An identity involving binomial coefficients

Question

Prove the following identity $$\displaystyle \sum_{i+j=m}\frac{(n-1) \binom{ai+n-1}{i} \binom{aj+1}{j}}{(ai+n-1)(aj+1)} = \frac{n\binom{am+n}{m}}{am+n}$$ where $i = 0,1,\cdots,m$ and $m, n$ are positive integers and $a$ is a positive integer or even a fraction.

One complete proof can be found in that famous book "Concrete Mathematics" by Ronald L. Graham, Donald E. Knuth, and Oren Patashnik which is widely used as a textbook in many computer science departments.

Another complete proof can be found in the book written by Egorychev which is directly using the inversion rule of residue, see page 49 in the book "Integral representation and the computation of combinatorial sums". If you are reading that book, keep in mind that the index under $\sum$ can be extended to infinity because the contour integral has no pole when $k>n$.

Answer 1

This is a nice example to apply the inversion rule of formal power series stated as rule 4 in G.P. Egorychevs Integral Representations and the Computation of Combinatorial Sums section 1.2.2.

I think it's worthwhile to present a complete proof. Here we use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a formal power series. We show

The following is valid for $m,n\geq 1$ and $a\in\mathbb{R}$ appropriate \begin{align*} \sum_{{i+j=m}\atop{i,j\geq 0}}&\frac{n-1}{ai+n-1}\binom{ai+n-1}{i}\frac{1}{aj+1}\binom{aj+1}{j} =\frac{n}{am+n}\binom{am+n}{m} \end{align*}

$$ $$

We obtain \begin{align*} \sum_{{i+j=m}\atop{i,j\geq 0}}&\frac{n-1}{ai+n-1}\binom{ai+n-1}{i}\frac{1}{aj+1}\binom{aj+1}{j}\\ &=\sum_{i=0}^m\frac{n-1}{ai+n-1}\binom{ai+n-1}{i}\frac{1}{am-ai+1}\binom{am-ai+1}{m-i}\\ &=\sum_{i=0}^\infty\left\{\binom{ai+n-1}{i}-a\binom{ai+n-2}{i-1}\right\}\\ &\qquad\qquad\cdot\left\{\binom{am-ai+1}{m-i}-a\binom{am-ai}{m-i-1}\right\}\tag{1}\\ &=\sum_{i=0}^{\infty}\left([z^i](1+z)^{ai+n-1}-a[z^{i-1}](1+z)^{ai+n-2}\right)\\ &\qquad\qquad\cdot\left([w^{m-i}](1+w)^{a(m-i)+1}-a[w^{m-i-1}](1+w)^{a(m-i)}\right)\\ &=[w^m](1+w)^{am}(1+w-aw)\\ &\qquad\qquad\cdot\sum_{i=0}^{\infty}\left(\frac{w}{(1+w)^a}\right)^i[z^i](1+z)^{n-2}(1+z-az)(1+z)^{ai}\tag{2}\\ &=[w^m](1+w)^{am}(1+w-aw)\left.\left((1+z)^{n-1}\right)\right|_{z=w}\tag{3}\\ &=[w^m](1+w)^{am+n}-a[w^{m-1}](1+w)^{am+n-1}\\ &=\binom{am+n}{m}-a\binom{am+n-1}{m-1}\\ &=\frac{n}{am+n}\binom{am+n}{m} \end{align*} and the claim follows.

Comment:

• In (1) we use the identity $$\frac{q}{pk+q}\binom{pk+q}{k}=\binom{pk+q}{k}-p\binom{pk+q-1}{k-1}$$ We also set the upper limit of the sum to $\infty$ without changing anything since we are only adding zero.

• In (2) we rearrange the sum, use the linearity of the coefficient of operator and the rule $[z^{n-k}]A(z)=[z^n]z^kA(z)$.

• In (3) we use the inversion rule

\begin{align*} \sum_{i=0}^\infty w^i [z^i]A(z)f^i(z)=\left.\left(A(z)\frac{f(z)}{f(z)-zf^{\prime}(z)}\right)\right|_{z=g(w)} \end{align*} with $g(w)$ the inverse of $w=\frac{z}{g(z)}$. We get from (2) \begin{align*} A(z)&=(1+z)^{n-2}(1+z-az)\\ f(z)&=(1+z)^a \end{align*} and obtain \begin{align*} A(z)\frac{f(z)}{f(z)-zf^{\prime}(z)}&=(1+z)^{n-2}(1+z-az)\frac{(1+z)^a}{(1+z)^a-az(a+z)^{a-1}}\\ &=(1+z)^{n-1} \end{align*} Since $w=\frac{z}{f(z)}=\frac{z}{(1+z)^a}$ we apply in (3) the substitution $z=w$.

