TITLE: Wrong reasoning yields $1=0$
QUESTION [5 upvotes]: I was thinking about this sequence
\begin{equation}\tag{1}
n=\sqrt{n^2-n+\sqrt{n^2-n+\sqrt{n^2-n+\sqrt{n^2-n+\sqrt{\ldots}}}}}
\end{equation}
It is well known that this converges to n. For example:
$$3=\sqrt{9}=\sqrt{6+3}=\sqrt{6+\sqrt{9}}=\sqrt{6+\sqrt{6+3}}=\sqrt{6+\sqrt{6+\sqrt{6+\ldots}}}$$
The problem is when $n=0$ and $n=1$, because when we apply formula $(1)$, in both cases we get 
$$\sqrt{0+\sqrt{0+\sqrt{0+\sqrt{0+\sqrt{\ldots}}}}}$$
This yields
$$0=1$$
Can you tell me where the mistake is in the reasoning?

REPLY [5 votes]: Think about how you would define the expression
\begin{equation}\tag{1}
\sqrt{n^2-n+\sqrt{n^2-n+\sqrt{n^2-n+\sqrt{n^2-n+\sqrt{\ldots}}}}}
\end{equation}
As others have already mentioned, you might use a sequence with recursion
$a_{k+1} = \sqrt{n^2-n+a_k}$. However this recursion formula by itself does not define a sequence. Rather, to define a sequence you must in addition give a recursion start $a_0$.
Now it is easy to see that if you use $a_0=n$, then for any $k>0$, $a_k=n$ as well, and thus the sequence trivially converges to $n$. However note that the expression $(1)$ does not imply a specific $a_0$. And that is the root of the problem.
Now consider specifically the case $n=1$. In this case, the recursion rule simplifies to $a_{k+1}=\sqrt{a_k}$.
Now consider $a_0=1$. Then it is easily seen that $a_k=1$ for all $k$. Thus the sequence converges to $1$.
However consider $a_0=0$. Then it is just as easily seen that $a_k=0$ for all $k$, and therefore the sequence converges to $0$.
Therefore the expression $(1)$ is not well defined unless you explicitly provide a rule how to choose $a_0$ (or provide another explicit way to interpret that expression; the point being that whatever you choose as meaning must be completely specified).