The following problem gives a geometric illustration to a sum of a simple convergent series. Two touching circles of radius 1 also share a common horizontal tangent. First, draw a circle that touches the given two and the horizontal line. Afterwards, proceed with a construction of a series of circles on top of each other, each touching the original two circles and its immediate predecessor. Let dn denote the diameter of the n-th circle.

The sum ∑dn = d1 + d2 + d3 + ... is a convergent series if ever there was one. Obviously, ∑dn = 1. The problem here is to determine dn and verify algebraically the identity

 (1) d1 + d2 + d3 + ... = 1

Let rn = dn/2 be the radius of the n-th circle. In ΔAO1B1, O1B1 = CD = 1, the radius of the big circles. O1A = r1 + 1 and AB1 = 1 - r1. The Pythagorean Theorem implies

 (2) (r1 + 1)2 = (1 - r1)2 + 12

Solving (2) for r1 gives r1 = 1/4, so that d1 = 1/2. The lowest point of the second circle is at 1 - d1 = 1/2. This we have an equation for r2:

(r2 + 1)2 = (1/2 - r2)2 + 12,

which gives r2 = 1/12 and d2 = 1/6. The lowest point of the third circle is at the distance 1 - d1 - d2 = 1/2 - 1/6 = 1/3 from the center line of the big circles, and its radius is found from

(r3 + 1)2 = (1/3 - r3)2 + 12,

to be r3 = 1/24. d3 = 1/12. The distance to the center line shrinks to 1/3 - 1/12 = 1/4. We see the emerging pattern:

The distance from the top of the n-th circle to the line connecting the centers of the big circles equals 1/(n+1).

We establish the truth of this statement by induction. The statement has already been verified for the first three circles. Assume it also holds for the k-th circle. Then, for rk+1 we have the following equation:

(rk+1 + 1)2 = (1/(k+1) - rk+1)2 + 12,

which gives rk+1 = 1/[2(k+1)(k+2)]. dk+1 = 1/[(k+1)(k+2)].

 1/(k+1) - 1/[(k+1)(k+2)] = (k+2)/[(k+1)(k+2)] = 1/[(k+1)(k+2)] = 1/(k+2).

And the proof is complete.

The magic emanates from the identity 1/[n(n+1)] = 1/n - 1/(n+1). The infinite series appears as

 (3) 1/2 + 1/6 + 1/12 + ... + 1/[n(n+1)] + ... =   (1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + ... + (1/n - 1/(n+1)) + ...

so that the sum of the first n terms is indeed 1 - 1/(n+1). All other terms cancel out. For this reason, the series in (3) is known as telescoping. (There is another example of such a series.)

### References

1. J. Cofman, What To Solve?, Oxford Science Publications, 1996. ### Telescoping situations

• Leibniz and Pascal Triangles
• Infinite Sums and Products
• Harmonic Series And Its Parts
• A Telescoping Series
• An Inequality With an Infinite Series
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• An Elementary Proof for Euler's Series
• $\sin 1^{\circ}+\sin {2^\circ}+\sin 3^{\circ}+\cdots+\sin 180^{\circ}=\tan 89.5^{\circ}$
• Problem 3824 from Crux Mathematicorum
• $x_n=\sin 1+\sin 3+\sin 5+\cdots+\sin (2n-1)$
• A Welcome Problem for the Year 2018
• 