aa r X i v : . [ m a t h . HO ] J un VISUALIZE GEOMETRIC SERIES
HÙNG VIÊ. T CHUA
BSTRACT . We review some “proofs without words” for the formula for geometricseries and find a common theme lurking behind them. We also answer negatively thequestion raised by Edgar on the existence of other proofs similar to Mabry’s and his.
1. I
NTRODUCTION
The well known formula for geometric series x + x + x · · · = 11 − x , (1.1)for any < x < , is very useful due to its ability to convert an infinite sum into afraction. A common proof of this formula is as follows: (1 − x )(1 + x + x + x + · · · + x n )= (1 + x + x + x + · · · + x n ) − ( x + x + x + x + · · · + x n +1 ) = 1 − x n +1 . Hence, we have (1 − x )(1 + x + x + x + · · · + x n ) = 1 − x n +1 . Dividing both sidesby − x , we arrive at x + x + x + · · · + x n = 1 − x n +1 − x . Letting n → ∞ and noting that lim n →∞ x n +1 = 0 because < x < , we have thedesired formula. Though the proof is elegant, it does not give a clear intuition behindthe formula. For this reason, a number of “proofs without words” have been proposed.The goal of this article is to review proofs without words by various authors, find thecommon theme in these proofs, and answer an open question by one of these authorsabout the existence of similar proofs.In 2001, the Viewpoints 2000 Group published Figure 1 to prove (1.1) without words.In 1999, Mabry [2] published Figure 2 as a visual proof of + ( ) + ( ) + · · · = .Though Mabry’s proof is only for a case of (1.1), the proof provides nice intuitionbehind the formula. For convenience, we assume that the area of the outmost triangleis . The idea is to find the colored area in two ways. The first way is to find the ratioof the colored area in each layer, which is / . The second way is to add up the areasof all colored triangles, which gives / / + (1 / + · · · . The two ways mustgive the same answer, so we have the desired identity. Continuing the work, in 2016,Edgar [1] proved that + ( ) + ( ) + · · · = using Mabry’s technique (see Figure3). At the end of his proof, Edgar asked: “Is it possible to determine which other seriesallow ananalogous proof without words?”. We will show that the answer is negative inSection 2. F IGURE
1. Viewpoints 2000 Group’s prooflayer 1layer 2 ··· F IGURE
2. Mabry’s proof.F
IGURE
3. Edgar’s proof2. D
EMYSTIFY P REVIOUS V ISUAL P ROOFS
In this section, we analyze the technique used by Mabry and Edgar thoroughly. Un-derstanding the technique is the key to answering the question posed by Edgar andallows us to connect it with the proof by the Viewpoints 2000 Group to find a commontheme.2.1.
Backbones of the beautiful proofs by Mabry and Edgar.
Mathematics Subject Classification.
ISUALIZE GEOMETRIC SERIES 3
C BA D layer 1layer 2layer 3 E F IGURE
4. Mabry’s and Edgar’s proof in a nutshellLet n be the total number of triangles in each layer and a be the number of coloredtriangles in each layer. Note that ≤ a < n . Also, let r be the ratio of the height ofeach layer over the distance from A to the base of that layer. The below analysis usesFigure 4, which is a sketch of Figures 2 and 3. We assume that △ ABC has area .We will compute the colored area, denoted by T , in two ways. Because in each layer,there are n triangles, a of which are colored, we know that T = a/n . Another way tocompute T is to find the total areas of all colored triangles as follows.The area of layer 1 is clearly − (1 − r ) . Hence, each triangle in layer 1 has area − (1 − r ) n , which is also the ratio of the area of each triangle in layer 1 over the area of △ ABC (because the area of △ ABC is ). Due to similarity, the area of each triangle in layer over the area of △ ADE is also − (1 − r ) n . Since the area of △ ADE is (1 − r ) , each triangle in layer has area − (1 − r ) n · (1 − r ) . In general, each triangle in layer n has area − (1 − r ) n · (1 − r ) n − . Therefore, the total colored area is (recall that in each layer, a triangles are colored) a · − (1 − r ) n + a · − (1 − r ) n (1 − r ) + a · − (1 − r ) n (1 − r ) + · · · . (2.1)As the two ways of computing T must produce the same result, we have a · − (1 − r ) n + a · − (1 − r ) n (1 − r ) + a · − (1 − r ) n (1 − r ) + · · · = an . (2.2) HÙNG VIÊ. T CHU
We now apply (2.2) to have the two identities proven by Edgar [1] and Mabry [2]. InMabry’s proof, we have ( n, a, r ) = (3 , , / . Plugging these numbers into (2.2) gives
14 + (cid:18) (cid:19) + (cid:18) (cid:19) + · · · = 13 . In Edgar’s proof, we have ( n, a, r ) = (5 , , / . Plugging these numbers into (2.2)gives
49 + (cid:18) (cid:19) + (cid:18) (cid:19) + · · · = 45 . More beautiful pictures?
