7. A person practices shooting at a target, starting 100 meters away, and performs the first shot. If this shot misses, he steps back 50 meters for the second shot, and continues this process. Before each shot, he steps back 50 meters until he hits the target. It is known that his first shot has a hit rate of $\frac{1}{4}$, and the hit rate is inversely proportional to the square of the distance. Therefore, the probability that he will eventually hit the target is $\qquad$
7. $\frac{1}{2}$.

Let the event "hit on the $n$-th shot" be denoted as $A_{n}$, with its probability being $P\left(A_{n}\right)$. Then, $P\left(A_{1}\right)=\frac{1}{4}$. According to the problem, we have $\frac{1}{4}=\frac{k}{100^{2}}$, solving for $k$ gives $k=2500$.
For the $n$-th shot, the distance to the target is $100+50(n-1)=50(n+1)$ meters, thus
$$
P\left(A_{n}\right)=\frac{2500}{(50(n+1))^{2}}=\frac{1}{(n+1)^{2}},
$$

Therefore, the probability of hitting within the first $n$ shots is
$$
\begin{aligned}
P_{n} & =1-P\left(\overline{A_{1}} \cdot \overline{A_{2}} \cdots \cdot \overline{A_{n}}\right)=1-P\left(\overline{A_{1}}\right) \cdot P\left(\overline{A_{2}}\right) \cdots \cdots P\left(\overline{A_{n}}\right) \\
& =1-\left(1-\frac{1}{4}\right) \cdot\left(1-\frac{1}{9}\right) \cdots\left(1-\frac{1}{(n+1)^{2}}\right) \\
& =1-\frac{1}{2} \cdot \frac{3}{2} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdots \cdots \frac{n}{n+1} \cdot \frac{n+2}{n+1} \\
& =1-\frac{1}{2} \cdot \frac{n+2}{n+1}=\frac{n}{2(n+1)} .
\end{aligned}
$$

Let $n \rightarrow \infty$, we get $\lim _{n \rightarrow \infty} P_{n}=\frac{1}{2}$. Therefore, the probability that this person can hit the target is $\frac{1}{2}$.