Answer 2

Suppose we seek to verify that

$$\sum_{k=0}^m \frac{q}{pk+q} {pk+q\choose k} {pm-pk\choose m-k} = {mp+q\choose m}.$$

Observe that

$${pk+q\choose k} = \frac{pk+q}{k} {pk+q-1\choose k-1}$$

so that $${pk+q\choose k} - p {pk+q-1\choose k-1} = \frac{q}{k} {pk+q-1\choose k-1} = \frac{q}{pk+q} {pk+q\choose k}.$$

This yields two pieces for the sum, call them $S_1$

$$\sum_{k=0}^m {pk+q\choose k} {pm-pk\choose m-k}$$

and $S_2$

$$- p \sum_{k=0}^m {pk+q-1\choose k-1} {pm-pk\choose m-k}.$$

For $S_1$ introduce the integrals

$${pk+q\choose k} = \frac{1}{2\pi i} \int_{|z|=\gamma} \frac{(1+z)^{pk+q}}{z^{k+1}} \; dz$$

and

$${pm-pk\choose m-k} = \frac{1}{2\pi i} \int_{|w|=\varepsilon} \frac{(1+w)^{pm-pk}}{w^{m-k+1}} \; dw.$$

The second one controls the range of the sum because the pole at zero vanishes when $k\gt m$ so we may extend $k$ to infinity, getting for the sum

$$\frac{1}{2\pi i} \int_{|w|=\varepsilon} \frac{(1+w)^{pm}}{w^{m+1}} \frac{1}{2\pi i} \int_{|z|=\gamma} \frac{(1+z)^{q}}{z} \sum_{k\ge 0} \frac{w^k}{z^k} \frac{(1+z)^{pk}}{(1+w)^{pk}} \; dz\; dw \\ = \frac{1}{2\pi i} \int_{|w|=\varepsilon} \frac{(1+w)^{pm}}{w^{m+1}} \frac{1}{2\pi i} \int_{|z|=\gamma} \frac{(1+z)^{q}}{z} \frac{1}{1-w(1+z)^p/z/(1+w)^p} \; dz\; dw \\ = \frac{1}{2\pi i} \int_{|w|=\varepsilon} \frac{(1+w)^{pm+p}}{w^{m+1}} \frac{1}{2\pi i} \int_{|z|=\gamma} (1+z)^{q} \frac{1}{z(1+w)^p-w(1+z)^p} \; dz\; dw.$$

We require convergence of the geometric series i.e. $|w(1+z)^p| \lt |z(1+w)^p|.$ This says that $\varepsilon/\gamma \lt |(1+w)^p/(1+z)^p|.$ We also have $|(1+w)^p/(1+z)^p| > (3/5)^p$ supposing that $\varepsilon,\gamma \lt \frac{1}{4}$. Hence we may take $\varepsilon = 1/Q$ and $\gamma = 2^p/Q$ with $Q\ge 2^{p+3}$, $Q$ large. Now $z=w$ is the only pole inside the contour and it is simple. The latter follows from the fact that when setting $z=w$ in the derivative

$$\left.(1+w)^p - pw(1+z)^{p-1}\right|_{z=w} \\ = (1+w)^p - pw (1+w)^{p-1} = (1+w)^{p-1} (1+w-pw).$$

we can choose $\varepsilon$ small enough such that $|1+w-wp| > 0$ so the pole is order one which yields

$$\frac{1}{2\pi i} \int_{|w|=\varepsilon} \frac{(1+w)^{pm+p}}{w^{m+1}} (1+w)^q \frac{1}{(1+w)^{p-1}} \frac{1}{1+w-pw} \; dw \\ = \frac{1}{2\pi i} \int_{|w|=\varepsilon} \frac{(1+w)^{pm+q+1}}{w^{m+1}} \frac{1}{1+w-pw} \; dw.$$

Following exactly the same procedure we obtain for $S_2$

$$-p \frac{1}{2\pi i} \int_{|w|=\varepsilon} \frac{(1+w)^{pm+q}}{w^{m}} \frac{1}{1+w-pw} \; dw.$$

Adding these two pieces now yields

$$\frac{1}{2\pi i} \int_{|w|=\varepsilon} \frac{(1+w)^{pm+q}}{w^{m}} \left(\frac{1+w}{w} - p\right) \frac{1}{1+w-pw} \; dw \\ = \frac{1}{2\pi i} \int_{|w|=\varepsilon} \frac{(1+w)^{pm+q}}{w^{m+1}} \; dw = {pm+q\choose m}.$$

Concerning the $p-1$ other poles we can factor

$$z(1+w)^p-w(1+z)^p = \sum_{q=0}^p {p\choose q} (zw^q-wz^q) = (z-w) + zw \sum_{q=2}^p {p\choose q} (w^{q-1}-z^{q-1}) \\ = (z-w)+ zw (w-z) \sum_{q=2}^p {p\choose q} \sum_{r=0}^{q-2} z^r w^{q-2-r}.$$

