$$\sqrt{2+\sqrt{2+\sqrt{2+}}}\ldots$$ How you could put this into a summation equation? I'm stuck. At one point thought it was this:

$$ \sum_{n=0}^{x} 2^{\frac{1}{2^x}} $$

but that would just be $2^{\frac{1}{2}}+2^{\frac{1}{4}}+2^{\frac{1}{8}}...0$, which is the value of each two. I thought it would make it work if I used fraction exponents instead of square roots. But I got nothing.

The very thing I want to know is, how do I put this in summation notation? Thanks in advance, and if I need to correct or explain anything further, I will.

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    $\begingroup$ Could you quote the problem in the puzzle book verbatim please? $\endgroup$ – anon Mar 16 '13 at 20:53
  • $\begingroup$ The puzzle book has nothing to do with this, just my own interest inspired by it. Nothing important, I might even remove that if it's confusing $\endgroup$ – Gᴇᴏᴍᴇᴛᴇʀ Mar 16 '13 at 20:55
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    $\begingroup$ Perhaps state a question somewhere. The one you want answered. $\endgroup$ – GEdgar Mar 16 '13 at 20:57
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    $\begingroup$ Do you mean $\sqrt{2+\sqrt{2+\sqrt{2+\ldots}}}$? I'm a bit confused by the ... in your expression. $\endgroup$ – Jared Mar 16 '13 at 20:59
  • $\begingroup$ Yeah sorry, I created it from memory. I'll correct it $\endgroup$ – Gᴇᴏᴍᴇᴛᴇʀ Mar 16 '13 at 21:02

Using your standard double-angle formulae:

$$\cos\dfrac{\pi}{2^m}= \frac{1}{2}\underbrace{\sqrt{2+\sqrt{2+ \sqrt{2+\cdots \sqrt{2}}}}}_m$$

Letting $m\to\infty$ immediately yields the result. If you wanted a series representation of the finite nested radical, we can use $\cos x:$

$$\underbrace{\sqrt{2+\sqrt{2+ \sqrt{2+\cdots \sqrt{2}}}}}_m=2\sum_{n\geq 0}\frac{(-1)^n}{(2n)!}\left(\frac{\pi}{2^m}\right)^{2n}$$

Note that this is an infinite series. I do not believe it is possible to write the expression as a finite sum: a nested radical is not a sum in the standard sense.

  • $\begingroup$ +1 By the way, $\displaystyle \sqrt{2 + \sqrt{2}} = \sqrt{1 + \frac{1}{\sqrt{2}}} + \sqrt{1 - \frac{1}{\sqrt{2}}}$. $\endgroup$ – J.H. Mar 16 '13 at 22:00
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    $\begingroup$ @J.H. Thanks Jack, and nice catch! $\endgroup$ – L. F. Mar 16 '13 at 22:41

Let $$a_{n}=\sqrt{2+\sqrt{2+\ldots+\sqrt{2}}}$$ Where there are $n$ $2$s. Then $a_{n+1}=\sqrt{a_{n}+2}$ and $a_{1}=\sqrt{2}$.

$a_{n}<2$ since if $a_{n}<2$, $a_{n+1}=\sqrt{a_{n}+2}<a_{n}<2$. Therefore $a_{n}$ is monotonic increasing and bounded above by $2$.

It follows $a_{n}$ tends to a limit $L$ which is at most $2$. $L$ must satisfy $L=\sqrt{L+2}$; in other words, $L=2$.


Define a sequence recursively by letting $a_0=\sqrt{2}$ and $a_{n+1}=\sqrt{2+a_n}$ for $n=0,1,2,\ldots$. Then $\sqrt{2+\sqrt{2+\sqrt{2+\ldots}}}=\lim_{n\rightarrow\infty}a_n$ if the limit exists (try to write out the first few terms of the sequence to verify this fact).

Intuitively, if the limit $L$ exists, the elements of the sequence must get so close to $L$ as $n$ gets large, so we can plug in $L$ for both $a_{n+1}$ and $a_n$ in the recursion relation above to obtain the equation $L=\sqrt{2+L}$. Solving this gives $L=2$.

To be more rigorous, we have to show the limit exists and justify the above intuition. Using induction, you can prove that $\{a_n\}$ is increasing and bounded above by $2$. The Monotone Convergence Theorem then says that the sequence must converge to some limit $L$. Then we know the following: $$ L=\lim_{n\rightarrow\infty}a_n=\lim_{n\rightarrow\infty}a_{n+1}=\lim_{n\rightarrow\infty}\sqrt{2+a_n}=\sqrt{2+\lim_{n\rightarrow\infty}a_n}=\sqrt{2+L} $$

Here we have used a few simple facts about limits:

  1. If the limit of $a_n$ exists, then the limit of $\{a_{n+1}\}$ exists and the two limits are equal (used for the second equality).
  2. If $f(x)$ is a continuous function, and $\lim_{n\rightarrow\infty}a_n=L$, the $\lim_{n\rightarrow\infty}f(a_n)=f(L)$. In our problem $f(x)=\sqrt{x}$, so we can pass the limit inside the square root (used for the fourth equality).

Back to the problem at hand, squaring both sides gives the quadratic $L^2-L-2=(L-2)(L+1)=0$ so that $L=2$ or $L=-1$. We know $L$ is always positive, so in fact $L=2$ (actually $L=-1$ doesn't solve $L=\sqrt{2+L}$).

  • $\begingroup$ I like this answer, but I'd simply consider $\displaystyle \lim_{n \to \infty} \cos\left(\frac{\pi}{2^n}\right)$. 😋 $\endgroup$ – J.H. Mar 16 '13 at 21:30

This does not fit any of the standard notations for a sum ($\sum$) or product ($\prod$). However, you could define your own notation.

To do this, look at how the summation operator is defined.

First, the empty sum: $\sum_{i=n}^{n-1} a_i = 0$.

Then the initial sum (if you do not like empty sums): $\sum_{i=n}^{n} a_i = a_n$.

Finally the induction step: If $n > m$ then $\sum_{i=m}^{n} a_i = a_n + \sum_{i=m}^{n-1} a_i$.

Let's emulate this for your nested radicals. I will call the operator $R$ for "radicals".

The initial step: $R_{i=n}^n(a_i) = \sqrt{a_n}$.

Then the induction step: Interestingly, you can go in two directions - nesting inward or nesting outwards. When doing sums or products, this does not matter, since the operations are commutative and associative. Here it does matter.

First way (nesting from the outside): If $n > m$ then $R_{i=m}^{n} a_i = \sqrt{a_n+ R_{i=m}^{n-1} a_i}$.

Second way (nesting from the inside): If $n > m$ then $R_{i=m}^{n} a_i = R_{i=m}^{n-1} a^{(n-1)}_i$ where $a^{(p)}_i = a_i$ if $i \ne p$ and $a^{(i)}_i = a_i+\sqrt{a_{i+1}}$.

The next step is to discover and prove properties (limit behavior, etc.) of the operator.

As to whether or not this is an improvement over discussing nested square roots, that's up to you.

Remember, pointless generalizations can at least sometimes get you credit in class or even a publication. With luck and skill, you might even make a career of it.


In terms of notation, I'd write something along these lines:

$$f=\sqrt {2 + f}$$


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