$\lim_{n\rightarrow \infty} a_n$ for $ a_{n+1} =a_n\cdot \frac{6+a_n}{3+2a_n}$

Question

let there be $a_1=3$,$a_{n+1} =a_n\cdot \frac{6+a_n}{3+2a_n}$
so $a_1=3,a_2=3,etc...$ let assume there is a limit L so $L =L\cdot \frac{6+L}{3+2L}$ L=0,3
But that is not a proof, how do I continue from here using induction?

Answer 1

Using proof by induction: Let's assume that for a value of $n>1$, $a_n=3$. If that is the case, then $a_{n+1}=3\frac{6+3}{3+6}=3$. Since we are given $a_1=3$, we can conclude that all $a_n = 3$ and the limit is thus $3$.

You may find the limit with algebra on $L=L\cdot\frac{6+L}{3+2L}$. First step is to divide both sides by $L$. To do so we must assume that $L\neq0$ (which, incidentally, would be the limit if $a_1$ was $0$). Then without much trouble you should come to $L=3$. The proof would be a bit different if $a_1$ was not 3 or 0 as you'd have to prove convergence.