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.

  • $\begingroup$ Why has this question been downvoted? The OP has clearly shown their efforts. $\endgroup$ – Toby Mak Mar 9 '19 at 8:05
  • 1
    $\begingroup$ 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. $\endgroup$ – Aidyn Mar 9 '19 at 8:11

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 $a_{k+1}$.


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