Is $\cup_{k=1}^\infty (r_k-\frac{1}{k}, r_k+\frac{1}{k}) = \mathbb{R}$? Let $r_k$ be the rational numbers in $\mathbb{R}$. 
(1).Is $\cup_{k=1}^\infty (r_k-\frac{1}{k^2}, r_k+\frac{1}{k^2}) = \mathbb{R}$?
(2).Is $\cup_{k=1}^\infty (r_k-\frac{1}{k}, r_k+\frac{1}{k}) = \mathbb{R}$?
(1).Because $m(\mathbb{R})=+\infty, \sum_{k=1}^\infty \frac{1}{k^2}<+\infty$, so $\mathbb{R} \setminus\cup_{k=1}^\infty (r_k-\frac{1}{k^2}, r_k+\frac{1}{k^2})\neq \Phi  $
(2) What about (2)?
 A: One can choose an enumeration $\{r_k\}_{k=1}^\infty$ of the rationals so that (2) fails.
Define $A = \{2^k : k \in \Bbb N\}$.
Let $\{r_k\}_{k=1}^\infty$ be an enumeration of the rationals such that
$$
\{k \in \Bbb N : r_k \in \Bbb Q \cap [0, \infty)\} = A.
$$
This is possible since $A$, $\Bbb N - A$, $\Bbb Q \cap [0, \infty)$ and $\Bbb Q \cap (-\infty, 0)$ are all countably-infinite.
Note that $[2, \infty)$ cannot be covered by $\bigcup_{r=1}^\infty (r_k - 1/k, r_k + 1/k)$. This is because only the intervals $(r_k - 1/k, r_k + 1/k)$ for $k \in A$ can cover elements in $[2, \infty)$, and the sum of the lengths of those intervals is finite.
A: I think (2) depends on how you enumerate the rationals. For example lets say you dont want $e$ in the image. Then enumerate the rationals so that if $q$ is a rational and 
$e-q \sim \frac{1}{n}$ the make sure that if $r_k=q$ we have $k > n$. 
(better is given in Ayman's answer and in the comments afterwards).
Conversely, there is an enumeration so that (2) is true, for example to cover the unit interval chose the rational sequence increasing so that you succsssively cover the interval, since the series diverges you will cover the whole interval, then do this over every interval.
I have given a very rough description of these constructions, they involve much back and forth to ensure one has a enumeration of all the rationls. 
A: For $k=2n+1(n \in N)$ we put $q_{2k}=\sum_{i=1}^{n-1}1/(2(2i+1))+1/(4(2n+1))$;
For $k=2n(n \in N)$ we put $q_{2k}=-\sum_{i=1}^{n-1}1/(4i)-1/(8n)$;
Let $f:N\setminus \{ 2k:k \in N\} \to Q \setminus \{q_{2k}:k\in N\}$ be a bijection. We put $q_n=f(n)$ for $n \in N\setminus \{ 2k:k \in N\}$. Then for the sequence $(q_n)_{n \in N}$ we have $R=\cup_{n \in N}(q_n-1/n,q_n+1/n)$. 
