6. As shown in Figure 2, a circle is inscribed in a square with side length 2, another square is inscribed in this circle, and then another circle is inscribed in the second square. This process continues. If $S_{n}$ represents the total area of the first $n$ circles, then $\lim S_{n}=(\quad)$.
(A) $2 \pi$
(B) $\frac{3}{2} \pi$
(C) $\frac{4}{3} \pi$
(D) $\pi$
6.A.

The diameter of the 1st circle is equal to the side length of the square, $r_{1}=1$; the radius of the 2nd circle $r_{2}=\frac{\sqrt{2}}{2} r_{1}=\frac{\sqrt{2}}{2}$; generally, the radius of the $n$th circle $r_{n}=\frac{\sqrt{2}}{2} r_{n-1}$, which is a geometric sequence with the first term 1 and common ratio $\frac{\sqrt{2}}{2}$, $r_{n}=\left(\frac{1}{2}\right)^{\frac{n-1}{2}}$. Therefore,
$$
\begin{array}{l}
S_{n}=\pi r_{1}^{2}+\pi r_{2}^{2}+\cdots+\pi r_{n}^{2} \\
=\pi\left(1+\frac{1}{2}+\frac{1}{2^{2}}+\cdots+\frac{1}{2^{n-1}}\right)=2 \pi\left[1-\left(\frac{1}{2}\right)^{n}\right] .
\end{array}
$$

Thus, $\lim S_{n}=2 \pi$.