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As noted in the comment, apparently I misunderstood the question. I did not notice the ellipsis in the picture in the OP and therefore thought that the process begins from the red circles and goes right-to-left.

So, I will retain the right-to-left-interpretation solution but will now add -- based on the same idea -- a simple and complete left-to-right-interpretation proof.

First here is the original, right-to-left solution, given while apparently misunderstanding the question:

The radii will be decreasing (and hence converging) iff the angle between the line through the centers of the red circles with the negative horizontal axis will be $<\pi/3$.

So, it is enough to show the angle between the line through the centers of the green circles with the negative horizontal axis is less than the angle between the line through the centers of the red circles with the negative horizontal axis.

But clearly $a<\pi/3<b$ in the picture below, so that $a<b$. $\quad\Box$

Now, with the ellipsis noticed, it is easy to adapt the idea used above, to the opposite, left-to-right direction, to a get a simple complete proof of the apparently desired result. Indeed, now let $a_n:=a<\pi/3<b=:b_n$, and also let $r_n$ and $r_{n+1}$ denote the respective radii of the green and red circles. Then the angle between the line through the centers of the green circles and the line through the centers of the red circles is $b_n-a_n$, so that $\sum_n(b_n-a_n)<\infty$ and hence $a_n,b_n\to\pi/3$. On the other hand, considering the two triangles in the picture above and using the first-order Taylor expansion of $\cos$ at $\pi/3$, we see that $$\frac{r_{n+1}}{r_n}=\frac{\cos a_n}{\cos b_n} =\exp\big((\sqrt3+o(1))\,(b_n-a_n)\big).$$ It remains to recall that $\sum_n(b_n-a_n)<\infty$. $\quad\Box$

As noted in the comment, apparently I misunderstood the question. I did not notice the ellipsis in the picture in the OP and therefore thought that the process begins from the red circles and goes right-to-left.

So, I will retain the right-to-left-interpretation solution but will now add -- based on the same idea -- a simple and complete left-to-right-interpretation proof.

First here is the original, right-to-left solution, given while apparently misunderstanding the question:

The radii will be decreasing (and hence converging) iff the angle between the line through the centers of the red circles with the negative horizontal axis will be $<\pi/3$.

So, it is enough to show the angle between the line through the centers of the green circles with the negative horizontal axis is less than the angle between the line through the centers of the red circles with the negative horizontal axis.

But clearly $a<\pi/3<b$ in the picture below, so that $a<b$. $\quad\Box$

Now, with the ellipsis noticed, it is easy to adapt the idea used above, to the opposite, left-to-right direction, to a get a simple complete proof of the apparently desired result. Indeed, now let $a_n:=a<\pi/3<b=:b_n$, and also let $r_n$ and $r_{n+1}$ denote the respective radii of the green and red circles. Then the angle between the line through the centers of the green circles and the line through the centers of the red circles is $b_n-a_n$, so that $\sum_n(b_n-a_n)<\infty$ and hence $a_n,b_n\to\pi/3$. On the other hand, considering the two isosceles triangles in the picture above and using the first-order Taylor expansion of $\cos$ at $\pi/3$, we see that $$\frac{r_{n+1}}{r_n}=\frac{\cos a_n}{\cos b_n} =\exp\big((\sqrt3+o(1))\,(b_n-a_n)\big).$$ It remains to recall that $\sum_n(b_n-a_n)<\infty$. $\quad\Box$

As noted in the comment, apparently I misunderstood the question. I did not notice the ellipsis in the picture in the OP and therefore thought that the process begins from the red circles and goes right-to-left.

So, I will retain the right-to-left-interpretation solution but will now add -- based on the same idea -- a simple and complete left-to-right-interpretation proof.

First here is the original, right-to-left solution, given while apparently misunderstanding the question:

The radii will be decreasing (and hence converging) iff the angle between the line through the centers of the red circles with the negative horizontal axis will be $<\pi/3$.

So, it is enough to show the angle between the line through the centers of the green circles with the negative horizontal axis is less than the angle between the line through the centers of the red circles with the negative horizontal axis.

But clearly $a<\pi/3<b$ in the picture below, so that $a<b$. $\quad\Box$

Now, with the ellipsis noticed, it is easy to adapt the idea used above, to the opposite, left-to-right direction, to a get a simple complete proof of the apparently desired result. Indeed, now let $a_n:=a<\pi/3<b=:b_n$, and also let $r_n$ and $r_{n+1}$ denote the respective radii of the green and red circles. Then the angle between the line through the centers of the green circles and the line through the centers of the red circles is $b_n-a_n$, so that $\sum_n(b_n-a_n)<\infty$ and hence $a_n,b_n\to\pi/3$. On the other hand, considering the two triangles in the picture above and using the first-order Taylor expansion of $\cos$ at $\pi/3$, we see that $$\frac{r_{n+1}}{r_n}=\frac{\cos a_n}{\cos b_n} =\exp\big((\sqrt3+o(1))\,(b_n-a_n)\big).$$ It remains to recall that $\sum_n(b_n-a_n)<\infty$. $\quad\Box$

As noted in the comment, apparently I misunderstood the question. I did not notice the ellipsis in the picture in the OP and therefore thought that the process begins from the red circles and goes right-to-left.

