What's an intuitive way to compute summation of this series? What's an intuitive way to compute 
$$\log(1)+\log (2)+\log (3)+\cdots+\log (n-1)+\log (n)$$
or 
for $n>a$
$$\log(a)+\log (a+1)+\log (a+2)+\cdots+\log (n-1)+\log (n) $$
I know the answer for general case is $$\log \left(\frac{n!}{(a-1)!}\right)$$

Note that
$$ \log(1)+\log (2)+\log (3)+....\log (n-1)+\log (n)=\log (n!) $$
and 
$$ \log(a)+\log (a +1)+\log (a + 2)+....\log (n-1)+\log (n)= \log \prod_{i=a}^{n}i $$
Note: You need the property

$$ \log(ab) = \log(a)+\log(b). $$

