Prove the following sequence converges to 2:

Question

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?

Answer 1

Verify by induction that $2 <a_n <3$. Then conclude that $a_{n+1}=\frac {a_n^{2}+6} 5 <a_n$ (using the fact that $(a_n-2)(a_n-3) <0$.

Now let $2.5<c<3$. Show by induction that $a_n <c$ for all $n$. It follows that $(a_n)$ (being monotone and bounded) is convergent and the limit $t$ is strictly less than $3$. But $5t=t^{2}+6$ so $t=2$ or $t=3$. This proves that $t=2$.

Answer 2

(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 $2$.