Q:

What is the limit of $a_{n+1}=\sqrt{7-(-1)^na_n}$?

Find the limit of the following recurrence relation:
  $$a_{n+1}=\sqrt{7-(-1)^na_n}, n\geq 0$$ with $a_0=0$.

I have thought that we can transform the relation to the following: $$a_{n+1}^2+(-1)^na_n=7$$ but I cannot take it further!!! Any help? (I also don't know how to prove the convergence of such a sequence)
Edit 1: I think that it is not a duplicate, because my relation has also the $(-1)^n$ part.

A:

One can show that the limit does not exist by contradiction. Suppose $\lim_na_n$ exists. Then we must have that $x=:\lim_na_{2n}$ and $y:=\lim_na_{2n+1}$ both exists. We show that $x\neq y$ and thus have a contradiction. 
By the recursive relation, one has
$$
a_{2n+1}=\sqrt{7-\sqrt{7+a_{2n-1}}},\quad
a_{2n}=\sqrt{7+\sqrt{7-a_{2n-2}}}
$$
which implies by taking the limit and the continuity of square root that
$$
y=\sqrt{7-\sqrt{7+y}}\qquad(1)\\
x=\sqrt{7+\sqrt{7-x}}\qquad(2)
$$
One can check that solutions to (1) and (2) can never be the same:

