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?
Answer
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 = $\sum_{i=1}^{n} \frac{1}{i}$
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