1. ## Fibonacci decimal

Find the value of this infinite decimal:
Code:
      0.01
0.001
0.0002
0.00003
0.000005
0.0000008
0.00000013
0.000000021
+     ...
-----------------
S  =  0.011235955 ...
That is: . $S \;=\;\sum^{\infty}_{n=1} \frac{F_n}{10^{n+1}}$

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

2. Originally Posted by Soroban
Find the value of this infinite decimal:
Code:
      0.01
0.001
0.0002
0.00003
0.000005
0.0000008
0.00000013
0.000000021
+     ...
-----------------
S  =  0.011235955 ...
That is: . $S \;=\;\sum^{\infty}_{n=1} \frac{F_n}{10^{n+1}}$

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

Well,
$F_n = \frac{1}{\sqrt{5}} \left ( \frac{1 + \sqrt{5}}{2} \right ) ^n - \frac{1}{\sqrt{5}} \left ( \frac{1 - \sqrt{5}}{2} \right ) ^n$

So S is the difference of two geometric series:
$S = \frac{1}{10\sqrt{5}}\sum_{n = 1}^{\infty} \left ( \frac{1 + \sqrt{5}}{2 \cdot 10} \right ) ^n - \frac{1}{10\sqrt{5}}\sum_{n = 1}^{\infty} \left ( \frac{1 - \sqrt{5}}{2 \cdot 10} \right ) ^n$

I have to get to dinner, so I'll let someone else finish. It's trivial from here anyway.

-Dan

3. Hey Soroban. I think I may have this one. Let's see. I encountered this proof as a homework assignment back in college. In a Seminars in Mathematics course.

[tex]\frac{1}{10}\sum_{n=1}^{\infty}\frac{F_{n}}{10^{n} }[\math]

Let $d=0.11235955056........=\frac{F_{1}}{10}+\frac{F_{ 2}}{10^{2}}+\frac{F_{3}}{10^{3}}+.........$

Now use $10d+d$ and we have:

$11d=F_{1}+\frac{F_{2}+F_{1}}{10}+\frac{F_{3}+F_{2} }{10}+\frac{F_{4}+F_{3}}{10}+....$

$=1+\frac{F_{3}}{10}+\frac{F_{4}}{10^{2}}+\frac{F_{ 5}}{10^{3}}+..........$

$=1+100d-10F_{1}-F_{2}$

$=1+100d-10-1$

$=100d-10$

So, we have:

$11d=100d-10$

$89d=10$

$d=\frac{10}{89}$

Don't forget to multiply by our 1/10 and get:

$\frac{1}{89}$

4. Welcome back, you were away for a long time!

5. Thanks PH. Yeah, it has been a while. I see you're up to 6200. Very prolific

How's the academic world treating you?.

6. Originally Posted by galactus
How's the academic world treating you?.
I am on vacation now. 2 Months to go.

7. Quote

/Quote

I'm pretty sure that F(4)/10 should be F(4)/100, etc.

8. You're so right, G. Thanks for the catch. I fix.

9. Originally Posted by topsquark
Well,
$F_n = \frac{1}{\sqrt{5}} \left ( \frac{1 + \sqrt{5}}{2} \right ) ^n - \frac{1}{\sqrt{5}} \left ( \frac{1 - \sqrt{5}}{2} \right ) ^n$

So S is the difference of two geometric series:
$S = \frac{1}{10\sqrt{5}}\sum_{n = 1}^{\infty} \left ( \frac{1 + \sqrt{5}}{2 \cdot 10} \right ) ^n - \frac{1}{10\sqrt{5}}\sum_{n = 1}^{\infty} \left ( \frac{1 - \sqrt{5}}{2 \cdot 10} \right ) ^n$

I have to get to dinner, so I'll let someone else finish. It's trivial from here anyway.

-Dan
Okay, I'll finish.

We have
$\sum_{n = 0}^{\infty}ar^n = a + \sum_{n = 1}^{\infty}ar^n = \frac{a}{1 - r}$

So
$\sum_{n = 1}^{\infty}ar^n =
\frac{a}{1 - r} - a$

Thus
$\sum_{n = 1}^{\infty} \left ( \frac{1 + \sqrt{5}}{2 \cdot 10} \right ) ^n = \frac{1}{1 - \left ( \frac{1 + \sqrt{5}}{2 \cdot 10} \right )} - 1$

and
$\sum_{n = 1}^{\infty} \left ( \frac{1 - \sqrt{5}}{2 \cdot 10} \right ) ^n = \frac{1}{1 - \left ( \frac{1 - \sqrt{5}}{2 \cdot 10} \right )} - 1$

So
$S = \frac{1}{10\sqrt{5}}\sum_{n = 1}^{\infty} \left ( \frac{1 + \sqrt{5}}{2 \cdot 10} \right ) ^n - \frac{1}{10\sqrt{5}}\sum_{n = 1}^{\infty} \left ( \frac{1 - \sqrt{5}}{2 \cdot 10} \right ) ^n$ $= \frac{1}{10\sqrt{5}} \left ( \frac{1}{1 - \left ( \frac{1 + \sqrt{5}}{2 \cdot 10} \right )} - \frac{1}{1 - \left ( \frac{1 - \sqrt{5}}{2 \cdot 10} \right )} \right )$

$S = \frac{1}{10\sqrt{5}} \left ( \frac{20}{19 - \sqrt{5}} - \frac{20}{19 + \sqrt{5}} \right )$

$S = \frac{2}{\sqrt{5}} \left ( \frac{19 + \sqrt{5}}{19^2 - 5} - \frac{19 - \sqrt{5}}{19^2 - 5} \right )$

$S = \frac{2}{\sqrt{5}} \left ( \frac{2 \sqrt{5}}{356} \right )$

$S = \frac{4}{356} = \frac{1}{89}$

$\frac{x}{1-x-x^{2}}\Rightarrow\frac{\frac{1}{10}}{1-(\frac{1}{10})-(\frac{1}{10})^{2}}=\frac{10}{89}$