Intuitive explanation why the radii converge in this sequence of eights?

The intuitive explanation is that the sequence approaches the hexagonal close packing, two staggered rows of identical circles:

Because the sequence of radii is monotonically increasing and bounded by this "perfect fit" geometry, it must converge to a finite limit.

Here is a proof that the radii converge without using any big formulas.

Let $r>0$, and consider the following setup in $\mathbb{R}^2$, where the two smaller circles have radius $r$:

The angles $\alpha(r),\beta(r)$ are decreasing with $r$, in particular $\alpha'(r),\beta'(r)<0$ for all $r\in(0,\infty)$. Let $\gamma(r)=\alpha(r)+\beta(r)-\pi/2$.

Now, in your figure, let $A_n,B_n$ be the centers of the $n^{\text{th}}$ circle, $r_n$ the $n^{\text{th}}$ radius and $\alpha_n$ the angle between the the $x$-axis and the ray $B_nA_n$ (so in the picture $\alpha_0=\pi$ and the sequence $(\alpha_n)$ looks decreasing in the first few steps). Note the function $f:(0,\pi)\to(0,\pi)$ that gives $\alpha_{n+1}$ in terms of $\alpha_n$ has fixed point $f(2\pi/3)=2\pi/3$ (see Carlo Beenaker's answer), and $f(\pi)<\pi$. Moreover, $f$ is increasing in $[2\pi/3,\pi]$; this is equivalent to saying that the angle $\theta(r)$ from the picture is increasing whenever it is defined, which is true because when $r$ increases, the point $F$ moves clockwise.

Thus, $\alpha_n$ is decreasing and converges when $n\to\infty$, and $r_{n+1}/r_n\to1$. But on the other hand, we have that $\alpha_{n+1}=\alpha_n+\gamma(r_{n+1}/r_n)$. As $\gamma'(1)<0$, there is $\varepsilon>0$ such that, for big enough $n$, we have $|\alpha_{n+1}-\alpha_n|>\varepsilon\left|1-\frac{r_{n+1}}{r_n}\right|$.

Thus, as $\sum_{n}|\alpha_{n+1}-\alpha_n|<\infty$, we also have $\sum_n\left|1-\frac{r_{n+1}}{r_n}\right|<\infty$, which implies $\prod_n\frac{r_{n+1}}{r_n}<\infty$, that is, the radii converge.

This proof is closely related to Saul RM's one, but I think it's a bit simpler to see what's going on.

For $0\leq \alpha<\pi/2$, let $f(\alpha)$ be the ratio of radii of circles when they are placed in the following configuration:

One can compute $f(\alpha)$ explicitly in terms of trigonometric functions, but all we need to know is that it is differentiable at $0$, which hopefully is believable. This in particular implies that for $\alpha$ small and some constant $C$ (which we can take to be $2f'(\alpha)$) we have $$f(\alpha)\leq f(0)+C\alpha=1+C\alpha.$$

Considering now the diagram in the OP, let $r_n$ be the radius of the $n$-th pair of circles, and $\alpha_n$ to be the angle between the lines connecting the centers of the $n$-th and $n+1$-st pair. By the definition of $f$, this means $r_n=r_{n-1}\cdot f(\alpha_{n-1})$, and so we find $$r_n=r_1\cdot\prod_{i=1}^{n-1}f(\alpha_i)\leq r_1\cdot\prod_{i=1}^{n-1}(1+C\alpha_i).$$ By a well-known connection between convergence of sums and products (easily shown with help of the inequality $1+x\leq e^x$) this product converges as $n\to\infty$ iff the sum $\sum_{i=1}^\infty\alpha_i$ converges. But in our case, the sum $\sum_{i=1}^{n-1}\alpha_i$ represents the angle between the line connecting the centers of the $n$-th pair of circles with the horizontal line, and it's not hard to convince yourself that this angle tends to $\pi/3$, giving the desired convergence.