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} = d_{1} + d_{2} + d_{3} + ... is a convergent series if ever there was one. Obviously,
(1) | d_{1} + d_{2} + d_{3} + ... = 1 |
Let r_{n} = d_{n}/2 be the radius of the n-th circle. In ΔAO_{1}B_{1}, O_{1}B_{1} = CD = 1, the radius of the big circles. O_{1}A = r_{1} + 1 and AB_{1} = 1 - r_{1}. The Pythagorean Theorem implies
(2) | (r_{1} + 1)^{2} = (1 - r_{1})^{2} + 1^{2} |
Solving (2) for r_{1} gives r_{1} = 1/4, so that d_{1} = 1/2. The lowest point of the second circle is at 1 - d_{1} = 1/2. This we have an equation for r_{2}:
(r_{2} + 1)^{2} = (1/2 - r_{2})^{2} + 1^{2},
which gives r_{2} = 1/12 and d_{2} = 1/6. The lowest point of the third circle is at the distance 1 - d_{1} - d_{2} = 1/2 - 1/6 = 1/3 from the center line of the big circles, and its radius is found from
(r_{3} + 1)^{2} = (1/3 - r_{3})^{2} + 1^{2},
to be r_{3} = 1/24. d_{3} = 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} + 1^{2},
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
- J. Cofman, What To Solve?, Oxford Science Publications, 1996.
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