Find the value of "n" in terms of "a" when $\sqrt[n]{a+{\sqrt[n]{a+{\sqrt[n]{a+{\sqrt[n]{a+\cdots}}}}}}} = a$ I saw on the internet that $\sqrt{2+{\sqrt{2+{\sqrt{2+{\sqrt{2+\cdots}}}}}}} = 2$
I want to find a solution for n in terms of "a" for a general formula 
$\sqrt[n]{a+{\sqrt[n]{a+{\sqrt[n]{a+{\sqrt[n]{a+\cdots}}}}}}} = a$ , $a\in\mathbb{R_{>0}}$
Here is my approach: 
assume $\sqrt[n]{a+{\sqrt[n]{a+{\sqrt[n]{a+{\sqrt[n]{a+\cdots}}}}}}} = a$ , 
then it follows that $\sqrt[n]{a+a} = a$.
Solve the equation, and we can get $n = \log_{a}{(2a)}$
The result will be that for $n = \log_{a}{(2a)}$ , $\sqrt[n]{a+{\sqrt[n]{a+{\sqrt[n]{a+{\sqrt[n]{a+\cdots}}}}}}} = a$ , $a\in\mathbb{R_{>0}}$
I know it is not formal to prove something based on an assumption, but I think the assumption is correct. Can people help me justify the proof? 
 A: You have $a = \sqrt[n]{a+\cdots}$
Raise both sides to the $n$-th power and subtract $a$:
$$a^n-a = a$$
$$a^n-2a = 0$$
$$a(a^{n-1}-2) = 0$$
$$a=0, a=2^{\tfrac{1}{n-1}}$$
The latter result exists for any $n\neq 1$.
Now, you can solve for $n$:
$$\log_2 a = \dfrac{1}{n-1}$$
$$n = 1+\log_a 2 = \log_a(2a)$$
A: So you have shown that if there exists such an $n$, then we must have $n = \log_a (2a)$.
To complete the argument, we need to justify that:


*

*When we take $n = \log_a(2a)$, the expression is well-defined; that is, the limit of the sequence $\sqrt[n]{a}, \sqrt[n]{a + \sqrt[n]{a}}, \sqrt[n]{a + \sqrt[n]{a + \sqrt[n]{a}}}, \dots$ exists.

*The limit actually does equal $a$.


I will assume $a>1$. It appears that both of the statements above are true for $a \in (0,1]$ as well, but things get messier, especially if $a \in (\frac12,1)$ and $n$ is negative.
For the first point, it's enough to show that the sequence is monotone and bounded. Let $x_k$ be the $k^{\text{th}}$ term of the sequence, so that $x_k = \sqrt[n]{a + x_{k-1}}$, with $x_0 = 0$. 
Then $f(x) = \sqrt[n]{a+x}$ has derivative $f'(x) = \frac{(a+x)^{\frac1n - 1}}{n}$ which is positive for $x \ge 0$, so it's increasing for positive $x$. So we can proceed by induction: $x_0 < x_1$ because $\sqrt[n]{a}>0$, and if $x_{k-1} < x_k$, then $f(x_{k-1}) < f(x_k)$, so $x_k < x_{k+1}$.
Moreover, we have $x_k < a$ for all $k$. This is, again, true by induction. We start with $x_0 = 0 < a$, and if $x_k < a$, then $x_{k+1} = \sqrt[n]{a + x_k} < \sqrt[n]{a+a} = (2a)^{\log_{2a}a} = a.$
So now we know that there is a limit $x$. We must have $f(x) = \sqrt[n]{a+x} = x$. In particular, $x \ge a$, because above we showed that if $x < a$, then $f(x) < a$, which contradicts $f(x) = x$. But because $x_k < a$ for all $a$, we must also have the limit $x$ satisfy $x \le a$. So we conclude that $x=a$.
