How to show that the limit of this sequence is $L=4$ (ex.8.11 Mathematical Analysis 2nd ed.- Apostol) I need to show that this sequence has limit $L=4$. I know it could be useful the principle of mathematical induction but I can't understand the way I should use this principle to prove the limit is $L=4$.
$$a_1=2;a_2=8;a_{2n+1}=\frac12(a_{2n}+a_{2n-1}); a_{2n+2}=\frac{a_{2n}\times a_{2n-1}}{a_{2n+1}}$$
I've also calculated the first terms of this sequence
$a_1=2;a_2=8;a_3=5;a_4=3,2;a_5=4,1;a_6=3,9...$
I am interested in understanding how I should think when I face this kind of problems.
Could you help me please? Any hint or help will be welcome.
 A: The starting numbers are a and b.
$$c:=\frac{a+b}{2}$$
$$d:=\frac{2ab}{a+b}$$
are the next two numbers.
So, the product of the two numbers is
cd = ab , so the product does not change in the process.
If L is the limit, we have $L^2 = ab = 16$ , so L = 4.
A: Since $a_{2n}a_{2n-1}=a_2a_1=16$ for every $n$, the sequence $b_n=a_{2n-1}$ solves $b_1=a_1=2$ and $b_{n+1}=\frac12\left(b_{n}+\frac{a_1a_2}{b_n}\right)=\frac12\left(b_n+\frac{16}{b_{n}}\right)$ for every $n\geqslant 1$. 
Let $u:x\mapsto\frac12\left(x+\frac{16}{x}\right)$, then $u(x)\geqslant4$ for every $x\gt0$, $u(4)=4$ and $4\leqslant u(x)\leqslant x$ on $x\geqslant4$, thus $b_n\geqslant4$ for every $n\geqslant2$ and $(b_n)_{n\geqslant2}$ is lower bounded by $4$ and nonincreasing. 
In particular, $b_n\to\ell$ when $n\to\infty$, where $\ell\geqslant4$ solves $u(\ell)=\ell$, that is, $\ell=4$. Finally, $a_{2n}=2b_{n+1}-b_n$ hence $a_{2n}\to2\ell-\ell=\ell$ as well. QED.
For other positive values of $(a_1,a_2)$, replace $16$ by $a_1a_2$ and $\ell=4$ by $\ell=\sqrt{a_1a_2}$, then the result still holds.
