Prove that the following sequence is convergent and find the limit:
$$\sqrt{7}, \sqrt{7-\sqrt{7}}, \sqrt{7-\sqrt{7+\sqrt{7}}}, ... $$
with $x_{n+2}=\sqrt{7-\sqrt{7+x_{n}}}$.
Notice that $$2=\sqrt{7-3}=\sqrt{7-\sqrt{7+2}}=\sqrt{7-\sqrt{7+\sqrt{7-\sqrt{7+2}}}} $$
$$=\sqrt{7-\sqrt{7+\sqrt{7-\sqrt{7+\sqrt{7-\sqrt{7+...}}}}}}$$
Q: Is this proof valid? If not why? Since this is a contest problem and the trick is well known (but not at that time i guess)
I already know the rigorous solution, but this is not my Q.
The problem is to find the limit of the sequence $$x_0=\sqrt{7},x_1=\sqrt{7-\sqrt{7}},x_{n+2}=\sqrt{7-\sqrt{7+x_{n}}}$$
You have a different sequence: $$x_0=2,x_1=\sqrt{7-\sqrt{7+2}},x_{n+2}=\sqrt{7-\sqrt{7+x_{n}}}$$ You have a good argument that this sequence is constant, so its limit is $2$.
Your sequence has the same recurrence relation as the given sequence. But why should it necessarily have the same limit? Maybe you can make an additional argument that these two sequences do have the same limiting behavior, but I think that is missing so far.
Also this last "expression" that you have, $\sqrt{7-\sqrt{7+\sqrt{7-\sqrt{7+\sqrt{7-\sqrt{7+...}}}}}}$ is meaningless by itself. In context of your answer, it is the limit of the second sequence above. But the standard interpretation of it would be the limit of the sequence from that Putnam problem. So we can also say there is a problem with giving that expression two meanings and taking them to mean the same thing.