Will this prime-factorization related sum converge, when summed over all the integers?

Question

This is related to this question of mine.

Define a function $h(n)$ that takes the sum of the reciprocals of $n$ minus each of its prime factors. In English, that's extremely confusing; hopefully an example will clarify:

$$h(12) = \frac{1}{12 - 2} + \frac{1}{12-2} + \frac{1}{12 - 3} = \frac{1}{10} + \frac{1}{10} + \frac{1}{9} = 0.3\overline{1}$$

Note that because $12$ has $2^2$ in its prime factorization, there are two $\frac{1}{10}$s in the sum. The exception: $h(p) = 0$ for all primes $p$. (Otherwise, division by $0$ would be necessary.)

In mathematical notation (using $f_i$ as the $i$th prime factor of $n$):

$$h(n) = \begin{cases} \text{if } n \text{ is prime}& 0 \\ \text{if } n \text{ is composite} &\displaystyle\sum_{i = 1}^{\Omega(n)} \frac{1}{n-f_i} \end{cases}$$

My question is:

Does $\displaystyle\sum_{n = 2}^{\infty} h(n)$ converge?

Clearly, this sum gets smaller and smaller on average as $n$ gets larger. But is it small enough that summing its results of all the positive integers will converge? The program I wrote has calculated that

$$\displaystyle\sum_{n = 2}^{10,000,000} h(n) \approx 46.48623444862424.$$

As an aside, the sequence of the numbers where $h(n)$ reaches a smaller nonzero value than any $h(m)$ for composite $m < n$ begins

$$4, 6, 8, 9, 10, 14, 15, 20, 21, 25, 33, 35, 39, 46, 49, 55, 65, 77, 85, 91, 95, 111, 115...$$

Don't know if it's in OEIS. Anyway, any help with solving my question would be appreciated, as I don't really know how to go about it -- thanks for your advice!

Answer 1

The sum diverges. Indeed it diverges even when restricted to integers of the form $n=2p$ with $p$ prime, since $$ \sum_p h(2p) = \sum_p \biggl( \frac1{2p-2} + \frac1{2p-p} \biggr) > \sum_p \frac1p $$ which Euler proved diverges.

Answer 2

Consider just the terms where $n=2k$ is even and greater than 2: \begin{align*} \sum_{n=4}^{2N} h(n) &> \sum_{k=2}^{N} h(2k)\\ &= \sum_{k=2}^{N} \bigg[ \frac1{2k-2}+\dots+\frac1{2k-k}\bigg]\\ &> \frac12 \sum_{k=1}^{N-1} \frac1{k} + \sum_{k=2}^{N} \frac1{k} \end{align*} Both these sums diverge as $N\to\infty$, as is well known.