Can $\prod_{i=1}^{\pi(n)} p_i^{\frac{1}{p_i-1}}$ be calculated? Is there a way to calculate this Product as a function of $n$?
$$\prod_{i=1}^{\pi(n)} p_i^{\frac{1}{p_i-1}}$$
where $p_i$ is the $i^{\text{th}}$ prime number, and $\pi(n)$ is the Prime-counting function;
Thanks.
 A: From Prime Number Theorem, we have $$\sum_{p\leq n}\log\left(p\right)\sim n$$
 so, taking log of this product and by partial summation we have, as $n\rightarrow\infty$
 $$\log\left(\underset{p\leq n}{\prod}p^{\frac{1}{p-1}}\right)=\sum_{p\leq n}\frac{\log\left(p\right)}{p-1}\sim\frac{n}{n-1}+\int_{2}^{n}\frac{t}{\left(t-1\right)^{2}}dt\sim\log\left(n\right)$$
 or, if you prefer, the asymptotic $$\sum_{p\leq n}\frac{\log\left(p\right)}{p}\sim\log\left(n\right)$$
  follows by Mertens theorem (http://en.wikipedia.org/wiki/Mertens%27_theorems). So $$\underset{p\leq n}{\prod}p^{\frac{1}{p-1}}\sim n$$
 as $n\rightarrow\infty$. Another (similar) way was to observe that$$\frac{\log\left(p\right)}{p-1}=\frac{\log\left(p\right)}{p}+O\left(\frac{1}{p^{\alpha}}\right)$$
 with $1<\alpha<2$
  so$$\sum_{p\leq n}\frac{\log\left(p\right)}{p-1}=\sum_{p\leq n}\frac{\log\left(p\right)}{p}+O\left(1\right)$$
 and using Mertens theorem, which states, as $n\rightarrow\infty$ $$\sum_{p\leq n}\frac{\log\left(p\right)}{p}=\log\left(n\right)+O\left(1\right)$$
 we have the same result.