A natural question is whether we can produce morebeautiful pictures like Mabry’s and Edgar’s for other geometric series of the forms v + v + v + · · · . (2.3)The answer is, unfortunately, negative. To show this, we prove two restrictions on r ,which are simultaneously satisfied only when r is / or / .Comparing (2.1) and (2.3) shows that the second term of the geometric series is thesquare of the first term; hence, we have a · − (1 − r ) n (1 − r ) = (cid:18) a · − (1 − r ) n (cid:19) , which is equivalent to (1 − r ) − (1 − r ) = an . (2.4)Because a/n < , we know that . < − √ < r < . (2.5)This is the first restriction on r .Furthermore, each triangle in layer 1 has area r , while the area of layer 1 is − (1 − r ) . Therefore, to have pictures like Mabry’s and Edgar’s, we need − (1 − r ) r ∈ N , which implies that r ∈ N . (2.6)Then the number of triangles in each layer is /r − , or n = 2 /r − . Plugging thisback to (2.4), we have (1 − r ) − (1 − r ) = a /r − , which is simplifed to a = 1 r + 1 − r . ISUALIZE GEOMETRIC SERIES 5
Due to (2.6) and a ∈ N , it follows that r = 1 /m for some m ∈ N . This is thesecond restriction on r . Combining this restriction with (2.5), we know that m ∈ { , } .Therefore, we answer negatively the question raised by Edgar [1] on whether there aremore similar “proofs without words” for other series.3. T HE COMMON THEME
At the first glance, the proof by the Viewpoints 2000 Group and the proof by Mabryare little related. However, there is a common theme lurking behind them. Figure 5 isobtained by repositioning the Viewpoints 2000 Group’s. For convenience, we substitute a = 1 and replace r by √ r . −√ r −√ r r rA BC layer 1layer 2layer 3F IGURE
5. Repositioning the Viewpoints 2000 Group’s pictureWe can now apply the Mabry’s technique to prove the formula for geometric series. Thecolored area is clearly + r + r + r + · · · . On the other hand, in each layer, thecolored area is r the area of the layer. Hence, the colored area in total is r thearea of △ ABC , which is − r ) . Therefore, we have
12 + 12 r + 12 r + 12 r + · · · = 11 + r · − r ) , which is simplified to r + r + r + · · · = 11 − r . HÙNG VIÊ. T CHU R EFERENCES [1] T. Edgar, Proof without words: sum of powers of , Math. Mag. no.3 (2016) 191.[2] R. Mabry, Proof without words: + ( ) + ( ) + · · · = , Math. Mag. no.1 (1999) 63.[3] The Viewpoints 2000 Group, Proof without words: geometric series, Math. Mag. no.4 (2001) 320. E-mail address : [email protected] D EPARTMENT OF M ATHEMATICS , U
NIVERSITY OF I LLINOIS AT U RBANA -C HAMPAIGN , U