Computing $\lim_{n \to \infty}\frac{n-H_{n}-\sum_{j\le n}\frac{\Omega\left(j\right)}{j}+\sum_{j\le n}\frac{\omega\left(j\right)}{j}}{n}$

Does the limit exist $$\lim_{n \to \infty}\frac{n-H_{n}-\sum_{j\le n}\frac{\Omega\left(j\right)}{j}+\sum_{j\le n}\frac{\omega\left(j\right)}{j}}{n}?$$

$n$ is a positive integer, $H_n$ is the $n$th harmonic number ($H_n = \sum_{j\le n}\frac{1}{j})$, $\Omega(j)$ denotes the sum of the distinct prime factors of $j$, and $\omega(j)$ denotes the number of distinct prime factors of $j$.

The expression in the limit is the expected value of a function $f(N)$ where $N$ is a uniform random variable on $\lbrace 1,2,\ldots, n\rbrace$ and $0< f(\cdot)<1$ is some function. If the limit exists, I don't expect the value to be 1 (so strictly less than 1 by virtue of $f$).

I have that $H_n \sim \log n + \gamma$, so I expect that the limit above is equal to $$\lim_{n \to \infty}\frac{n-\sum_{j\le n}\frac{\Omega\left(j\right)}{j}+\sum_{j\le n}\frac{\omega\left(j\right)}{j}}{n}.$$ However, I am unsure of how to deal with the sums remaining sums in the numerator. The closest bound that I have is $\omega(n)\le (1+o(1))\frac{\log n}{ \log \log n}$.

Solution 1:

Although $\Omega(j)$ often denotes the number of prime factors counted with multiplicity, let's follow OP's definition in this answer:

$$ \Omega(j)=\sum_{p|j}p $$

This indicates that

$$ \sum_{j\le n}{\Omega(j)\over j}=\sum_{j\le n}\frac1j\sum_{p|j}p=\sum_{p\le n}p\sum_{\substack{j\le n\\p|j}}\frac1j $$

Let $j=pr$, so the rightmost sum becomes $r\le n/p$:

\begin{aligned} \sum_{j\le n}{\Omega(j)\over j} &=\sum_{p\le n}\sum_{r\le n/p}\frac1r \\ &=\sum_{p\le n}\log\frac np+\mathcal O\{\pi(n)\} \\ &=\pi(n)\log n-\vartheta(n)+\mathcal O\left(n\over\log n\right) \end{aligned}

By the prime number theorem we know that $\vartheta(n)\sim\pi(n)\log n\sim n$, so we have

$$ \sum_{j\le n}{\Omega(j)\over j}=o(n) $$

where $\pi(n)$ denotes the number of primes $\le n$ and $\vartheta(n)=\sum_{p\le n}\log p$. For the second sum, we can use similar tricks to handle the sum over $\omega(j)$:

\begin{aligned} \sum_{j\le n}{\omega(j)\over j} &=\sum_{p\le n}\frac1p\sum_{r\le n/p}\frac1r \\ &=\sum_{p\le n}\frac1p\log\frac np+\mathcal O\left\{\sum_{p\le n}\frac1p\right\} \\ &=\sum_{p\le n}{\log n\over p}-\sum_{p\le n}{\log p\over p}+\mathcal O(\log\log n) \\ &=(\log n)\log\log n-\log n+\mathcal O(\log\log n)=o(n) \end{aligned}

Since $H_n=\mathcal O(\log n)=o(n)$, we conclude that OP's limit exists and equals to one.

Appendix

In the derivation process, we make use of following asymptotic formulas related to prime numbers:

$$ \pi(n)=\sum_{p\le n}1\sim{n\over\log n} $$

$$ \vartheta(n)=\sum_{p\le n}\log p\sim n $$

$$ \sum_{p\le n}{\log p\over p}=\log n+\mathcal O(1) $$

$$ \sum_{p\le n}\frac1p=\log\log n+B_1+\mathcal O\left(1\over\log n\right) $$

The first two of them follow from the prime number theorem, and the last two of them are Mertens' theorems.