Wrong reasoning yields $1=0$

Question

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?

Answer 1

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 $n < 1/2$.

Answer 2

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.

Answer 3

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 $x=1-n$.

Answer 4

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 $L = 1 - n$.

Answer 5

• 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.

Answer 6

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).