I got this question from "An Introduction to Analytic Number Theory". I am looking for a just a hint. The question is to prove that, for all $n$ greater than 1, $$\sum_{k=1}^{n}\frac{1}{k}$$ is never an integer. My idea to prove this is to do induction on $n$ and show that each sum is written in lowest for as $$\frac{a}{n!}$$ for some integer $a$. Starting with $n=2$ as the base case. $\frac{3}{2}=\frac{3}{2!}$ which is in its lowest form and is not an integer. So now for our inductive step. We must show that because $$\sum_{k=1}^{\infty}\frac{1}{k}=\frac{a}{n!}$$ with $\gcd(a,n!)=1$ the same is true for $n$. First, we will add $\frac{1}{n+1}$ to $\frac{a}{n!}$. We have:
$$\frac{a}{n!}+\frac{1}{n+1}=\frac{a(n+1)+n!}{(n+1)!}.$$ Now, I need to show that $$\gcd\left( a(n+1)+n!,(n+1)! \right)$$ From here is the part I need a hint for. I think I have to use Bezout's lemma. Can you help?
