Assume that $n \in \mathbb{N}$ and $a,b \in \mathbb{N}$ such that $\gcd(a,b)=1$ , i want to show that:
$$\frac{1}{a}+\frac{1}{a+b}+\cdots+\frac{1}{a+nb} \notin \mathbb{N}$$
but i can't,i think we should consider the set {$a,a+b,...,a+bn$} and use chebyshev's theorem but i dont know how shoud i use that??!!are there any one help me??
for example in special case, i got $b=1,$ so each number of {$a,a+1,...,a+n$} is of the form $2^{r}m$ which $b$ is odd and $r>-1$ and we can discuss about $r$.
What I write next is not really a response but a strategic one to tackle the problem. A statement $\frac{1}{a}+\frac{1}{a+b}+\cdots+\frac{1}{a+nb} \neq p $ for all $p\in \mathbb{N}$ is equivalent to the statement $$ p\cdot \prod_{1\leq k\leq n}(a+kb)- \sum_{0\leq i\leq n}\prod_{\substack{0\leq k\leq n \\ k\neq i }}(a+kb)\neq 0 $$ for all $p\in\mathbb{N}$.So it is enough to prove that the equation $$ x\cdot \prod_{1\leq k\leq n}(a+kb)- \sum_{0\leq i\leq n}\prod_{\substack{0\leq k\leq n \\ k\neq i }}(a+kb)= 0 $$ has no integer roots for all $n\in\mathbb{N}$.