$36 \cdot 42$ A bug (of negligible size) starts at the origin in the coordinate plane. First, it moves 1 unit to the right to $(1,0)$. Then it turns $90^{\circ}$ counterclockwise and walks $\frac{1}{2}$ unit to $\left(1, \frac{1}{2}\right)$. If it continues in this manner, turning $90^{\circ}$ counterclockwise each time and walking half the distance of the previous move, the point closest to the bug is
(A) $\left(\frac{2}{3}, \frac{2}{3}\right)$.
(B) $\left(\frac{4}{5}, \frac{2}{5}\right)$.
(C) $\left(\frac{2}{3}, \frac{4}{5}\right)$.
(D) $\left(\frac{2}{3}, \frac{1}{3}\right)$.
(E) $\left(\frac{2}{5}, \frac{4}{5}\right)$.
(31st American High School Mathematics Examination, 1980)
[Solution 1] If the bug walks infinitely, the algebraic sum of its horizontal movement, i.e., the limit of the geometric series, is
$$
1-\frac{1}{4}+\frac{1}{16}-\cdots=\frac{1}{1-\left(-\frac{1}{4}\right)}=\frac{4}{5} .
$$

Similarly, the algebraic sum of its vertical movement approaches
$$
\frac{1}{2}-\frac{1}{8}+\frac{1}{32}-\cdots=\frac{2}{5} .
$$

Therefore, the bug is closest to $\left(\frac{4}{5}, \frac{2}{5}\right)$, so the answer is $(B)$.
[Solution 2] These line segments can be viewed as a complex geometric series with the first term $a_{1}=1$ and common ratio $r=\frac{i}{2}$. Thus,
$$
\sum_{k=1}^{\infty} a_{k}=\frac{a_{1}}{1-r}=\frac{2}{2-i}=\frac{4+2 i}{5},
$$

In coordinate terms, the limit is the point $\left(\frac{4}{5}, \frac{2}{5}\right)$. Hence, the answer is $(B)$.