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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 $A_nB_n$. 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). 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, $\alpha_n\to2\pi/3$ 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.

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.

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 $A_nB_n$. 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). 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.

is increasing, and has only fixed point $2\pi/3$ (see Carlo Beenaker's answer).

Thus, $\alpha_n\to2\pi/3$ 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.

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 $A_nB_n$. 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). 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, $\alpha_n\to2\pi/3$ 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.

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 $A_nB_n$. 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). Moreover, $f$ is increasing; this is equivalent to saying that the angle $\theta(r)$ from the picture is increasing whenever it's defined, which is true because when $r$ increases, the point $F$ moves clockwise.

is increasing, and has only fixed point $2\pi/3$ (see Carlo Beenaker's answer).

Thus, $\alpha_n\to2\pi/3$ 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.

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 $A_nB_n$. 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). 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.

is increasing, and has only fixed point $2\pi/3$ (see Carlo Beenaker's answer).

Thus, $\alpha_n\to2\pi/3$ 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.