Function of $ \sqrt{2+\sqrt{2+\sqrt{2+}}}\ldots $

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.

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 $a_{n}<2$, $a_{n+1}=\sqrt{a_{n}+2}<\sqrt{2+2} = 2$, and $\sqrt{a_n+2} > a_n$. 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 $L=\sqrt{2+L}$).