# A recurrence relation

1. Aug 8, 2011

### asmani

Hi all

Suppose that $a_1=\sqrt5$, $a_{n+1}=a_n^2-2$ and $g_n=\frac{a_1a_2...a_n}{a_{n+1}}$.
Evaluate $\lim_{n\rightarrow \infty } g_n$.

I have seen some information in this link. Besides, the sequence gn seems as a good rational approximation for $\sqrt5$. I know that the answer is 1, But I can't find any nice solution. Any hint is strongly appreciated.

2. Aug 8, 2011

### Mentallic

I have yet to study recurrence relations for myself, but I played around with it for a bit and it seems like it was headed in the right direction.

$$g_n=\frac{a_1a_2...a_n}{a_n^2-2}$$

$$=\frac{a_n^2\left(\frac{a_1a_2...a_{n-1}}{a_n}\right)-2\left(\frac{a_1a_2...a_{n-1}}{a_n}\right)+2\left(\frac{a_1a_2...a_{n-1}}{a_n}\right)}{a_n^2-2}$$

$$=\frac{\left(a_n^2-2\right)\left(\frac{a_1a_2...a_{n-1}}{a_n}\right)+2\left(\frac{a_1a_2...a_{n-1}}{a_n}\right)}{a_n^2-2}$$

$$=\frac{a_1a_2...a_{n-1}}{a_n}+2\frac{a_1a_2...a_{n-1}}{a_na_{n+1}}$$

$$=g_{n-1}+\frac{2}{a_n^2}g_n$$

Hence we can express gn in terms of gn-1, and the factor multiplying gn-1 seems to say a lot about what gn is as n gets very large.

3. Aug 8, 2011

### asmani

Thanks. I can't see what does it say about gn. Can you help me on this?

I found out how to show that:
$$a_n=\frac{\varphi ^{2^n}-\varphi ^{-2^n}}{\varphi ^{2^{n-1}}-\varphi ^{-2^{n-1}}}$$
And then the rest is easy. But I'm still looking for a more elegant solution.

4. Aug 8, 2011

### Mentallic

Hmm, after looking at it a bit more I realize it's really not telling us as much as I originally thought.

After solving for gn in

$$g_n=g_{n-1}+\frac{2}{a^2_n-2}g_n$$ to obtain

$$g_n=\frac{a_n^2}{a_n^2-2}g_{n-1}$$

I was simply arguing that the factor approaches 1 as n gets large, but this doesn't really tell us about the limit of gn...

5. Aug 8, 2011

### tiny-tim

hi asmani!
reverse-engineering that expression, we get

an = sinh(C2n)/sinh(C2n-1) = 2cosh(C2n-1)​

sooo … try defining an = 2coshbn

6. Aug 8, 2011

### asmani

Thanks. I think it's better not to simplify the fraction to get:
$$g_n=a_1a_2\cdots a_n\frac{1}{a_{n+1}}=\frac{\sinh(C2^1) }{\sinh(C2^0)}\frac{\sinh(C2^2) }{\sinh(C2^1)}\cdots \frac{\sinh(C2^n) }{\sinh(C2^{n-1})}\frac{\sinh(C2^n) }{\sinh(C2^{n+1})}$$
$$=\frac{\sinh^2(C2^n) }{\sinh(C2^0)\sinh(C2^{n+1})}$$
Where C=Log(φ).
And now calculating the limit is easy.

7. Aug 8, 2011

### Mentallic

Sorry about wasting your time asmani, I should've left it to the big guys

8. Aug 9, 2011

### asmani

No, I appreciate your participation in this thread.