Finding the Limit of a Fibonacci Series

[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:

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

 

[chuckd1356]

Can you explain?

 

[mathman]

What does f_n have to do with anything?

 

[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.

 

[HallsofIvy]

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

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

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 n_th partial sum divided by n. That's a whole different matter.

 

[matt grime]

What the hell is anyone talking about here?

 

[chuckd1356]

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 n_th partial sum divided by n. That's a whole different matter.

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

 

[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.)

 

[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:
$$\lim_{n\to\infty}\frac{f_{1}+f_{2}+...+f_{n}}{n}=\frac{14}{3}$$

 

[CRGreathouse]

Yes, 14/3 is right.