General formula for the numerators? Suppose that $a$ is a natural number. The numerator of $\dfrac {1}{a}$ is $1$. The numerator of $\dfrac {1}{a} + \dfrac {1}{a+1}$ is $2a+1$ [Note: Here for our purpose we don't cancel common factors of the numerator and denominator]. The numerator of $\dfrac {1}{a} + \dfrac {1}{a+1} + \dfrac {1}{a+2}$ is $3a^2+6a+2$. And so on. I tried a lot to come up with a general formula for the numerator of the sum of $n$ consecutive terms but I couldn't. I think there should exist some formula for it same as we have a formula for the coefficients of a binomial expansion. Any idea?   
Saying in a more mathematical language: 
The numerator of $\dfrac {1}{a} + \dfrac {1}{a+1} + \dots + \dfrac {1}{a+n}$ is $b_1+b_2a+ \dots + b_na^n$. What are the $b_i$'s? 
 A: The expression is:
$\frac{1}{a} + \frac{1}{a+1} + \frac{1}{a+2} ... $
The numerator is going to be sum of the product of all the denominators with one term missing. 
$(a+1) * (a+2) *.... (a+n)$ [a missing]
$+ a*(a+2)... (a+n)... $ [a+1 missing]
.
.
$+ a*(a+1)... (a+n-1)$ [a+n missing]
Each term can be expressed in terms of factorials.
$= \sum_{i = a}^{a+n} \frac{(a+n)!}{(a-1)!i}$ 
$ = \frac{(a+n)!}{(a-1)!} \sum_{i = a}^{a+n} \frac{1}{i}$
I don't think there is a closed form for the numerator though.
EDIT: Thanks to Rick Decker's useful comment, I remembered the forgotten Harmonic numbers, so the sum of 1/i becomes H(a+n) - H(a - 1)
Where H(n) is the $n^{th}$ Harmonic number = $\sum_{i=1}^{n} \frac{1}{i}$
A: The $b_i$ are simple multiples of unsigned Stirling numbers of the first kind.  In a common notation, the coefficient of $x^k$ in the combined sum of $n$ terms is $(k+1)\left[{n \atop k+1}\right]$ 
Take the triangle of values in Wikipedia
which starts 
n \ k    0    1    2    3    4   

0        1
1        0    1   
2        0    1    1    
3        0    2    3    1
4        0    6   11    6    1

multiply the values in the triangle by the index at the top and drop the initial column and the indices to get a revised triangle 
              1   
              1    2    
              2    6    3
              6   22   18    4

This new triangle corresponds to the coefficients in  


*

*$\dfrac{1}{a} = \dfrac{1}{a}$

*$\dfrac {1}{a} + \dfrac {1}{a+1} =\dfrac {1+2a}{a(a+1)}$

*$\dfrac {1}{a} + \dfrac {1}{a+1} + \dfrac {1}{a+2}=\dfrac {2+6a+3a^2}{a(a+1)(a+2)}$

*$\dfrac {1}{a} + \dfrac {1}{a+1} + \dfrac {1}{a+2}+ \dfrac {1}{a+3}=\dfrac {6+22a+18a^2+4a^3}{a(a+1)(a+2)(a+3)}$


and so on
