What about the reciprocals of primes?

Method 1: $1 < 1/2 + 1/3 + 1/5 + 1/7 = A(7) < A\left (e^{e^3}\right)$ .

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Method 2: I'll prove a stronger result, $A(n) > 2.5$ for sufficiently large $n$ .

Assume all sums and product over $p$ are over the set of prime numbers.

Claim: $\displaystyle A(n) = \sum_{p \leq n}\dfrac1p > \ln(\ln n) - \dfrac12$ .

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Proof: We start with $\prod_{p\leq n } \left( 1 - \dfrac1p \right)^{-1} = \prod_{p\leq n} \sum_{k=0}^\infty \dfrac1{p^k} \geq \sum_{k=1}^n \dfrac1k > \int_1^n \dfrac1x \, dx = \ln n .$

And

$\begin{aligned} \ln \left [ \prod_{p\leq n } \left(1 - \dfrac1p \right)^{-1}\right ] &= &\sum_{p \leq n} \ln \left( 1 - \dfrac1p \right)^{-1} = \sum_{p \leq n} \sum_{k=1}^\infty \dfrac1{k p^k} \\ &<& \sum_{p \leq n} \dfrac1p + \sum_{p \leq n} \dfrac1{2p^2} \left( \sum_{k=0}^\infty \dfrac1{p^k} \right) \\ &=& \sum_{p \leq n} \dfrac1p + \dfrac12 \sum_{p \leq n} \dfrac1{p(p-1)} < \sum_{p \leq n} \dfrac1p + \dfrac12 \sum_{n=2}^\infty \dfrac1{p(p-1)} \\ &<& \sum_{p \leq n} \dfrac1p + \dfrac12 \\ \sum_{p \leq n} \dfrac1p &>& \ln \left [ \prod_{p\leq n } \left(1 - \dfrac1p \right)^{-1}\right ] - \dfrac12 > \ln (\ln n) - \dfrac12 \quad \square \\ \end{aligned}$

In this case, $x = e^{e^3}$ . Thus, $A\left( e^{e^3} \right) > \ln\left (\ln \left(e^{e^3} \right)\right) - \dfrac12 = 2.5 > 1$ .

The answer is $\boxed{\text{Yes, it is true}}$ .

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Source: ProofWiki.Org

Note that $\displaystyle \lim_{n\to\infty} \left( \sum_{p \leq n} \dfrac1p - \ln(\ln n) \right ) = M \approx 0.261$ , where $M$ denotes the Meissel–Mertens constant .