0
$\begingroup$

I found a series that is $a_{n+1}=\frac{a_n^2 + 2}{2a_n}$ yet I'm not sure. can someone give me a more umm solid example? thanks.

$\endgroup$
  • $\begingroup$ Do you mean $a_{n+1}=\frac{a_n^2 + 2}{2a_n}$ or $a_{n+1}=\frac{a_n^2 + 2}{2}a_n$. (Or some of other possible interpretations of your question.) The expressions in my comment can by typeset as $a_{n+1}=\frac{a_n^2 + 2}{2a_n}$ or $a_{n+1}=\frac{a_n^2 + 2}{2}a_n$. $\endgroup$ – Martin Sleziak Dec 17 '14 at 16:18
  • $\begingroup$ the first one, yet I did manage to find a solution, the Fibonacci series that converges to the golden ratio. $\endgroup$ – Marie Dec 17 '14 at 21:34