Finding $n$ such that $\sum^{n}_{k=0} \frac{2}{p_k} = \left ( \prod^{n}_{j=0} p_j^{-1}\right) p_x$

Question

Let $p_n$ denote the $n$th prime.

Is it possible to find $n$ such that

$$\sum^{n}_{k=0} \frac{2}{p_k} = \left ( \prod^{n}_{j=0} p_j^{-1}\right) p_x$$

any other way than calculating both the product and sum and comparing?

Answer 1

$$\sum_{k=0}^n\frac{2}{p_k}=2\frac{\sum_{k=0}^np_k}{\prod_{j=0}^np_j},$$ which reduces the question to asking whether

$$2\sum_{k=0}^np_k=p_x$$

which is clearly impossible since the l.h.s. is even (perhaps $p_1=2$ is possible depending on your indexing, say $p_0:=1, \ n=0$).