Consider the geometric series,
for 0 < r < 1. The goal is to find a closed-form expression for s_r.
Consider now Figure 1. Given that |AB|=|AD|=1 and |DE|=r, the rest of the figure can be constructed (the lines AC and BF are parallel and the rest of the lines, with the exception of BC, are perpendicular to AC). It is important to note that the four-sided figures ABED, DEHG, GHKJ, and so on, are all similar to each other, and we see that the length |AC| is exactly the quantity we are looking for.
Note now how the triangle ABC is similar to the triangle FEB, leading to
If we then evaluate each side of the equality-sign, we get
Quite elegant, I think.
The figure also shows that the sum converges for all 0 < r < 1 since the proof described above can be carried out whenever BC crosses AC the “right” way. Note, however, that the sum does not converge only for these values of r – in fact, it converges whenever |r| < 1 for complex r (easily seen by considering the first N terms of the sum and then letting N \rightarrow \infty).
(I don’t know who to attribute this proof to, unfortunately. I saw it in a magazine for elementary school teachers.)
Update 2009-08-22: Apparently, the proof was discovered by Benjamin G. Klein and Irl C. Bivens, and it appears on page 120 of Proofs without Words: Exercises in Visual Thinking by Roger B. Nelson (thanks for the reference, David Radcliffe).




How do we get |HG| = r^2?
| 2009-08-22 @ 14:33
@andi: The figure DEHG is a scaled version of ABED and the scaling factor is r (which makes them fit together, side by side).
| 2009-08-22 @ 20:09
Cool proof. (I think you mean “attribute” and not “contribute” though in that last sentence)
| 2009-08-22 @ 15:14
@Scott: Thanks, it has been corrected.
| 2009-08-22 @ 20:09
Very nice. I will show this to my students :-D
| 2009-08-22 @ 15:58
This is a very nice proof. It was discovered by Benjamin G. Klein and Irl C. Bivens, and it appears on page 120 of “Proofs without Words: Exercises in Visual Thinking” by Roger B. Nelson.
| 2009-08-22 @ 17:15
@David Radcliffe: Thanks a lot, I have updated the post with your reference.
| 2009-08-22 @ 20:09
How do you prove that segment BEH is a straight line?
| 2009-08-22 @ 19:25
@JL: Because the figures DEHG and ABED are similar, that is, the corresponding angles within each figure are equal. Hence, the angle ABE and DEH are equal.
| 2009-08-22 @ 20:09
Great proof. However you should start it with proving that the continuation of BE line will hit the continuation of AD line in such a point C where |AC| is the sum of the progression.
As Andi and JL has pointed out, this is not immediately obvious (although very easy to prove).
| 2009-08-23 @ 01:30
|AC| should be a parabola, unless I’m mistaken. As the exponent for r reaches infinity, the value for that specific term also approaches 0, but never meets it.
Technically, there should be an approximate sign over the equality sign.
| 2009-08-23 @ 02:30
JL: That is an excellent question. First you have to note that 1/r is the same as r/r^2 which is the same as r^2/r^3 etc for r>0. From this the similarity of ABED and DEHG etc follows. And from this similarity it follows that angles formed by BED and EHG are the same. This in turn implies that BEH lie on the same line.
| 2009-08-28 @ 02:04
r^2 line can be drawn on the line .just draw a line r^2 perpendicular on AC then move r^2 along A to C .SINCE r^2 is less than r it will fit itself in the triangle somewhere.
| 2010-06-14 @ 23:57