Is the sum $\sum_{d\mid n}\frac1{d+1}$ never integral?

Question

Recall that a positive integer $n$ is a perfect number if and only if $$\frac{\sigma(n)}n=\sum_{d\mid n}\frac1d=2.$$

QUESTION: Is my following conjecture true?

Conjecture. (i) We have $\sum_{d\mid n}\frac1{d+1}\not\in\mathbb Z$ for all $n=1,2,3,\ldots$. Moreover, for any positive integers $k$ and $m$, all the numbers $$\sum_{d\mid n}\frac1{(d+m)^k}\ \ (n=1,2,3,\ldots)$$ have pairwise distinct fractional parts, and none of them is an integer.

(ii) For any integer $k>1$, all the numbers $$\sum_{d\mid n}\frac1{d^k}\ \ (n=1,2,3,\ldots)$$ have pairwise distinct fractional parts.

I formulated this conjecture in October 2015 on the basis of my computation. Your comments are welcome!

Answer 1

For a given set of primes $Q=\{q_1,\dots,q_k\}$, to each prime $p\not\in Q$ we may associate the lattice $$ L=L_{q_1,\dots,q_k,p}=\{(a_1,\dots,a_k)\in\mathbb{Z}^k: \prod_{i=1}^kq_i^{a_i}\equiv 1 \bmod p\}. $$ and the coset $$ H_m=H_{m;q_1,\dots,q_k,p}=\{(a_1,\dots,a_k)\in\mathbb{Z}^k: \prod_{i=1}^kq_i^{a_i}\equiv -m \bmod p\}. $$ Then for all $n$ of the form $$ n=\prod_{i=1}^kq_i^{e_i}, $$ $e_i>0$ for all $1\leq i\leq k$, if there exists a prime $p$ such that the box $$ E_n=\{(a_1,\dots,a_k)\in\mathbb{Z}^k: 0\leq a_i\leq e_i, \mbox{for all }1\leq i\leq k\}, $$ intersects $H_m$ at exactly one element, (thus exactly one divisor $d$ of $n$ satisfies $p|(d+m)$), such $n$ satisfies $$ \sum_{d|n}\frac{1}{d+m}\not\in\mathbb{Z}. $$