# Why the limit doesn't exist?

1. Dec 10, 2007

### al-mahed

Hi,

I would like some help in this problem:

given the fibonacci sequence 1,2,3,5,8,13....

we know that a_n/a_(n-1) --> phi when n --> $$\infty$$

I have tried a new proof in these terms:

My goal is to prove that when n --> $$\infty$$, a_n/a_(n-1) = a_(n+1)/a_n in the limit

I have notice that a_n^2= a_(n-1)*a_(n+1) $$\pm$$ 1 for some values

Dividing both sides of the equation by a_n*a_(n-1)

$$\frac{a_n}{a_n_-_1}$$ = $$\frac{a_n_+_1}{a_n}$$$$\pm$$ $$\frac{1}{a_n*a_n_-_1}$$

I want to find the limits, because the second term in the right side of the equation --> 0 when n --> $$\infty$$

But I don't know if the equation above is valid for all n, so I'll prove this by induction as follows:

supose that the expression above is valid for some n

by definition a_(n+1)=a_n + a_(n-1)

a_n^2= a_(n-1)*[a_n + a_(n-1)] $$\pm$$ 1 ==> a_n^2 - a_(n-1)^2= a_(n-1)*a_n $$\pm$$ 1 ==>

==> [a_n + a_(n-1)]*[a_n - a_(n-1)] = a_(n-1)*a_n $$\pm$$ 1

by definition a_n - a_(n-1)= a_(n-2) ==>

==> [a_n - a_(n-1)]* a_(n-2) = a_(n-1)*a_n $$\pm$$ 1

by definition a_n = a_(n-1) + a_(n-2) ==>

==> [a_n - a_(n-1)]* a_(n-2) = a_(n-1)*[a_(n-1) + a_(n-2)] $$\pm$$ 1 ==>

==> a_(n-2)*a_n + a_(n-1)*a_(n-2) = a_(n-1)^2 + a_(n-1)*a_(n-2) $$\pm$$ 1 ==>

==> a_(n-1)^2 = a_(n-2)*a_n $$\pm$$ 1

compare the two expressions

a_n^2= a_(n-1)*a_(n+1) $$\pm$$ 1

a_(n-1)^2 = a_(n-2)*a_n $$\pm$$ 1

most generally we have

$$\{a}{_i}{^2}$$} = $$\{a_i_-_1*a_i_+_1}$$} $$\pm$$ 1, with i=2,3,4,5,6...,n

proving by induction (sorry about the english and notation)

Hence,

Lim$$\{_n_-_>_i_n_f_i_n_i_t_y}$$$$\frac{a_n}{a_n_-_1}$$} = Lim$$\{_n_-_>_i_n_f_i_n_i_t_y}$$$$\frac{a_n_+_1}{a_n}$$$$\pm$$ $$\frac{1}{a_n*a_n_-_1}$$}

Lim$$\{_n_-_>_i_n_f_i_n_i_t_y}$$$$\pm$$ $$\frac{1}{a_n*a_n_-_1}$$} = 0

in the limit $$\frac{a_n}{a_n_-_1}$$ = $$\frac{a_n_+_1}{a_n}$$

Why this do not prove the convergence?

2. Dec 10, 2007

### BoTemp

I'm not sure what you're trying to prove. When you say convergence, are you trying to prove that in the $$\lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n}$$ = 1?

That would imply that the sequence of numbers goes to some final value, although the sum of those numbers would still diverge since that number isn't 0.

The sequence grows uniformly, a_n+1 > a_n, so a_n+1/a_n > 1 always. What you've done is prove that

$$\lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n}$$ = constant

With further math you'll be able to solve for that constant as phi, the golden ratio. You can set up a quadratic equation, solve for it, and get phi.

Starter:
$$a_{n+1} = a_n + a_{n-1}$$ divide both sides by a_n

$$\frac{ a_{n+1} }{a_n} = 1 + \frac{ a_{n-1} }{a_n} }$$

Take the limit as n goes to infinity, define
$$\lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n} = x$$, and go.

3. Dec 11, 2007

### al-mahed

BoTemp, I am not so good in english. What i'm trying to prove is that the sequence of the terms a_n/a_(n-1), ... is convergent to a value, not the sum of these ratios (not series, like 1 + 1/2 + 1/3 + ...).

