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 d_n denote the diameter of the n-th circle.

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

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

Let r_n = d_n/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 r_k+1 we have the following equation:

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

which gives r_k+1 = 1/[2(k+1)(k+2)]. d_k+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.

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