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Let $(a_n)_{n=1}^\infty$ be a decreasing sequence of positive real numbers such that $\sum_{n=1}^\infty a_n$ converges. I am interested in the sum $\sum_{p}a_p$, where $p$ ranges over the primes. This subsum obviously converges, but I am interested in how quickly it converges. More precisely, I would like to know what I can say about the asymptotic size of $R(x):=\sum_{p\geq x}a_p$. Because $\sum_{n\geq x}a_n=o(1)$, it seems as though we should have $R(x)=o\left(\frac{1}{\log x}\right)$. In fact, I would like to be able to prove that $$\sum_{m=1}^\infty \frac{R(m)}{m}$$ converges. I am not sure if this is true, though. Any help would be greatly appreciated.

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What you need is that $\pi(x) \sim \sum_{n < x} \frac{1}{\ln n}$ for saying that since $a_n > 0$ and is non-increasing, then $\sum_{p > x} a_p \sim \sum_{n > x} \frac{a_n}{\ln n}$.

Summing by parts : $$\sum_{p > x} a_p = \sum_{n > x} a_n 1_{n \in P} = \sum_{n > x} \pi(n) (a_n-a_{n+1}) =\sum_{n > x} ((1+o(1))\sum_{k < n} \frac{1}{\ln k}) (a_n-a_{n+1}) \\= \sum_{n > x} (\frac{1}{\ln n}+ o(\frac{1}{\ln n})) a_n =(\sum_{n > x} \frac{a_n}{\ln n})(1+o(1)) $$ Where some of those $o(1)$ means as $n \to \infty$, the others as $x \to \infty$ (use that $a_n > 0$ and non-increasing for bounding the former in term of the latter)

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