the stretched elastic cord one centimeter from the pole. The monster rabbit, seeing this, again leaps into the air and lands another kilometer away from the pole (i.e., a total of two kilometers from the pole). Undaunted, the flea bravely leaps into the air again, landing on the elastic cord one centimeter further along. Once again the rabbit jumps another kilometer and the flea jumps another centimeter along the cord. If this continues indefinitely, will the flea ever catch up to the rabbit? (Assume the earth is flat and extends infinitely far in all directions.)"</font></i><div><br></div><div>Let us call $x$ the number of the flea’s jumps, $F(x)$ the function which inolves the flea’s motion and $R(x)$ the function which involves the rabbit’s one. We have that: \[ R(x) = kx, k = 1000m = 10^{5}cm \] $F(x)$ is equal to the position of the flea after the strech of the elastic cord more a centimeter. Called $\overline{OF_0}$ the flea’s distance before the rabbit’s jump, and $\overline{OR_0}$ the initial rabbit’s distance, $\overline{OR_1}$ the rabbit’s distance after his jump, and $\overline{OF_1}$ the resulting flea’s distance after the rabbit’s jump. Since the cord is elasic it spaces proportionally his points. Thus: \[ \overline{OF_0}:\overline{OR_0} = \overline{OF_1}:\overline{OR_1} \] The flea’s distance after rabbit’s jump results to be: \[ \overline{OF_1}=\frac{\overline{OF_0} \cdot \overline{OR_1}}{\overline{OR_0}} \] Since $\overline{OR_0} $ is $k(x-1)$ and $\overline{OR_1}$ is $kx$ in the flea’s motion function: \[ F(x) = \frac{F(x-1)\cdot kx}{Fk(x-1)}+ 1 \] \[ F(x) = \frac{F(x-1)\cdot x}{Fk(x-1)}+ 1 \] \[ F(x) = F(x-1) \cdot \frac{x}{x-1}+1 \] \[ F(x) = F(x-1) \cdot \frac{x}{x-1}+1 \] \[ F(x) = F(x-1) + \frac{F(x-1)}{x-1}+1 \] That implies $F(x)$ is an increasing function. Since $\frac{1}{x-1}$ for great values of $x$ tends to $0$, we have that some portions of $F(x)$ may look like a striaight line that we will call $f$. Thus $f$ and $R(x)$ will meet if $f$’s slope is grater than $R(x)$ ’s one, which is $100000$. By rewriting the equation, it is easyer to get the right insight to solve the problem. Infact: \[\frac{1}{x-1}+\frac{1}{x-2}+\cdots+\frac{1}{x-x+1}\cdot\frac{F(1)\cdot(x-x+1)}{x-x+1-1} +1 &gt; 9999 \] \[ 1+ \frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{x-1} &gt;9999 \] It is known the value of the harmonic series $\sum^{+\infty}_{n = 0}{\frac{1}{n}} = \infty $. Thus, for an appropriate value of $n$, this series can assume any value. Thus, with some patience, the flea will reach the rabbit.</div>