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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?

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    Perhaps a more intuitive way to write the numerator (adding up to $\frac1{a+n}$) could be $$ a(a+1)\cdots(a+n-1) + a(a+1)\cdots(a+n-2)(a+n) + \cdots\ + a(a+2)\cdots (a+n-1)(a+n) + (a+1)(a+2)\cdots(a+n) $$where each term has one of the factors missing. I don't know if it can be sorted out any nicer. – Arthur Apr 08 '16 at 00:37
  • @Arthur - Yes of course. That's the definition! But what's this summation in polynomial form, is my question? What's the point of getting the coefficients of (a+b)^n as c(n,k) when we know we can write the sum as (a+b)^n?! –  Apr 08 '16 at 00:39
  • If $N_n$ designates the nth numerator, then we have immediately the recurrence relationship $N_{n}=(a+n)N_{n-1}+1$. from which for example one can deduce that the constant term is $n!$. – Jean Marie Apr 08 '16 at 00:41
  • @JeanMarie it's not exactly n!, it's $\frac{(a+n)!}{(a-1)!}$ – sanketalekar Apr 08 '16 at 00:44

2 Answers2

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

sanketalekar
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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

Henry
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  • There is also a recurrence, so for example the values in the bottom row can be found from the previous row with $6=0+3 \times 2$, while $22=\frac21 \times 2+3\times 6$, and $18=\frac32 \times 6+3\times 3$ and $4=\frac43 \times 3+3 \times 0$ – Henry May 16 '18 at 06:58