I know the formulas to find the golden ratio (1 + 5^1/2)/2, I would like to know if that proof is correct, if I could conclude that there is a limit or not.

thanks

4. Dec 11, 2007

### HallsofIvy

"My goal is to prove that when n --> , a_n/a_(n-1) = a_(n+1)/a_n in the limit"

That's simply adjusting the indices. Let j= n-1. Then an-1= aj and an= aj+1. The sequence an/an-1 is exactly the same as aj+1/aj which is, in turn, exactly the same as an+1/an just labled differently. In the limit as n goes to infinity, j also goes to infinity and the two sequences have the same limit.

5. Dec 11, 2007

### al-mahed

a_n/a_(n-1) $$\neq$$ a_(n+1)/a_n for all n that you can test by yourself, is quite different to say
$$\lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n}$$ = $$\frac{a_n}{a_{n-1}}$$

6. Dec 11, 2007

### EnumaElish

I may be misunderstanding, but the limit value cannot include subscripts "n" or "n-1."

7. Dec 11, 2007

### al-mahed

what you mean EnumaElish?

8. Dec 11, 2007

### EnumaElish

$$\lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n} = \frac{a_n}{a_{n-1}}$$

the right-hand side cannot have a_n or a_{n-1}. The limit must be independent of n.

Did you mean $\lim_{n \rightarrow \infty} \left({a_{n+1}}\left/{a_n}\right.\right) = \lim_{n \rightarrow \infty} \left({a_n}\left/{a_{n-1}}\right.\right)$? [1]

If so, [1] does not prove convergence because it presumes that a limit exists in the first place. If there is no limit then [1] has no meaning.

Last edited: Dec 11, 2007
9. Dec 11, 2007

### al-mahed

I meant [1] as you said, sorry my mistake.

I see what you mean, my dificult is:

if [1] does not prove the convergence, why the procedure below do?

$$\frac{ a_{n+1} }{a_n} = 1 + \frac{ a_{n-1} }{a_n} }$$ ==>

==> phi = 1 + 1/phi ==> phi = (1 + 5^1/2)/2

Here I'm suposing that there is a limit like the other solution, right?

10. Dec 11, 2007

### EnumaElish

Since a(n) > 0 for all n > 0 and a(n+1) = a(n) + a(n-1), {a(n)} is an increasing sequence for all n > 1.

Let r(n+1) = a(n+1)/a(n), then r(n+1) > 1 for all n > 1.

Suppose lim r(n) = infinity. Then lim 1/r(n) = 0. But since lim r(n+1) = 1 + lim 1/r(n), this implies infinity = 1, a contradiction. Therefore I can conclude that lim r(n) = L < infinity.

Then you can solve L = 1 + 1/L.

11. Dec 11, 2007

### al-mahed

"Therefore I can conclude that lim r(n) = L < infinity."

as 0 < a(n-1)/a_n < 1 ==> 1 < a(n-1)/a_n + 1 < 2 ==> 1 < a(n+1)/a_n < 2

and if the values oscillate between 1 and 2? you prove that the limit do not diverges to infinity, but this not implies that there is a limit, am I right?

f(x) = sen(x), lim sen(x) is not infinite when x --> infinity, but there is no limit, the function oscillate between -1 and 1

12. Dec 11, 2007

### EnumaElish

Yes, you are right; which makes it clear that one cannot presume a limit exists.

13. Dec 12, 2007

### al-mahed

I see, you're right...

We could prove convergence starting with this fact: 1 < a(n+1)/a_n < 2 , showing (I presume) that the sequence is bounded

to prove that it is monotonic, we should work with $$\frac{a_n}{a_n_-_1}$$ = $$\frac{a_n_+_1}{a_n}$$$$\pm$$ $$\frac{1}{a_n*a_n_-_1}$$

for n even $$\frac{a_n}{a_n_-_1}$$ = $$\frac{a_n_+_1}{a_n}$$ - $$\frac{1}{a_n*a_n_-_1}$$

for n odd $$\frac{a_n}{a_n_-_1}$$ = $$\frac{a_n_+_1}{a_n}$$ + $$\frac{1}{a_n*a_n_-_1}$$

a_n < a_(n+1) < a_(n+2) < a_(n+3) < ... ==> $$\frac{1}{a_n*a_n_-_1}$$ --> 0 ==> the sequence $$\frac{a_n}{a_n_-_1}$$ increase in monotonic fashion, bounded above for n odd and decrease in monotonic fashion, bounded below for n even