Prove geometrically that $\sum\limits_{n=1}^\infty\,\dfrac{1}{n(n+1)}=1$. $\mathrm{S_1}$ and $\mathrm{S_2}$ are two circles of unit radius touching at point $P$; $T$ is a common tangent to $\mathrm{S_1}$ and $\mathrm{S_2}$ touching them at $X$ and $Y$; $\mathrm{C}{_1}$ is the circle touching $\mathrm{S_1}$, $\mathrm{S_2}$, and $T$. $\mathrm{C}{_n}$ is the circle touching $\mathrm{S_1}$ , $\mathrm{S_2}$, and  $\mathrm{C}{_{n-1}}$ for $n > 1.$ By computing diameters of $\mathrm{C}{_n}$ prove that $$\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\ldots =1\,.$$ 
 A: By using Descartes circle theorem we find that the curvature of circle $C_1$ is $4$, of $C_2$ is $12$, and in general, the curvature of circle $C_n$ is $2n(n+1)$.The diameter of circle $C_n$ is $\frac1{n(n+1)}$. The $C_n$ circles are stacked on top of each other which implies the sum of all the diameters is the distance from the common tangent $T$ to the intersection point $P$. This is the radius of $S_1$ and $S_2$ which is $1$. Q.E.D.
The Wikipedia article gives
 $\ k_4 =  K_\pm(k_1, k_2, k_3) := k_1 + k_2 + k_3 \pm 2\sqrt{k_1k_2 + k_2k_3 + k_3k_1}, \ $ the formula to compute the curvature of one of the four curvatures of four mutually tangent circles in terms of the other three curvatures.
We begin with $\ K_1 = 4 = K_+(1, 1, 0) \ $ since the tangent line has curvature $0$. Then $\ K_2 = 12 = K_+1, 1, 4), \quad K_3 = 24 = K_+1, 1, 12), \ $
and so on. You can check that $\ K_+(1, 1, 2n(n+1)) = 2(n+1)(n+2). \ $ By definition, the curvature is the reciprocal of the radius, and thus the radius of circle $C_n$ is $\ \frac1{2n(n+1)}. \ $
A: Here is an alternative way to compute the diameter $d_n$ of $C_n$.  Once you obtain $d_n$, you use the same argument as Somos presented to show that $\sum\limits_{n=1}^\infty\,d_n=R$, where $R$ is the radius of $S_1$ and $S_2$ (here $R$ is set to be $1$).
Consider the inversion $i$ with center $P$ and radius $\rho:=PX=PY=\sqrt{2}R$.  Then, $l_1:=i(S_1)$ and $l_2:=i(S_2)$ are two parallel lines perpendicular to $XY$ at $X$ and $Y$, respectively.  The image $i(XY)$ of the line $XY$ is the circle $\Gamma$ passing through $X$, $Y$, and $P$ (noting that $\Gamma$ toughes both $l_1$ and $l_2$).
Now, let $\gamma_n$ denote the image $i(C_n)$ of the circle $C_n$, for $n=1,2,3,\ldots$.  Write $PP_1$ for the diameter of $\Gamma$ with one endpoint being $P$.  It can be easily seen that there are points $P_2,P_3,\ldots$ on the ray $PP_1$ such that $2R=PP_1=P_1P_2=P_2P_3=\ldots$ and each $P_nP_{n+1}$ is a diameter of $\gamma_n$.  This means $PP_n=2Rn$.   Now, write $Q_n$ for $i(P_n)$.  Then,
$$PQ_n=\frac{\rho^2}{PP_n}=\frac{2R^2}{2Rn}=\frac{R}{n}\,.$$
That is, if $d_n$ is the diameter of $\gamma_n$, then
$$d_n=Q_nQ_{n+1}=PQ_n-PQ_{n+1}=\frac{R}{n}-\frac{R}{n+1}=\frac{R}{n(n+1)}\,.$$
