TITLE: Prove the following sequence converges to 2:
QUESTION [1 upvotes]: I've been tasked a question in a worksheet that asks:

(c) Deduce that $(a_n), n\in N$ converges and determine the limit
  for the sequence $$5a_n+1=a_n^2+6$$ with $a_1=5/2$.

I've worked out that the series is strictly decreasing and $2<a_n<3$ hence the limit I'm looking for is clearly $2$ however I was wondering how I could deduce this? 
Would I use the epsilon definition (if so how would I go about doing so as I have no input equation for $a_n$?) or is there a theorem that implies that there is a strictly decreasing bounded sequence has a limit of the lower bound?

REPLY [1 votes]: (I am assuming you meant $5a_{n+1}$ on the left side of your equation.) After having figured out that it converges you can usually find the limit by rewriting it like this:
$5a_\infty=a_\infty ^2+6$ and solve for $a_\infty$. One of the solutions is $2$.
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