Fibonacci limit

1. Jan 24, 2007

tehno

Let:
$$a_{1}=a_{2}=1;a_{n+2}=a_{n+1}+a_{n};n\geq 1$$

Let $f_{n}$ be the last digit in decimal notation
of Fibonacci number $a_{n}$.
Find:

$$\lim_{n\to\infty}\frac{a_{1}+a_{2}+...+a_{n}}{n}$$

Last edited: Jan 24, 2007
2. Jan 24, 2007

chuckd1356

Can you explain?

3. Jan 24, 2007

mathman

What does fn have to do with anything???

4. Jan 24, 2007

chuckd1356

By theory, there is no limit to Fibonacci, unless I'm mistaken.

The sequence wouldn't be a sequence if there was a limit.

5. Jan 24, 2007

HallsofIvy

Staff Emeritus
That doesn't quite make sense. A fair part of Calculus courses is devoted to limits of sequences! Of course, the Fibonacci sequence is increasing without upperbound so it has no limit. But the question is about the nth partial sum divided by n. That's a whole different matter.

6. Jan 24, 2007

matt grime

What the hell is anyone talking about here?

7. Jan 24, 2007

chuckd1356

That's what I was getting at, thanks for clarifying!

8. Jan 24, 2007

CRGreathouse

The question is about the final digits, which are periodic with period 60. The sum of the 60 values is ***, so the average value at the limit is ***/60.

(It's not hard to calculate this, so I left it as an exercise. I can check it if you think you have an answer.)

9. Jan 25, 2007

tehno

correction (+ solution)

Let:
$$a_{1}=a_{2}=1;a_{n+2}=a_{n+1}+a_{n};n\geq 1$$

Let $f_{n}$ be the last digit in decimal notation
of Fibonacci number $a_{n}$.

Find:

$$\lim_{n\to\infty}\frac{f_{1}+f_{2}+...+f_{n}}{n}$$

My apology for the confusion I made.

EDIT:
Yes the key for the solution is "***/60".
IOW ,$f_{1}=f_{61},f_{2}=..etc.$
I get:
$$\lim_{n\to\infty}\frac{f_{1}+f_{2}+...+f_{n}}{n}=\frac{14}{3}$$

Last edited: Jan 25, 2007
10. Jan 25, 2007

CRGreathouse

Yes, 14/3 is right.