About summation with p-adic valuation Here is a problem i came across. 
Prove that $$\sum_{i=1}^{n}\frac{1}{i}$$ is not an integer for $n \geq 2$. The book olympiad number theory by Justin Stevens says that after writing  $\frac{1}{i}$ as $\frac{\frac{n!}{i}}{n!}$, we need to proceed with $p-$adic valuation. He consider $v_2(\frac{n!}{i})$ and somehow gets an equality $v_2(\sum_{i=1}^{n}\frac{n!}{i})=v_2(\frac{n!}{2^{\lfloor log_2n \rfloor}})$. I tried many times but cannot understand how it happened. Please sirs can you explain me?

$v_2(m)$ is the number of times $2$ divides $m$. 
This means that $v_2(m)$ is the number $k$
such that $m\equiv0\pmod {2^k}$ but
$m\not\equiv0\pmod {2^{k+1}}$
If we have a finite sum $A=a_1+\cdots+a_s$ of integers with $v_2(a_1)<v_2(a_j)$
for all $j>1$ then $v_2(A)=v_2(a_1)$. This is because $a_1+\cdots+a_s
\equiv0\pmod{2^k}$ for $k=v_2(a_1)$ but $a_1\not\equiv0\pmod{2^{k+1}}$ and
$a_2+\cdots+a_s\equiv0\pmod{2^{k+1}}$ so that $A\not\equiv0\pmod{2^{k+1}}$.
Now consider the sum $n!/1+n!/2+\cdots+n!/n$.
Take $2^i$ to be the greatest power of $2$ which is $\le i$.
Then $v_2(j)<v_2(2^i)$ for all $j\in\{1,\ldots, n\}$ except $j=2^i$.
Therefore $v_2(n!/j)>v_2(n!/2^i)$ for all $j\in\{1,\ldots, n\}$ except $j=2^i$. (Of course all the $n!/j$ are integers.) Therefore
$$v_2(n!/1+n!/2+\cdots+n!/n)=v_2(n!/2^i)=v_2(n!)-i<v_2(n!)$$
as long as $n\ge2$. Thus $n!$ cannot divide the integer
$n!/1+n!/2+\cdots+n!/n$, equivalently $1/1+1/2+\cdots+1/n$ isn't an integer.
