# Show that two series both converge or diverge (number theory)

Suppose $F(n)$ is defined for $n \geq 2$. If $F(n)$ is positive and $\dfrac{F(n)}{\log n}$ is non-increasing, prove that the two series $$\sum_{p \text{ prime}} F(p),\ \sum_{n = 2}^\infty \frac{F(n)}{\log n}$$ both converge or diverge.

I have tried writing $$\sum_{\substack{p \leq N\\p \text{ prime}}} F(p) = \sum_{n=2}^N (\theta (n) - \theta (n-1)) \frac{F(n)}{\log n},$$ where $$θ(x) = \sum_{\substack{p \leq x\\p \text{ prime}}} \log p,$$ to somehow use monotonicity for $\theta (n)$ (I found an upper bound for $\theta (n)$, which is $\theta (x) < cx$ for some $c$) but I have gotten nowhere. Any ideas on how to proceed/conclude?

Writing the characteristic function of the primes as $\dfrac{\theta(n) - \theta(n-1)}{\log n}$ is a good start. But one doesn't use the monotonicity of $\theta(n)$, it's the monotonicity of $F(n)/\log n$ that one uses, together with upper and lower bounds for $\theta(n)$. A summation by parts yields

$$\sum_{n = 2}^N \bigl(\theta(n) - \theta(n-1)\bigr)\frac{F(n)}{\log n} = \theta(N) \frac{F(N+1)}{\log (N+1)} + \sum_{n = 2}^N \theta(n)\biggl(\frac{F(n)}{\log n} - \frac{F(n+1)}{\log (n+1)}\biggr) \tag{1}$$

using $\theta(1) = 0$. All terms on the right hand side are non-negative (that's also the case on the left, but we don't use that), and so we make the right hand side smaller/larger if we replace $\theta(n)$ with something smaller/larger. We use Chebyshev's result that there are $c_1, c_2 > 0$ such that

$$c_1\cdot n \leqslant \theta(n) \leqslant c_2\cdot n$$

for all $n \geqslant 2$ and replace $\theta(n)$ with $c_k\cdot n$ on the right hand side of $(1)$. Then reversing the summation by parts gives

\begin{align} c_kN\frac{F(N+1)}{\log (N+1)} + \sum_{n = 2}^N c_kn\biggl(\frac{F(n)}{\log n} - \frac{F(n+1)}{\log (n+1)}\biggr) &= \sum_{n = 2}^N \bigl(c_k n - c_k(n-1)\bigr) \frac{F(n)}{\log n} \\ &= c_k\sum_{n = 2}^N \frac{F(n)}{\log n}. \end{align}

Thus we find

$$c_1 \sum_{n = 2}^N \frac{F(n)}{\log n} \leqslant \sum_{p \leqslant N} F(p) \leqslant c_2 \sum_{n = 2}^N \frac{F(n)}{\log n}. \tag{2}$$

From $(2)$ it is clear (since $c_k > 0$) that convergence of $\sum_p F(p)$ implies the convergence of $\sum_{n = 2}^{\infty} \frac{F(n)}{\log n}$ and vice versa.