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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?

Larry
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2 Answers2

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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 $n$:

$$\log_2 a = \dfrac{1}{n-1}$$

$$n = 1+\log_a 2 = \log_a(2a)$$

SlipEternal
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  • This starts by assuming the implied limit of the expression $\sqrt[n]{a + \sqrt[n]{a + \dots}}$ exists, so it's not any more of a justification than the argument in the question. (In fact, I don't see the difference between your argument and Larry's.) – Misha Lavrov Sep 04 '18 at 21:40
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    You are correct. I did not think it needed justification. The OP did not seem familiar with Banach Contractions, so I was attempting to avoid the issue. I merely wanted to show that there are values for $n$ where $a$ does not exist. We obviously also cannot have $n=0$, which I did not state. I was trying to keep things simple. While you did not state it in your proof, you basically showed that the function $f(x)$ that you defined is a Banach Contraction, and therefore it has a fixed point. – SlipEternal Sep 05 '18 at 13:34
  • Right, I do not know what Banach Contractions is. – Larry Sep 16 '18 at 21:49
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    Right. I didn't want to go into it because the existing accepted answer (while correct) leaves out a lot of things. For example, it is important that the sequence start with $x_0=0$. While a similar proof can be used for any starting point, it would not be the exact same proof. For instance if $x_0\ge 2^{\tfrac{1}{n-1}}$ you would need a slightly different induction. Either the sequence is constant or strictly decreasing. The latter would be far more difficult to show. – SlipEternal Sep 17 '18 at 11:35
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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 $x=a$.

Misha Lavrov
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