Can you get a closed-form for $\prod_{p\text{ prime}}\left(\frac{p+1}{p-1}\right)^{\frac{1}{p}}$? When I use the Taylor expansion series for $$\log(1+x)^{1+x}+\log(1-x)^{1-x}$$ with $x=\frac{1}{p}$, $p$ prime, I believe that I can deduce 
$$\sum_{p\text{ prime}}\left(\frac{1}{p^2}+\frac{1}{2\cdot3p^4}+\frac{1}{3\cdot5p^6}+\frac{1}{4\cdot7p^8}+\cdots\right)=\log\frac{6}{\pi^2}+\log\prod_{p\text{ prime}}\left(\frac{p+1}{p-1}\right)^{\frac{1}{p}}.$$
If previous computations are right, and can be justified I want to ask the following  

Question. Can you compute 
  $$\prod_{p\text{ prime}}\left(\frac{p+1}{p-1}\right)^{\frac{1}{p}}?$$

Thanks in advance.
My goal is learn, thus if you can give the details about how are justified the more important steps to deduce my computations it is the best. Also, how can we justify the convergence of such infinite product?
 A: The prime products that are 'easy' to evaluate are usually those that can somehow be related somehow to the Euler product for the Riemann $\zeta$ function
$$\frac{1}{\zeta(s)} = \prod_p \left[1 - \frac{1}{p^s}\right]\tag{1}$$
Your product is not on this form, or anything resembling it, so that makes evaluating it much harder so I doubt that there is a known closed form for it. Note for example that there is not even a known closed form for the much simpler product $\prod_p\left[1- \frac{2}{p^2}\right]$. However we can derive a very good approximation for your product.
For $x\ll 1$ we have the approximation $(1+x)^m \approx 1+mx$. By taking $m = x = \frac{1}{p}$ we obtain
$$\left(\frac{1+\frac{1}{p}}{1-\frac{1}{p}}\right)^{\frac{1}{p}}\approx \frac{1+\frac{1}{p^2}}{1-\frac{1}{p^2}}~~~~\text{for large }p$$
The product of the term above can be evaluated since $\prod_{p} \left[1-\frac{1}{p^2}\right] = \frac{6}{\pi^2}$ and $\prod_{p} \left[1+\frac{1}{p^2}\right] = \frac{15}{\pi^2}$. These products can be derived from the Euler product $(1)$ as $\frac{1}{\zeta(2)}$ and $\frac{\zeta(2)}{\zeta(4)}$ respectively. Now by multiplying togeather the first $N$ terms in your product and using the approximation above for the remaining terms we obtain the simple approximation
$$\prod_p \left(\frac{p+1}{p-1}\right)^{\frac{1}{p}} \approx \frac{5}{2}\prod_{n=1}^N\left(\frac{p_n+1}{p_n-1}\right)^{\frac{1}{p_n}}\frac{p_n^2-1}{p_n^2+1}$$
which gets better and better the larger we take $N$. For example if $N=3$ we get $\frac{72}{65} 2^{2/15} 3^{7/10}\simeq 2.62145$ which are within $0.04\%$ of the true answer $2.62239915779\ldots$ found by numerically evaluating the sum for $N=10000$.
