What is the sum of reciprocal of product of $n$ primes?

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Question : What is the sum of the series given below?

$$\frac{1}{2} + \frac{1}{2 \cdot 3} + \frac{1}{2 \cdot 3 \cdot 5} + \cdots \frac{1}{2 \cdot 3 \cdot 5 \cdot 7 \cdot \cdots \cdot \sqrt{n}}$$

in the denominator these are the product of primes . What is the bound on this sum?

Is this $\le c\frac{1}{\sqrt{n}}$ ? where $c$ is some constant I know the given sum above is $\le \sqrt{n}/2$ but I need a tighter bound.

Question 2: Is this series converges to constant ?

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The sum is $>1/2$. You can't have a bound that $\to 0$ like $c/\sqrt n$.

Answer for $2$: the series $$\sum_{n=1}^\infty\frac1{\prod_{k=1}^np_n}$$ is convergent by comparison with $\sum_{n=1}^\infty\frac1{n!}$:

$$\frac1{\prod_{k=1}^np_n}\le\frac1{n!}.$$