# SMO problem: Sequence and series.

Problem on Series and Sequences (SMO Test):

For each positive integer $$n \ge 1$$ , we define the recursive relation given by $$a_{n+1}=\cfrac{1}{1+a_n}$$.

Suppose that $$a_1=a_{2012}$$.Find the sum of the squares of all possible values of $$a_1$$.

How to do this? I do not know how to approach it at all. What does recursive relation mean?

• recursive relation means each term (except the initial) is a function of the preceding terms Commented Mar 7, 2021 at 12:52
• Oh so like: $a_{2012} = \cfrac {1}{1+a_{2011}}$ right, that's the function?
– user880107
Commented Mar 7, 2021 at 12:55
• Yes, $a_{2012}$ is a function of $a_{2011}.$ Do you know how to show this converges and/or what the possible limits are? Commented Mar 7, 2021 at 12:56
• No:( I am on an elementary level, I almost know nothing about limits.
– user880107
Commented Mar 7, 2021 at 12:58

Hint :

Lets calculate first few terms.

$$a_2=\frac{1}{1+a_1}$$ $$a_3=\frac{1}{1+a_2}=\frac{1+a_1}{2+a_1}$$ $$a_4=\frac{1}{1+a_3}=\frac{2+a_1}{3+2a_1}$$ $$a_5=\frac{1}{1+a_4}=\frac{3+2a_1}{5+3a_1}$$ $$a_6=\frac{1}{1+a_5}=\frac{5+3a_1}{8+5a_1}$$

It can be shown by induction that $$a_n=\frac{F_n+F_{n-1}\, a_1}{F_{n+1}+F_n \, a_1} \tag{n \ge 2}$$

where $$F_n=0,1,1,2,3,5,8,13,\ldots$$ are Fibonacci numbers.

Then $$a_{2012}=a_1$$ is a quadratic in $$a_1$$ and using Vieta's, sum of squares of its roots can be found.

$$A_{n+1}=\frac{1}{1+A_n}\implies A_{n+1} A_n+A_{n+1}=1~~~(1)$$ Let $$A_n=\frac{B_{n-1}}{B_{n}}$$, in (1) then $$B_{n+1}=B_n+B_{n-1},$$ which is like Fibonacci sequence. Next, let $$B_n =x^n$$ to get $$x=1+x^{-1} \implies x=(1\pm\sqrt{5}) /2=a,b$$. So $$B_n= P a^n +Q b^n \implies A_n=\frac{a^{n-1}+R~ b^{n-1}}{a^{n}+R ~ b^{n}}.$$ The constant $$R$$ can be determined by a given condition.