How to prove the summation $\sum\limits_{i=0}^b {b \choose i}(-1)^i\frac{1}{a+i+1}=\frac{a!b!}{(a+b+1)!}$

Question

I am working on the Beta function integration. After the integral, I want to prove the summation: $$\sum_{i=0}^b {b \choose i}(-1)^i\frac{1}{a+i+1}=\frac{a!b!}{(a+b+1)!}$$ Can anyone give an idea?

Answer 1

Using $\bf{Brian M. Scott}$ explanation, Here I am start summation from $i=0$

We can write $$\displaystyle \int_{0}^{1}x^{a+i} = \frac{1}{a+i+1}$$

So $$\displaystyle \sum_{i=0}^b {b \choose i}(-1)^i\frac{1}{a+i+1} = \sum^{b}_{i=0}(-1)^{i}\cdot \binom{b}{i}\int_{0}^{1}x^{a+i}dx = \int_{0}^{1}\left[x^{a}\cdot \sum_{i=0}^b(-1)^{i}\cdot \binom{b}{i}x^{i}\right]$$

Now Let $$\displaystyle I(a,b)= \int_{0}^{1}x^{a}\cdot (1-x)^{b}dx$$

Now Using Integration by parts

Now Let $$\displaystyle I(a,b) = -\left[\frac{x^{a+1}(1-x)^{b}}{a+1}\right]_{0}^{1}+\frac{b}{a+1}\int_{0}^{1}x^{a+1}\cdot (1-x)^{b-1}dx$$

So we get $$\displaystyle I(a,b) = \frac{b}{a+1}I(a+1,b-1)$$

Repeating this gives us

$$\displaystyle I(a,b) = \frac{b}{a+1}\cdot \frac{b-1}{a+2}.........\frac{1}{a+b}I(a+b,0)$$

So We can write as $$\displaystyle I(a,b) = \frac{b!}{\frac{(a+b)!}{a!}}\times \frac{1}{a+b+1} = \frac{1}{a+b+1}\cdot \frac{1}{\binom{a+b}{a}}$$

So we get $$I(a,b) = \frac{a!\cdot b!}{(a+b+1)!}$$

Answer 2

It took a while, but after suitably rearranging the proposed identity I came up with a combinatorial argument.

First observe that we can rewrite it as

$$(a+1+b)\binom{a+b}a\sum_{k=0}^b\frac{(-1)^k}{a+1+k}\binom{b}k=1\;.\tag{1}$$

Now

$$\begin{align*} \frac{a+1+b}{a+1+k}\binom{a+b}a\binom{b}k&=\frac{(a+1+b)!b!}{(a+1+k)a!b!k!(b-k)!}\\ &=\frac{(a+1+b)!}{(a+1+k)a!k!(b-k)!}\\ &=\frac{(a+1+b)!}{(a+1+k)!(b-k)!}\cdot\frac{(a+k)^{\underline k}}{k!}\\ &=\binom{a+1+b}{a+1+k}\binom{a+k}k\;, \end{align*}$$

where $x^{\underline k}=x(x-1)\ldots(x-k+1)$ is the falling factorial. Thus, we can further rewrite $(1)$ as

$$\sum_{k=0}^b(-1)^k\binom{a+1+b}{a+1+k}\binom{a+k}k=1\;.\tag{2}$$

Imagine that we have a pool of $a+1+b$ players numbered $0$ through $a+b$ in order of increasing height, and we want to form a team of $a+1$ players, making the shortest member of the team the captain. The product

$$\binom{a+1+b}{a+1+k}\binom{a+k}k$$

is the number of ways to make a preliminary selection of $a+1+k$ players, make the shortest of them captain, and then discard $k$ of the $a+k$ taller players to leave a team of $a+1$ with its shortest member as captain.

Now let’s break up the teams according to the identity of the captain. The captain must have one of the numbers $0$ through $b$, since there must be $a$ taller players on the team. Suppose that the captain is number $j$, where $0\le j\le b$. There are $\binom{a+b-j}a$ teams with $j$ as captain, and there are $\binom{a+b-j}{a+k}$ ways to make a preliminary selection of $a+1+k$ players with $j$ as the shortest of them. Each of the $\binom{a+b-j}a$ teams with captain $j$ can be extended to a pool of $a+1+k$ players with $j$ still the shortest in $\binom{b-j}k$ ways. Thus, the alternating sum in $(2)$ counts each of these $\binom{a+b-j}a$ teams

$$\sum_{k=0}^b(-1)^k\binom{b-j}k=\begin{cases} 0,&\text{if }j<b\\ 1,&\text{if }j=b \end{cases}$$

times. There is only one possible team with $b$ as captain, the team whose numbers are $b$ through $a+b$; it is counted once in that sum, and each of the other teams is counted a net of $0$ times. This establishes the identity $(2)$.

Answer 3

hint: $\dfrac{1}{a+i+1} = \displaystyle \int_{0}^1 x^{a+i}dx$, and use integral of sum equals sum of integral with the addition that: $\displaystyle \sum_{i=1}^b (-1)^ix^{a+i} = x^a\cdot \displaystyle \sum_{i=1}^b \binom{b}{i}(-x)^i$