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

Let the event "hit on the $n$-th shot" be denoted as $A_{n}$, with its probability being $P\left(A_{n}\right)$. Then,
$$
P\left(A_{1}\right)=\frac{1}{4} \text {. }
$$

The distance from the target during the $n$-th shot is
$$
100+50(n-1)=50(n+1)(\text{m}) \text {, }
$$

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

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 $\frac{1}{2}$.