Applonius Circle/ Ford Circle / Infinite GP / Circle Packing 
All the smaller circles are mutually tangent and continue to infinity. What is sum of radii of all the smaller circles?

This is similar to the Pappus chain in the arbelos and can be answered with circle inversion the same way as that problem is analyzed at Cut The Knot.

Inversion of the figure through a circle centered at $A$ through $C$ (dashed red) takes the semicircles on $AB$ and $AC$ to parallel lines perpendicular to $AC$. The images of all of the inscribed circles are tangent to both lines and so are all congruent.
Given $AB=k, BC=1$, the radius of inversion is $AC=k+1$. Hence the image of $B$ satisfies
$$
AB\cdot AB' = AC^2 = (k+1)^2\\
AB' = \frac{(k+1)^2}{k}
$$
and the common radius of the image circles is
$$
r' = CM=\frac{AB'-AC}{2} = \frac{(k+1)^2-k(k+1)}{2k} = \frac{k+1}{2k}
$$
Let the inscribed circles have centers at $O_1,O_2,\ldots$ with radii $r_1,r_2,\ldots$, and their images have centers at $O_1', O_2',\ldots$ with common radius $r'$. Then since all image circles are tangent along the line $MO_1'$ it is immediate that $MO_n' = (2n-1)r'$.
Take a common tangent to the $n$th circle and its image meeting them at $T_n$ and $T_n'$ respectively. Then $AT_nO_n$ is similar to $AT_n'O_n'$ and hence
$$
r_n = r' \frac{AT_n}{AT_n'}
$$
but $T_n$ and $T_n'$ are also inverses and so satisfy
$$
AT_n \cdot AT_n' = AC^2 = (k+1)^2
$$
and by the Pythagorean theorem
$$
(AT_n')^2 + (r')^2 = (AO_n')^2 = AM^2+(MO_n'^2) \\
(AT_n')^2 = -\left(\frac{k+1}{2k}\right)^2+\left(\frac{(k+1)(2k+1)}{2k}\right)^2+\left(\frac{(2n-1)(k+1)}{2k}\right)^2 \\
(AT_n')^2 = \frac{(k+1)^2}{4k^2}\left(-1+(2k+1)^2+(2n-1)^2\right)
$$
and putting it all together
$$
\begin{align}
r_n &= r' \frac{AT_n}{AT_n'} \\
&= \frac{k+1}{2k} \frac{(k+1)^2}{(AT_n')^2} \\
&= \frac{2k(k+1)}{4k(k+1)+(2n-1)^2}
\end{align}
$$
Then with help from WolframAlpha the sum of the radii is
$$
\begin{align}
\sum_{n=1}^\infty r_n &= 2k(k+1) \sum_{n=1}^\infty \frac{1}{(2n-1)^2+4k(k+1)} \\
&= 2k(k+1) \frac{\pi \tanh(\pi \sqrt{4k(k+1)}/2)}{4\sqrt{4k(k+1)}}\\
&= \frac{\pi}{4}\sqrt{k(k+1)} \tanh(\pi\sqrt{k(k+1)})
\end{align}
$$
QED.
Let $P_1$ be the foot of the perpendicular from $O_1$ to $AC$. As a check we can observe that the semicircle on $AP_1$ that is the inverse of the ray $MO_1'$ (show green dashed in the diagram) passes through the points of tangency between the inscribed circles and near their centers, so is an approximation (and lower bound) for twice the sum of the radii.
Since $AP_1O_1$ is similar to $AMO_1'$
$$
AP_1 = AM\frac{r_1}{r'} = \frac{2k(k+1)}{2k+1}
$$
and half the arc length gives the bound
$$
\sum_{n=1}^\infty r_n > \frac{\pi AP_1}{4} = \frac{\pi k(k+1)}{4k+2}
$$
for example, when $k=50$ this bound gives $\sum r_n>39.6587\cdots$, whereas the exact value from the expression above is $\sum r_n=39.6606\cdots$.
