Fibonacci legerdemain

From rabbits to pineapples and from the Parthenon to sea shells, the Fibonacci numbers seem to make miraculous appearances in quite a few man made or natural structures, often via the closely associated “divine proportion” Φ=1.6180339887…

The two first Fibonacci numbers are 1 and 1. To find the third Fibonacci number, add the first and the second to get 2. To find the fourth Fibonacci number, add the second and the third to find 3. All other infinite Fibonacci numbers are produced in the same recursive manner, so that each Fibonacci number after the second is equal to the sum of the two preceding Fibonacci numbers:

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, …

Though simple, the Fibonacci sequence is far from being trivial and uninteresting, exhibiting quite a few remarkable properties and often showing up unexpectedly in various problems in Mathematics. Ιt is a standard exercise to prove the remarkable fact that the limit of the ratio of successive terms of the Fibonacci sequence is the golden ratio

\Phi =\frac{1+\sqrt{5}}{2}

which means that the ratios of successive terms provide ever improving approximations of the irrational number Φ.

A quite strange property of the Fibonacci sequence involves the inverse of 89, which is the eleventh Fibonacci number:

\frac{1}{89}=0.0\overline{1123595506179775280898876404494382022471910}...

Let’s use the terms of the famous Fibonacci sequence in the following “Christmas tree” arrangement:

0 .0 1

0 . 0 0 1

0 . 0 0 0 2

0 . 0 0 0 0 3

0 . 0 0 0 0 0 5

0 . 0 0 0 0 0 0 8

0 . 0 0 0 0 0 0 1 3

0 . 0 0 0 0 0 0 0 2 1

………………………..

In the n-th row of the arrangement, the n-th Fibonacci number is placed in the n+1 decimal place and this way a new sequence of decimal fractions is created. Adding up all those numbers, one at a time, we get successively:

0 . 0 1

0 . 0 1 1

0 . 0 1 1 2

0 . 0 1 1 2 3

0 . 0 1 1 2 3 5

0 . 0 1 1 2 3 5 8

0 . 0 1 1 2 3 5 9 3

0 . 0 1 1 2 3 5 9 5 1

………………….

making a new “Christmas tree” arrangement, which grows from the “ground” up. As we add one more term each time, the sum becomes larger yet it does not get infinitely large. Instead, it seems to approach the inverse of 89, the eleventh Fibonacci number. With every new term that is added, this partial sum at the base of the “Christmas tree” bears an ever increasing resemblance to the inverse of 89, indicating that possibly, if the sum of the infinite terms could ever be completed, the result would be exactly equal to the inverse of 89.

But how can that be and why 89?

All kinds of legerdemain are based upon the masking of simple facts and similarly part of the “mathematical legerdemain” in this “Fibonacci magic trick” is to mask the numbers being added as decimal fractions nicely arranged in a Fibonacci “Christmas tree”. But what we are actually doing is forming the series

S=1\cdot \frac{1}{10^{2}}+1\cdot \frac{1}{10^{3}}+2\cdot \frac{1}{10^{4}}+3\cdot \frac{1}{10^{5}}+5\cdot \frac{1}{10^{6}}+...

which is similar to a decimal representation, with Fibonacci numbers in the place of digits. Infinite series such as the above are often calculated by observing that they contain scaled down copies of themselves. To see why this is the case here too, let’s first break each Fibonacci number (after the second) of the series into the sum of its two preceding Fibonacci numbers:

S=1\cdot \frac{1}{10^{2}}+1\cdot \frac{1}{10^{3}}+(1+1)\cdot \frac{1}{10^{4}}+(1+2)\cdot \frac{1}{10^{5}}+(2+3)\cdot \frac{1}{10^{6}}+...

Now rearrange the terms of the series to reveal two scaled down copies of the sum S within S itself:

S=1\cdot \frac{1}{10^{2}}+\frac{1}{10^{2}}\cdot (1\cdot \frac{1}{10^{2}}+1\cdot \frac{1}{10^{3}}+2\cdot \frac{1}{10^{4}}+3\cdot \frac{1}{10^{5}}+5\cdot \frac{1}{10^{6}}+...)+\frac{1}{10}\cdot(1\cdot \frac{1}{10^{2}}+1\cdot \frac{1}{10^{3}}+2\cdot \frac{1}{10^{4}}+3\cdot \frac{1}{10^{5}}+5\cdot \frac{1}{10^{6}}+...)

which means that

S=\frac{1}{100}+\frac{1}{100}\cdot S+\frac{1}{10}\cdot S

It is now quite easy to solve this equation with respect to S and get

S=\frac{1}{89}

There’s actually nothing too special about the Fibonacci sequence in that respect and one could make a similar “Christmas tree” arrangement using for example an arithmetic progression (a sequence such as every term after the first is found from its previous one by adding a fixed number d called “common difference”) with positive integer terms instead. For example, the arrangement

0 . 0 1

0 . 0 0 2

0 . 0 0 0 3

0 . 0 0 0 0 4

0 . 0 0 0 0 0 5

0 . 0 0 0 0 0 0 6

0 . 0 0 0 0 0 0 0 7

0 . 0 0 0 0 0 0 0 0 8

………………………..

involves the arithmetic sequence 1, 2, 3, 4, 5, 6, … , produces a sum S which again contains a scaled down copy of itself and is calculated similarly as above:

S=\frac{1}{81}

Using that result, it is quite straightforward to prove that the corresponding “Christmas tree” sum for any arithmetic progression with first term a_{1} and difference d is

S=\frac{9a_{1}+d}{810}

A geometric progression is a sequence such as every term after the first is found from its previous one by multiplying by the same non-zero number r, called “common ratio”. For a geometric progression with positive integer terms, the corresponding “Christmas tree” is rather trivial as it reduces to the sum of the terms of a geometric progression with common ratio equal to

\frac{r}{10}

and remains finite only if

0< r< 10

Then the “Christmas tree” sum is

S=\frac{a_{1}}{100-10r}

For example, the arrangement

0 . 1 6

0 . 0 3 2

0 . 0 0 6 4

0 . 0 0 1 2 8

0 . 0 0 0 2 5 6

0 . 0 0 0 0 5 1 2

0 . 0 0 0 0 1 0 2 4

0 . 0 0 0 0 0 2 0 4 8

………………………..

involves the geometric progression with first term equal to 16 and a common ratio equal to 2. The rows of the tree make a geometric progression with common ratio

\frac{r}{10}=\frac{1}{5}

The corresponding “Christmas tree” sum is equal to 0.2.