What is the value of the infinite nested radical $\sqrt{0+\sqrt{0+\sqrt{0+\sqrt{\cdots}}}}$?

Question

The value is usually taken to be the limit of the partial sums as the number of terms increases beyond limit. In this case each partial sum is trivially zero, so the value of the infinite nested radical is to be taken zero. However, if we let the value to be $x$, then $x^2 = 0+ x$, which implies $x=1$. So, what is the value, $0$ or $1$? And why should we prefer $0$ but not $1$?

Answer 1

Let $x_{n+1}=\sqrt{0+x_n}.$

You easily get $$x_n=x_0^{1/2^n}$$

The limit of this expression is $1$ when $x_0>0.$

But the partial expressions here have $x_0=0,$ the only case that matters.

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More generally, if $x_{n+1}=f(x_n)$ with $f$ a continuous function, then if $x_n\to x$ then $f(x)=x.$ But there might be multiple solutions to $f(x)=x.$ Which solution. Whether you get a limit, and if so, which limit, might depend on $x_0.$

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If $a>0,$ the expression:

$$x=\sqrt{a+\sqrt{a+\dots}}$$ gives two solutions, too:

$$x^2-a=x\\x=\frac{1\pm \sqrt{1+4a}}2$$

The reason for two solutions when $a>0$ is that squaring both sides introduces the possibility that $x<0.$

So here, the sequence $x_{n+1}=\sqrt{a+x_n}$ always converges to the positive root, if it converges.