So, I will retain the right-to-left-interpretation solution but will now add -- based on the same idea -- a simple and complete left-to-right-interpretation proof.

First here is the original, right-to-left solution, given while apparently misunderstanding the question:

The radii will be decreasing (and hence converging) iff the angle between the line through the centers of the red circles with the negative horizontal axis will be $<\pi/3$.

So, it is enough to show the angle between the line through the centers of the green circles with the negative horizontal axis is less than the angle between the line through the centers of the red circles with the negative horizontal axis.

But clearly $a<\pi/3<b$ in the picture below, so that $a<b$. $\quad\Box$

Now, with the ellipsis noticed, it is easy to adapt the idea used above, to the opposite, left-to-right direction, to a get a simple complete proof of the apparently desired result. Indeed, now let $a_n:=a<\pi/3<b=:b_n$, and also let $r_n$ and $r_{n+1}$ denote the respective radii of the green and red circles. Then the angle between the line through the centers of the green circles and the line through the centers of the red circles is $b_n-a_n$, so that $\sum_n(b_n-a_n)<\infty$ and hence $a_n,b_n\to\pi/3$. On the other hand, considering the two triangles in the picture above and using the first-order Taylor expansion of $\cos$ at $\pi/3$, we see that $$\frac{r_{n+1}}{r_n}=\frac{\cos a_n}{\cos b_n} =\exp\big((\sqrt3+o(1))\,(b_n-a_n)\big).$$ It remains to recall that $\sum_n(b_n-a_n)<\infty$. $\quad\Box$

As noted in the comment, apparently I misunderstood the question. I did not notice the ellipsis in the picture in the OP and therefore thought that the process begins from the red circles and goes right-to-left.

Here is the original solution, given while apparently misunderstanding the question:

The radii will be decreasing (and hence converging) iff the angle between the line through the centers of the red circles with the negative horizontal axis will be $<\pi/3$.

So, it is enough to show the angle between the line through the centers of the green circles with the negative horizontal axis is less than the angle between the line through the centers of the red circles with the negative horizontal axis.

But clearly $a<\pi/3<b$ in the picture below, so that $a<b$. $\quad\Box$

Now, with the ellipsis noticed, it is easy to adapt the idea used above, in the opposite direction, to a get a simple complete proof of the apparently desired result. Indeed, now let $a_n:=a<\pi/3<b=:b_n$, and also let $r_n$ and $r_{n+1}$ denote the respective radii of the green and red circles. Then the angle between the line through the centers of the green circles and the line through the centers of the red circles is $b_n-a_n$, so that $\sum_n(b_n-a_n)<\infty$ and hence $a_n,b_n\to\pi/3$. On the other hand, considering the two triangles in the picture above and using the first-order Taylor expansion of $\cos$ at $\pi/3$, we see that $$\frac{r_{n+1}}{r_n}=\frac{\cos a_n}{\cos b_n} =\exp\big((\sqrt3+o(1))\,(b_n-a_n)\big).$$ It remains to recall that $\sum_n(b_n-a_n)<\infty$. $\quad\Box$

As noted in the comment, apparently I misunderstood the question. I did not notice the ellipsis in the picture in the OP and therefore thought that the process begins from the red circles and goes right-to-left.

So, I will retain the right-to-left-interpretation solution but will now add -- based on the same idea -- a simple and complete left-to-right-interpretation proof.

First here is the original, right-to-left solution, given while apparently misunderstanding the question:

The radii will be decreasing (and hence converging) iff the angle between the line through the centers of the red circles with the negative horizontal axis will be $<\pi/3$.

So, it is enough to show the angle between the line through the centers of the green circles with the negative horizontal axis is less than the angle between the line through the centers of the red circles with the negative horizontal axis.

But clearly $a<\pi/3<b$ in the picture below, so that $a<b$. $\quad\Box$

Now, with the ellipsis noticed, it is easy to adapt the idea used above, to the opposite, left-to-right direction, to a get a simple complete proof of the apparently desired result. Indeed, now let $a_n:=a<\pi/3<b=:b_n$, and also let $r_n$ and $r_{n+1}$ denote the respective radii of the green and red circles. Then the angle between the line through the centers of the green circles and the line through the centers of the red circles is $b_n-a_n$, so that $\sum_n(b_n-a_n)<\infty$ and hence $a_n,b_n\to\pi/3$. On the other hand, considering the two triangles in the picture above and using the first-order Taylor expansion of $\cos$ at $\pi/3$, we see that $$\frac{r_{n+1}}{r_n}=\frac{\cos a_n}{\cos b_n} =\exp\big((\sqrt3+o(1))\,(b_n-a_n)\big).$$ It remains to recall that $\sum_n(b_n-a_n)<\infty$. $\quad\Box$