# Prove that the given sequence converges

Prove that the given sequence $$\{a_n\}$$ converges:

$$a_1 > 0, a_2 > 0$$

$$a_{n+1} = \frac{2}{a_n + a_{n-1}}$$ for $$n \geq 2$$

As I observed, this sequence does not seem to be monotonic and that it could be bounded since the values of $$a_1$$ and $$a_2$$ are arbitrary positive numbers.

If the limit of the sequence existed, it would be equal to 1 by letting the limit of $$a_n$$ be x as n goes to infinity, and solving the equation x = $$\frac{2}{x + x}$$ => x = 1 or -1, from which we choose x = 1 since x must be positive.

The only idea that came to my mind is bounding the sequence using two other sequences that could be shown to converge to 1 (Let these sequences be $$b_n$$ and $$c_n$$):

$$b_n <= a_n <= c_n$$

If we could find such sequences,and prove that they converge to 1, the problem would be solved. So, I tried to bound the sequence from both sides, and try to show that the limits are equal to 1, but failed to find such sequences. I found that it is a little difficult to analyze sequences of the form presented in the problem since the sequence fluctuates a lot.

I am not sure how to start off, any ideas or tricks for such problems would be appreciated.

• Why has this question been downvoted? The OP has clearly shown their efforts. – Toby Mak Mar 9 '19 at 8:05
• This was my mistake. This is my first time posting a question here, I did not specify details of my effort, and edited it afterwards. – Aidyn Mar 9 '19 at 8:11

## 1 Answer

We can use the fact that for any $$x>1$$, if $$\frac{1}{x}\leq a,b \leq x$$ for some $$a\neq b$$, then $$\frac{1}{x}< \frac{2}{a+b} < x$$. This means that subsequent terms of the sequence will also be within the bound given by $$\frac{1}{x}$$ and $$x$$.

Now, we need to show that this boundary keeps shrinking as the sequence progresses. This can be argued as follows: For any two consecutive terms, $$a_k$$ and $$a_{k+1}$$ in the sequence, this bound will be given by either the pair $$a_k,\frac{1}{a_k}$$ or $$a_{k+1},\frac{1}{a_{k+1}}$$ (whichever gives the larger range). For the next two terms, the bound will similarly be given by either $$a_{k+2},\frac{1}{a_{k+2}}$$ or $$a_{k+3},\frac{1}{a_{k+3}}$$. However, from the inequality above, both $$a_{k+2}$$ and $$a_{k+3}$$ fall within the boundary determined by the previous two terms ($$a_k$$ and $$a_{k+1}$$), which means the new range will be smaller than that for $$a_k$$ and $$a_{k+1}$$.