We are given an infinite nested radical as follows
$$\sqrt{1+\sqrt{2+\sqrt{3+\cdots}}},$$
which can be proven that it is convergent, namely, is equal to a real constant.
For this purpose, we consider applying the monotone convergence principle. Since its monotone property is obvious, what we need is to show it is bounded above.
First, we claim
$${\frac{n}{2^{2^{n-1}}}\leq 1(\forall n)}$$
$Proof~~~~$ It's easy to show $2^{n-1}\geq n.$ For $n=1,2$, it holds trivially. For $n \geq 3$, by Newton's binominal theorem, we have $$2^n=(1+1)^n=\binom n0+\binom n1+\cdots+\binom n{n-1}+\binom nn\geq 1+n+n+1\geq 2n.$$ Thus $2^{n-1}\geq n$. Furthur, $2^{2^{n-1}}\geq 2^n \geq 2^{n-1}\geq n,$ which implies ${\dfrac{n}{2^{2^{n-1}}}\leq 1(\forall n)}$.
Next, we claim
$${ \sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}<\infty}.$$
This is a well-known result. In fact, substituting $1$ among here for any $a>0$, it's also true. For the recursion $x_1=\sqrt{a},x_n=\sqrt{a+x_{n-1}}(n\geq2)$, we can show that $a_n\leq a$, just by mathematical induction.
Now, go back to the present problem. Notice that
\begin{align*} \sqrt{1+\sqrt{2+\sqrt{3+\cdots}}}&=\sqrt{2}\cdot\sqrt{\frac1 {2}+\sqrt{\frac2{2^2}+\sqrt{\frac3{2^4}+\cdots}}}\\ &\leq\sqrt{2}\cdot\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}\\ &<\infty, \end{align*}
which is desired.
But, my question is that, can we evaluate it? In another word, can the limit be expressed as an elementary closed form?
