Nested Radicals Involving Primes

Question

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.

Answer 1

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

$$s_{10}(0)=2.10359748\dots\le s_{\infty}\le2.10359778\dots=s_{10}(227502.84\dots).$$

Answer 2

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 $$2.0708...=\sqrt{2+\sqrt{3}+\sqrt{5}}<\alpha<\frac{1+\sqrt{5}}{\sqrt{2}}=2.2882...$$