# Another interesting fraction

#### soroban

##### Well-known member

$$\text{We have: }\:\dfrac{1}{89} \;=\;0.01123595\,\,.\,.\,.$$

$$\text{The decimal is formed like this:}$$

. . $$0.0{\bf1}$$
. . $$0.00{\bf1}$$
. . $$0.000{\bf2}$$
. . $$0.0000{\bf3}$$
. . $$0.00000{\bf5}$$
. . $$0.000000{\bf8}$$
. . $$0.000000{\bf{13}}$$
. . $$0.0000000{\bf{21}}$$
. . $$0.00000000{\bf{34}}$$
. . . . . . $$\vdots$$

$$\displaystyle\text{It seems that: }\:\frac{1}{10}\sum^{\infty}_{n=1} \frac{F_n}{10^n} \;=\;\frac{1}{89}$$

. . $$\text{where }F_n\text{ is the }n^{th}\text{ Fibonacci number.}$$

$$\text{Care to prove it?}$$

#### chisigma

##### Well-known member

$$\text{We have: }\:\dfrac{1}{89} \;=\;0.01123595\,\,.\,.\,.$$

$$\text{The decimal is formed like this:}$$

. . $$0.0{\bf1}$$
. . $$0.00{\bf1}$$
. . $$0.000{\bf2}$$
. . $$0.0000{\bf3}$$
. . $$0.00000{\bf5}$$
. . $$0.000000{\bf8}$$
. . $$0.000000{\bf{13}}$$
. . $$0.0000000{\bf{21}}$$
. . $$0.00000000{\bf{34}}$$
. . . . . . $$\vdots$$

$$\displaystyle\text{It seems that: }\:\frac{1}{10}\sum^{\infty}_{n=1} \frac{F_n}{10^n} \;=\;\frac{1}{89}$$

. . $$\text{where }F_n\text{ is the }n^{th}\text{ Fibonacci number.}$$

$$\text{Care to prove it?}$$
The Fibonacci's numbers have been studied for something like 800 years and, among the others 'discoveries' there is the the generating function that can be directly derived from the difference equation $\displaystyle f_{n+2}= f_{n+1}+f_{n},\ f_{0}=0,\ f_{1}=1$...

$\displaystyle g(x)=\sum_{n=1}^{\infty} f_{n}\ x^{n} = \frac{x}{1-x-x^{2}}$ (1)

Setting in (1) $x=\frac{1}{10}$ You have...

$\displaystyle \sum_{n=1}^{\infty} f_{n}\ 10^{- n} = \frac{10}{89}$ (2)

Kind regards

$\chi$ $\sigma$

#### chisigma

##### Well-known member
The Fibonacci's numbers have been studied for something like 800 years and, among the others 'discoveries' there is the the generating function that can be directly derived from the difference equation $\displaystyle f_{n+2}= f_{n+1}+f_{n},\ f_{0}=0,\ f_{1}=1$...

$\displaystyle g(x)=\sum_{n=1}^{\infty} f_{n}\ x^{n} = \frac{x}{1-x-x^{2}}$ (1)

Setting in (1) $x=\frac{1}{10}$ You have...

$\displaystyle \sum_{n=1}^{\infty} f_{n}\ 10^{- n} = \frac{10}{89}$ (2)
... but much more 'elegant' is what You obtain setting in (1) $x=\frac{1}{2}$...

$\displaystyle \sum_{n=1}^{\infty} \frac{f_{n}}{2^{n}} = 2$

$\chi$ $\sigma$