Hence we need to show that

$$1 - zw \sum_{q=2}^p {p\choose q} \sum_{r=0}^{q-2} z^r w^{q-2-r}.$$

does not have any roots inside $|z|=\gamma.$ Exchangeing summations and re-writing,

$$\sum_{r=0}^{p-2} z^r \sum_{q=r+2}^p {p\choose q} w^{q-2-r} = \frac{1}{zw}.$$

Choosing $Q$ large and using that $\varepsilon,\gamma \ll 1$ we can make the RHS as large as we want in modulus yet the LHS is a finite sum bounded above by a constant in $p$ and hence equality does not occur in this domain.

Remark Mon Jan 25 2016.

An alternate proof from the integral

$$\frac{1}{2\pi i} \int_{|w|=\varepsilon} \frac{(1+w)^{pm}}{w^{m+1}} \sum_{k\ge 0} \frac{w^k}{(1+w)^{pk}} \frac{1}{2\pi i} \int_{|z|=\gamma} \frac{(1+z)^{q}}{z^{k+1}} (1+z)^{pk} \; dz\; dw$$

Now put

$$u = \frac{z}{(1+z)^p} \quad\text{and introduce}\quad g(u) = z.$$

Note that the origin in $z$ gets mapped to the origin in $u.$

We then have $$du = \left(\frac{1}{(1+z)^p} - p\frac{z}{(1+z)^{p+1}}\right) \; dz = \left(\frac{u}{g(u)} - \frac{pu}{1+g(u)}\right) \; dz$$

and $$dz = \frac{1}{u}\frac{g(u) (1+g(u))}{1 + g(u) - p g(u)} \; du.$$

This yields

$$\frac{1}{2\pi i} \int_{|w|=\varepsilon} \frac{(1+w)^{pm}}{w^{m+1}} \sum_{k\ge 0} \frac{w^k}{(1+w)^{pk}} \\ \times \frac{1}{2\pi i} \int_{|u|=\gamma} \frac{1}{g(u) u^{k}} (1+g(u))^q \frac{1}{u}\frac{g(u) (1+g(u))}{1 + g(u) - p g(u)} \; du\; dw$$

or $$\frac{1}{2\pi i} \int_{|w|=\varepsilon} \frac{(1+w)^{pm}}{w^{m+1}} \left.(1+g(u))^q \frac{1+g(u)}{1 + g(u) - p g(u)} \right|_{u=w/(1+w)^p} \; dw.$$

Now observe that $g(w/(1+w)^p) = w$ by definition so we get

$$\frac{1}{2\pi i} \int_{|w|=\varepsilon} \frac{(1+w)^{pm}}{w^{m+1}} (1+w)^q \frac{1+w}{1 + w - p w} \; dw \\ = \frac{1}{2\pi i} \int_{|w|=\varepsilon} \frac{(1+w)^{pm+q+1}}{w^{m+1}} \frac{1}{1 + w - p w} \; dw.$$

This is exactly the same as before and the rest of the proof continues unchanged. Concerning $g(u)$ it is analytic in a neighborhood of the origin and has a Taylor series there whose coefficients can be calculated for $n\ge 1$ (the value for $n=0$ must be set to zero):

$$\frac{1}{n} \frac{1}{2\pi i} \int_{|u|=\gamma} \frac{1}{u^n} g'(u) \; du = \frac{1}{n} \frac{1}{2\pi i} \int_{|z|=\gamma} \frac{(1+z)^{pn}}{z^n} \; dz \\ = \frac{1}{n} {np\choose n-1} = \frac{1}{n(p-1)+1} {np\choose n}.$$

The radius of convergence of the series was shown to equal $(p-1)^{p-1}/p^p$ at the following link which is MSE 4983957. Note also that there is a potential pole in the last integral involving $g(u)$ when $1+g(u)-pg(u)=0$ or $g(u) = 1/(p-1)$ but this gives $u=1/(p-1)/(p/(p-1))^p = (p-1)^{p-1}/p^p$ and we take $\gamma$ less than the radius of convergence, so this pole is not inside the contour, making for a zero contribution. Note furthermore that our choice of $Q$ excludes the pole at $w=1/(p-1)$ from inside the contour.

Answer 3

Follow the same reasoning when proving this identity, $$\displaystyle \sum_{i=0}^m\frac{q }{pi+q}\binom{pi+q}{i} \binom{pm-pi}{m-i} = \binom{mp+q}{m}$$ You can just read the book yourself. Too difficult to post the book here!

it can be proved using the inversion rule of residue, see page 49 Integral representation and the computation of combinatorial sums