1
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.