How can I substitute here?

I don't understand how to find the answer to this. Any help will be appreciated!

What is the value of $x$ if

$x= \sqrt{1+2\sqrt{1+4\sqrt{1+16\sqrt{1+256\sqrt{1+\cdots}}}}}$ ?

#Algebra

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown	Appears as
`italics` or `_italics_`	italics
`bold` or `__bold__`	bold
- bulleted - list	bulleted list
1. numbered 2. list	numbered list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1 paragraph 2	paragraph 1 paragraph 2
`[example link](https://brilliant.org)`	example link
`> This is a quote`	This is a quote
# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"	# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

I think that the expression does not converge at all. To show that, we will proof $n\sqrt{1 + n^2\sqrt{1 + n^4\sqrt{...}}} > n^k$ for all natural numbers $n$ and $k$ . Here is the proof by induction on $k$ :

START: k = 1 The values of all square roots in the expression are greater than 1, so we have: $n\sqrt{1 + ... } > n = n^1$

STEP from k to k +1:

We assume $n\sqrt{1 + n^2\sqrt{1 + n^4\sqrt{...}}} > 2^k$ for all $n$ and a particular $k$ .

This also applies to $n^2$ , so we have: $n^2\sqrt{1 + n^4\sqrt{...}} > (n^2)^k = n^{2k}$

We conclude: $n\sqrt{1 + n^2\sqrt{1 + n^4\sqrt{...}}} > n\sqrt{1 + n^{2k}} > n\sqrt{n^{2k}} = n \times n^k = n^{k+1}$

The start of induction and the step of induction show the assumption for all $k$ .

We can apply this statement for $n = 2$ :

$2\sqrt{1 + 4\sqrt{1 + 16\sqrt{...}}} > 2^k$

This will exceed any real number for large $k$ and does not converge. Therefore, the whole expression does not converge, the value of $x$ is undefined.

Finnley Paolella - 1 year ago

Thanks a lot @Finnley Paolella! Just asking of curiosity, do you know of any expressions like this(I mean continuing infinitely0), which look like they won't converge, but they do?