# Looking for the sum an asympototic expansion of $\sum_{d|n}\frac{\Lambda \left({d}\right)}{d}$

I can not find any examples concerning this sum $$\sum_{d|n}\frac{\Lambda \left({d}\right)}{d}$$ where $$\Lambda \left({n}\right)$$ is the Von Mangoldt function. I suspect that there is no closed form solution in which case the real value that I need is the asymptotic expansion to the best possible order. The average over $$N$$ as $$N \rightarrow \infty$$ is about 0.566, that is $$\lim_{N \rightarrow \infty}\frac{1}{N} \sum_{n = 1}^{N} \sum_{d|n}\frac{\Lambda \left({d}\right)}{d} \rightarrow 0.566$$ so this suggests that $$\sum_{d|n}\frac{\Lambda \left({d}\right)}{d} \sim 0.566$$ + other terms.

I was summing over $$d|2n$$ instead of $$d|n$$ as stated.

$$f(n)=\sum_{d|n}\Lambda(d)/d$$ itself does not possess an asymptotic expansion since it can be very large and very small infinitely many times. When $$n$$ is a prime, we have $$f(n)=\log n/n=o(1)$$. However, when $$n=\prod_{p\le x}p$$, we have
$$\log n=\sum_{p\le x}\log p\sim x$$
$$f(n)=\sum_{p\le x}{\log p\over p}=\log x+O(1)\sim\log\log n.$$
• $f(n)\ne 2$ when $n=2^k$. Jun 30, 2023 at 10:28
We have \begin{align*} \sum_{n\le N} \sum_{d\mid n} \frac{\Lambda(d)}d &= \sum_{d\le N} \frac{\Lambda(d)}d \sum_{\substack{n\le N \\ d\mid n}} 1 \\ &= \sum_{d\le N} \frac{\Lambda(d)}d \biggl\lfloor \frac Nd \biggr\rfloor \\ &= \sum_{d\le N} \frac{\Lambda(d)}d \biggl( \frac Nd + O(1) \biggr) \\ &= N \sum_{d\le N} \frac{\Lambda(d)}{d^2} + O\biggl( \sum_{d\le N} \frac{\Lambda(d)}d \biggr) \\ &= N \sum_{d=1}^\infty \frac{\Lambda(d)}{d^2} + O\biggl( N \sum_{d>N} \frac{\Lambda(d)}{d^2} + \sum_{d\le N} \frac{\Lambda(d)}d \biggr) \\ &= N \sum_{d=1}^\infty \frac{\Lambda(d)}{d^2} + O(\log N). \end{align*} Therefore the limiting constant in question is $$\sum_{d=1}^\infty \frac{\Lambda(d)}{d^2} = -\frac{\zeta'(2)}{\zeta(2)} \approx 0.569961,$$ which I've checked a couple of different ways (I'm not sure why the OP is getting around $$0.8$$).