Wrong reasoning yields $1=0$ 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?
 A: 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).
A: *

*The expression on the right hand side is not well-defined since one can not tell if such object exists or not:
\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}
On has to give a precise definition of the $\sqrt{\cdots}$ part. "Keep going on like this", for instance, is not a mathematical definition. 

*
"That is, well known, converges to n."

Excuse my ignorance, I don't think it is a "well known" fact. And I don't even think it is a correct mathematical statement. Again, one needs to define "converges" carefully. 

*
For example:
  $$3=\sqrt{9}=\sqrt{6+3}=\sqrt{6+\sqrt{9}}=\sqrt{6+\sqrt{6+3}}=\sqrt{6+\sqrt{6+\sqrt{6+\ldots}}}$$

"For example" is not a proof. 
A: If we look at $(1)$, say we set
$$x=\sqrt{n^2-n+\sqrt{n^2-n+\sqrt{n^2-n+\sqrt{\dots}}}}$$
Because this goes on forever, we clearly have that
$$x^2-(n^2-n)=x$$
Solving this using the quadratic formula yields $x=1-n$ or $x=n$. At this point, we simply choose whatever makes since (i.e. if $n>1$, then $x$ is clearly positive so we take $x=n$). In our case of $n=1$, it makes sense to have $x=1-n$.
A: We can write $S_{k+1} = \sqrt{n^2 - n + S_k}$. 
Assuming the sequence converges to some $L$ (which you can show since it's increasing and bounded by $n$) you can solve to get 
$$L^2 - L - (n - n^2) = 0,
$$
 which gives solutions of $L = n$ or $L = 1 - n$.
A: If the nested square root has any meaning at all, it is as the limit of a sequence of finitely nested expressions.  You need to specify which ones.
For example, you might specify some initial function $a_0(n)$, and let
$a_{j+1}(n) = \sqrt{n^2 - n + a_{j}(n)}$.  You then want to prove something about the limit of $a_j(n)$ as $j \to \infty$.  It's easy to see that if the 
limit exists it must be $n$ or $1-n$, these being possible only if they are nonnegative (if we want everything real not allowing negative square roots), because the function $f(t) = \sqrt{n^2-n+t}$ has only those fixed points, but that doesn't say that the limit exists.
Note also that $f'(n) = 1/(2 |n|)$, while $f'(1-n) = 1/(2 |1-n|)$, so the fixed point $n$ is stable if $n > 1/2$ while $1-n$ is stable if $n < 1/2$. 
A: Formulae $1$ naturally does not converge to $1$. If a formula works for almost all $n$, it need not work for all $n$, right? So it does not make sense  to apply the formula  at $n=1$, although it converges everywhere else.
To give another example, suppose that I am given this  sequence:$$
\sqrt{n^2 - bn + b\sqrt{n^2-bn + b\sqrt{n^2-bn + b\sqrt{n^2-bn + \ldots}}}}
$$
Then, it is easy to see that this sequence converges to $n$ for almost all $n$, since we just have $n = \sqrt{n^2-bn+bn}$. However, at $n=b$, we see that this just simplifies to $0$. Now, we would like to say that $b=0$, but then this is not true, is it? (We could have chosen $b$ randomly, you did it for $b=1$).
Hence, the up shot of the whole thing is that the formula doesn't work in certain cases, although it may work in comprehensive generality.  
