It is possible to show that the limit of the following fraction is a constant $$ \frac {e ^ {1/2} \Big (1 + e ^ {1/3} \big (1 + e ^ {1/5} (1 + e ^ {1/7} (\dots (1 + e ^ {1 / p (n)})} {p(n)} \sim c $$ How can I determine the value of the constant $ c $?
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From the prime number theorem we have that $ \pi (x) \sim \text {li} (x): = \int_0 ^ x \frac {dt} {\ln (t)} $, where $ \pi (x ) $ is the prime counting function. $ \pi (n) $ can be thought as the inverse of the function $ p (n) $ which gives the $n$-th prime number, such that $ p (n) \sim \text {ali} (x) = \text {li} ^ {- 1} (x) $. Calling $ \text {ali} (x) $ the inverse function of the logarithmic integral: $$\text{li}(\text{ali}(x))=x \ \ \Rightarrow \ \ \frac{d}{dx}\Big[\text{li}(\text{ali}(x))\Big]=1\Rightarrow \text{ali}'(x)=\ln(\text{ali}(x))$$ $$\text{ali}''(x)=\frac{d}{dx}\Big[\ln(\text{ali}(x))\Big]=\frac{\text{ali}'(x)}{\text{ali}(x)} \Rightarrow \text{ali}'(x)=\text{ali}(x) \text{ali}''(x)\sim p(x) \ \text{ali}''(x) $$
Solving the differential equation $ y '(x) = p (x) y' '(x) $ we obtain:
$$y(x)=c_2+c_1\int_{1}^{x}\exp \Big(\int_1^{t}\frac{du}{p(u)}\Big)dt$$
Interpreting the integral as the area under a broken line, we can write for the integer $ n $:
$$y(n)=c_2+c_1\sum_{v=1}^n \exp\Big(\sum_{w=1}^{v}\frac{1}{p(w)}\Big)\sim p(n)$$
$$y(4)=c_2+c_1 \Big( e^{\frac{1}{2}}+e^{\frac{1}{2}+\frac{1}{3}}+e^{\frac{1}{2}+\frac{1}{3}+\frac{1}{5}}+e^{\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}}\Big)=c_2+c_1 \Big( e^{1/2}\Big(1+e^{1/3}\Big(1+e^{1/5}\Big(1+e^{1/7}\Big)\Big)\Big)\Big)\approx 7$$
This means that $ \big (y (n) -c_2 \big) / p (n) \sim c_1 $. Setting $ c_2 = 0 $, the trend of $ y (n) / p (n) $ is as follows:
The orange line represents $ \frac {4} {\pi} $ but there is no reason for it to be the limit, I did it for clickbait :-\
The limit equals $e^M$, where $M$ is the Meissel–Mertens constant: \begin{align*} M&=\lim_{n\to\infty}\left(\sum_{p\leqslant n}\frac1p-\log\log n\right) \\&\approx0.261497212847642783755426838608695859051566648+ \\e^M&\approx\color{blue}{1.29}8873321409030320833690880357295240469110387+ \\4/\pi&\approx\color{red}{1.27}3239544735162686151070106980114896275677166- \end{align*}
To show the above, one may use $p(n)\asymp n\log n$ (essentially, it is the prime number theorem) and $M_n:=\sum_{p\leqslant n}(1/p)\asymp\log\log n+M+\mathcal{O}(1/\log n)$ (this is Mertens' second theorem). These give \begin{align*} M_{p(n)}&\asymp\log\log n+M+\mathcal{O}\left(\frac{\log\log n}{\log n}\right), \\e^{M_{p(n)}}&\asymp e^M\log n\left[1+\mathcal{O}\left(\frac{\log\log n}{\log n}\right)\right], \\\sum_{k=1}^n e^{M_{p(k)}}&\asymp e^M \log(n!)+\mathcal{O}(n\log\log n), \end{align*} and the result follows after the division by $p(n)\asymp n\log n\asymp\log (n!)$.