#### To determine

**To find:** The expression for the diameter
cn
.

#### Answer

The expression for the diameter
cn
is
dn=1n(n+1)
.

#### Explanation

**Given:**

The two circles *C* and *D* of radius 1 touch at *P*.

**Calculation:**

Assume that the center of the circle *D* is *R* and the center of the circle
C1
is *A* and the triangle *PRA* is a right triangle

By the Pythagoras theorem,

(AP)2+(PR)2=(AR)2(1−d12)2+12=(1+d12)21−d1+d124+1=1+d1+d124

Simplify further,

1−d1+=1+d22d2=1d2=12

Assume that the center of the circle *D* is *R* and the center of the circle
C2
is *B* and the triangle *PRB* is a right triangle

By the Pythagoras theorem,

(PB)2+(PR)2=(BR)2(1−(d1−d22))2+12=(1+d22)2(1−(12+d22))2+1=(1+d22)21−2(12+d22)+(12+d22)2=1+d2+d224

That is,
d2=16
.

Assume that the center of the circle *D* is *R* and the center of the circle
C3
is *C* and the triangle *PRB* is a right triangle

By the Pythagoras theorem

(PC)2+(PR)2=(CR)2(1−(d1+d2+d32))2+12=(1+d32)2(1−(12+16+d32))2+1=(1+d32)2(1−(23+d32))2+1=(1+d32)2

That is,
d3=112
.

The diameters
d1,d2 and d3
can be expressed as follows,

d1=11×2d2=12×3d3=13×4

Therefore, it can be conclude that
dn=1n(n+1)
.

Assume that the center of the circle *D* is *R* and the center of the circle
Cn+1
is *N* and the triangle *PRN* is a right triangle

By the Pythagoras theorem,

(NR)2+(PR)2=(NR)2(1−(d1+d2+d3+⋯dn+dn+12))2+12=(1+dn+12)2(1−(∑i=1ndi+dn+12))2+1=(1+dn+12)2(1−(∑i=1n1i(i+1)+dn+12))2+1=(1+dn+12)2

Further simplification,

(1−(1−1n+1+dn+12))2+1=(1+dn+12)2(1n+1−dn+12)2+1=(1+dn+12)21(n+1)2−dn+1n+1+dn+124+1=1+dn+1+dn+124dn+1=1(n+1)(n+2)

Hence, conclude that
dn=1n(n+1)
for every
n
.

Since,
di=1i(i+1)
for every
i≤n
.

∑i=1n1i(i+1)=∑i=1n1i−1i+1=1−1n+1

As
n→∞
, the partial sum
∑i=1ndi
approaches to 1.

Therefore, the expression for the diameter
cn
is
dn=1n(n+1)
.