Nested Radicals Involving Primes How do you evaluate $\sqrt { 2+\sqrt { 3+\sqrt { 5+\sqrt { 7+\sqrt { 11+ \dots }  }  }  }  } $ ?
This question appears to be rather difficult as there is no way to perfectly know what $p_{ n }$ is , if $p_{ n }$ denotes the $n$th prime.
It is simple to show that the value above is convergent. Bertrand`s Postulate implies that $p_{ n } \le 2^n$, which implies that  $\sqrt { 2+\sqrt { 3+\sqrt { 5+\sqrt { 7+\sqrt { 11+ \dots }  }  }  }  } \le \sqrt { 2+\sqrt { 4+\sqrt { 8+\sqrt { 16+\sqrt { 32+\dots }  }  }  }  } $, which is convergent, as seen here.
So it is pretty clear that  $\sqrt { 2+\sqrt { 3+\sqrt { 5+\sqrt { 7+\sqrt { 11+ \dots }  }  }  }  } $ is convergent.
However, in which fashion can you evaluate the value above? If there is no exact way to evaluate it, is it irrational or rational?
The value seems to be about $2.10359749633989726261993..$ as seen here.
Any help would be appreciated.
 A: For the primes $(p_k):=(2,3,5,\dots)$ we have
$$\tag{1}0\le\dfrac{p_{n+1}}{p_1p_2\cdots p_n}\le2.$$
The inequality is trivial if $p_n\lt p_{n+1}\lt p_1p_2\cdots p_n$. Otherwise the number $1+p_1p_2\cdots p_n$ is a prime and the inequality holds.
For $a,x\gt0$ we have
$$0\lt\sqrt{a+x}={\sqrt{a}+\int_a^{a+x}\dfrac{d\xi}{2\sqrt{\xi}}}\le
{\sqrt{a}+\int_a^{a+x}\dfrac{d\xi}{2\sqrt{a}}}={\sqrt{a}+\dfrac{x}{2\sqrt{a}}}.$$
We obtain
$$\tag{2}\sqrt{p_1}\lt{\sqrt{p_1+\sqrt{p_2+\sqrt{p_3+\sqrt{p_4+\sqrt{\dots+\sqrt{p_n}}}}}}}\le
{\sqrt{p_1}+\dfrac{1}{2\sqrt{p_1}}\sqrt{p_2+\sqrt{p_3+\sqrt{p_4+\sqrt{\dots+\sqrt{p_n}}}}}}\le
{\sqrt{p_1}+\dfrac{1}{2}\sqrt{\dfrac{p_2}{p_1}}+\dfrac{1}{2^2\sqrt{p_1p_2}}\sqrt{p_3+\sqrt{p_4+\sqrt{\dots+\sqrt{p_n}}}}}\le
{\sqrt{p_1}+\dfrac{1}{2}\sqrt{\dfrac{p_2}{p_1}}+\dfrac{1}{2^2}\sqrt{\dfrac{p_3}{p_1p_2}}+\dfrac{1}{2^3\sqrt{p_1p_2p_3}}\sqrt{p_4+\sqrt{\dots+\sqrt{p_n}}}}\le
{\sqrt{2}+\dfrac{1}{2}\sqrt{2}+\dfrac{1}{2^2}\sqrt{2}+\dfrac{1}{2^3}\sqrt{2}+\dots}\le{2\sqrt{2}}$$
and the limit
$$\sqrt{2}\lt\lim_{n\to\infty}{\sqrt{p_1+\sqrt{p_2+\sqrt{p_3+\sqrt{p_4+\sqrt{\dots+\sqrt{p_n}}}}}}}\le{2\sqrt{2}}$$
exists. Though a strictly monotonic bounded sequence converges we prefer to know the error of the sum
$$s_{n-1}:={\sqrt{p_1+\sqrt{p_2+\sqrt{p_3+\sqrt{p_4+\sqrt{\dots+\sqrt{p_{n-1}}}}}}}}\;.$$ 
Similar to above we have
$${\sqrt{p_n}}\le\sqrt{p_n+\sqrt{p_{n+1}+\sqrt{\dots}}}\le{\sqrt{p_n}+\dfrac{1}{2}\sqrt{\dfrac{p_{n+1}}{p_{n}}}+\dfrac{1}{2^2}\sqrt{\dfrac{p_{n+2}}{p_{n}p_{n+1}}}+\dots}$$
and with the inequality $(1)$ we obtain
$${\dfrac{\sqrt{p_n+\sqrt{p_{n+1}+\sqrt{\dots}}}}{2^{n-1}\sqrt{p_1p_2\cdots p_{n-1}}}}\le
{\dfrac{1}{2^{n-1}}\sqrt{\dfrac{p_{n}}{p_1p_2\cdots p_{n-1}}}+\dfrac{1}{2^{n}}\sqrt{\dfrac{p_{n+1}}{p_1p_2\cdots p_{n}}}+\dots}\le
{\dfrac{\sqrt{2}}{2^{n-1}}+\dfrac{\sqrt{2}}{2^{n}}+\dots}\le\dfrac{\sqrt{2}}{2^{n-2}}$$
or
$$\sqrt{p_n+\sqrt{p_{n+1}+\sqrt{\dots}}}\le2\sqrt{2p_1p_2\cdots p_{n-1}}.$$
We define 
$$\tag{3}{s_{n-1}(\sigma):=\sqrt{p_1+\sqrt{p_2+\sqrt{\dots+\sqrt{p_{n-1}+\sigma}}}}}.$$
and obtain
$$s_{n-1}(0)\le s_{\infty}\le s_{n-1}(2\sqrt{2p_1p_2\cdots p_{n-1}}).$$
We get
$$s_5(0)=2.1028\dots\le s_{\infty}\le2.1046\dots=s_5(135.94\dots)$$
and
$$s_{10}(0)=2.10359748\dots\le s_{\infty}\le2.10359778\dots=s_{10}(227502.84\dots).$$
A: Since convergence has been established, let's note
$$\alpha=\sqrt{p_1+\sqrt{p_2+\sqrt{p_3+...}}}$$
A very raw estimation may be obtained from
$$\alpha\sqrt{2}=\sqrt{2p_1+2\sqrt{p_2+\sqrt{p_3+...}}}=\sqrt{2p_1+\sqrt{4p_2+\sqrt{16p_3+...}}}>...$$
using Bertrand's postulate
$$...>\sqrt{p_2+\sqrt{2p_3+\sqrt{8p_4+...}}}>\sqrt{p_2+\sqrt{p_3+\sqrt{p_4+...}}}=\alpha^2-p_1=\alpha^2-2$$
$$\alpha^2-\alpha\sqrt{2}-2<0$$
this gives an estimate of
$$2.0708...=\sqrt{2+\sqrt{3}+\sqrt{5}}<\alpha<\frac{1+\sqrt{5}}{\sqrt{2}}=2.2882...$$
