Sum of these quotient can not be integer Suppose $a$ and $b$ are positive integers such that are relatively prime (i.e., $\gcd(a,b)=1$).
Prove  that, for all $n\in \mathbb{N}$, the sum
$$                    
\frac{1}{a}+\frac{1}{a+b}+\frac{1}{a+2b}+\cdots+\frac{1}{a+nb}
$$
is not an integer.
I think I have tried many ways I could, but none led me to the complete answer. Do you have any idea?  
 A: Partial answer:
If $b$ is odd, then the problem is easy: pick $k$ maximal so that $2^k$ divides some $a+jb$ for $1 \leq j \leq n$. Such a $j$ must exists, since $n \geq 2$ (the case $n=1$ is trivial, and false when $a=n=1$) and at least one of $a, a+b$ must be even.
Now, if $2^k$ divides two of the terms, it is easy to find in between one term divisible by $2^{k+1}$ which contradicts the maximality of $k$.
Therefore $2^k$ divides exactly one of the terms. The conclusion follows immediately by bringing the other fractions to the same denominator, and observing that the denominator is divisible by at most $2^{k-1}$.
A: A good way to demonstrate it could be to find an integer K such that, multiplying the sum by K, the result is not an integer. Finding this K would clearly imply that the sum is not an integer.
For example, we could choose as K a value obtained starting from the product $a(a+b)(a+2b)(a+3b)...(a+nb)$ and eliminating one factor (or more factors) in the sequence. Multiplying the whole sum by this K, all elements of the sum except one (or except those corresponding to the eliminated terms) become integers. Showing that the product of the remaining term/terms with K is not an integer would provide the proof. 
