i have this numerical calculation problem :
$$\prod \limits\_{i=121443371}^{455052511} 1+\frac{1}{p(i)} \leq 1.06406506887043952285362856325019948 $$ 
such that $p(i)$ is the $i$-th prime number


i want to check whether the left side of the inequality is truly less than the number on the right side or not ?


i used Wolfram Mathematica 10.0 to do the calculations but it stopped for low memory problems !


could anyone give me some programs that can do these calculations with high accuracy ? 



Numerical Calculations
In 1939 Rosser proved that $p\_n>n\log n$. In 1999 Dusart proved
$p\_n>n(\log n+\log\log n-1)$ and also
$$p\_n\ge n\Bigl(\log n+\log\log n-1+\frac{\log\log n-2.25}{\log n}\Bigr),\qquad 
n\ge 2.$$
Finally Dusart in 2010 proved
$$p\_n\le n\Bigl(\log n+\log\log n-1+\frac{\log\log n-2}{\log n}\Bigr),\qquad 
n\ge 688383.$$
Each of these inequalities gives a computable bound for your product 
If $b\_n\le p\_n\le a\_n$ we obtain
$$\prod\_{n=121443371}^{455052511}\Bigl(1+\frac{1}{b\_n}\Bigr)\ge \prod\_{n=121443371}^{455052511}\Bigl(1+\frac{1}{p\_n}\Bigr)\ge 
\prod\_{n=121443371}^{455052511}\Bigl(1+\frac{1}{a\_n}\Bigr).$$
(In the above cases Euler-MacLaurin allow to compute this bound).
With Mathematica and the two last inequalities we obtain
$$1.064094593275173279\ge P\ge 1.064054235887163321,$$
where $P$ is your product.
Your product is between the last two numbers. But the bound is also there.


Dusart have obtained a better bound in which 2.25 is substituted by 2.1, but 
still this gives 
$$1.064070372735214460\ge P\ge 1.064054235887163321,$$
that is not sufficient. 


It is possible Riemann hypothesis implies your inequality because in this
case we have a much better bound of $p\_n>ali(n)-\frac{1}{\pi}\sqrt{n}(\log n)^{5/2}$
for $n>11$, where $ali(x)$ is the inverse of logarithmic integral function $li(x)$.
But in this case the computation is more complicate.


