for any positive integer $a,b,n$,and $(a,b)=1$,Is $\frac{1}{a+b}+\frac{1}{a+2b}+\cdots+\frac{1}{a+nb}$ non integer,and How to prove that? It's easy to prove that both $\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}$ and $\frac{1}{3}+\frac{1}{5}+\cdots+\frac{1}{2n+1}$ are nonintegers by multiply $2^k$and $3^k$,
and how about the $\frac{1}{a+b}+\frac{1}{a+2b}+\cdots+\frac{1}{a+nb}$?
 A: Suppose first, that $b=1$ or $2$. Then you are done.
Consider now the remaining case $b\geq 3$. One can use a lemma of Erdős:
Lemma: One of $a+b,\ a+2b,\ a+nb$ is divisbile by a prime power $p^\alpha$ such that $p^\alpha>n$.
Suppose that $a+kb$ is divisible by $p^\alpha>n$. According to the lemma we can do that. Then there cannot exist $l$ such that $a+lb$ is also divisbile by $p^\alpha$. As a first step we show this.
Contrary suppose there exists such. From $(a,b)=1$ and $p^\alpha\mid a+kb$ we have $(b,p)=1$, because if we do not, then $p^\alpha>n$ implies $p\mid a$ which contradicts $(a,b)=1$. That is $b^{-1}$ exists modulo $p^\alpha$. Hence,
$$
(a+kb)-(a+lb)=b(k-l) \equiv 0 \pmod{p^\alpha}
$$
yields
$$
k-l\equiv 0 \pmod{p^\alpha},
$$
which is clearly impossible due to $p^\alpha>n$ and $0<|k-l|\leq n$.
So for now we know that there is only $a+kb$ which is divisible by $p^\alpha$. Then in the fraction
$$
\dfrac{\dfrac{(a+b)(a+2b)\dots(a+nb)}{a+b}+\dots+\dfrac{(a+b)(a+2b)\dots(a+nb)}{a+nb}}{(a+b)(a+2b)\dots(a+nb)}
$$
we have that all of the terms except
$$
\dfrac{(a+b)(a+2b)\dots(a+nb)}{a+kb}
$$
are divisible by $p^\alpha$ while the exception is divisible by a power less than $p^\alpha$. That is, the enumerator is divisible by a power less than $p^\alpha$. In the denominator $p^\alpha$ divides the product which shows that
$$
\dfrac{1}{a+b}+\dots+\dfrac{1}{a+nb}
$$
cannot be an integer. I think everything is all right with the proof, but please check.
Edit: I was aksed this question - with the Lemma of Erdős as a tool to answer - on one of my number theory courses during the undergraduate years as a bonus homework. I thought that maybe it is also a result of Erdős (because the applicability of the Lemma) or some other mathematician. It seems that quite lot of people considered the problem and that my proof nearly the same as it was of Erdős. I give you the reference
http://www.renyi.hu/~p_erdos/1932-02.pdf
However, it is in hungarian. You will find a more elementary proof of it on MathOverflow https://mathoverflow.net/questions/39326/reference-request-unit-fraction-equally-spaced-denominators-not-integer
