Question about product of generating functions in a proof that for positive integer $n$, $\sum_{k=0}^n(-1)^k\binom nk\binom{2n-k}n=1$.

Question

I was recently looking at a copy of the Mathematics Magazine from 2004 and was reading Q944 (here). It asks this:

Show that for positive integer $n$, $$\sum_{k=0}^n(-1)^k\binom nk\binom{2n-k}n=1.$$

The solution is here. Basically, if we let $S_n$ be the sum, we find that we can write $$S_n=\sum_{k=0}^na_{n-k}b_k,$$ where $a_k=(-1)^k\binom nk$ and $b_k=\binom{n+k}n$. Then we can find generating functions for $a_k$ and $b_k$. In particular, we find that $$\sum_{k=0}^na_kx^k=(1-x)^n$$ and $$\sum_{k=0}^\infty b_kx^k=\frac1{(1-x)^{n+1}}.$$

So far, all of this makes sense to me. Now for the final step of the solution, we note that $$\sum_{n=0}^\infty S_nx^n=\sum_{n=0}^\infty\left(\sum_{k=0}^na_{n-k}b_k\right)x^n=(1-x)^n\cdot\frac1{(1-x)^{n+1}}.$$

This last step doesn't really make a lot of sense to me. After all, aren't the generating functions for $a_k$ and $b_k$ dependent on $n$? And so, for example, don't we get that $a_1$ means different things depending on what $n$ is?

Sorry if I'm not being clear here--I'm having a bit of trouble formulating exactly what my confusion is. But, basically, if somebody could explain the last step in a bit more detail, that'd be fantastic.

Answer 1

I’d carry out the last step a bit differently. What the first part shows is that $S_n$ is the coefficient of $x^n$ in the product

$$(1-x)^n\cdot\frac1{(1-x)^{n+1}}\;,\tag{1}$$

something that is often written

$$S_n=[x^n]\left((1-x)^n\cdot\frac1{(1-x)^{n+1}}\right)$$

with the $[x^n]$ operator. Clearly, then,

$$\begin{align*} S_n&=[x^n]\left(\frac{(1-x)^n}{(1-x)^{n+1}}\right)\\ &=[x^n]\left(\frac1{1-x}\right)\\ &=[x^n]\sum_{k\ge 0}x^k\\ &=1\;. \end{align*}$$

The $n$ in $(1)$ really does depend on which $S_n$ we’re computing, but $(1)$ simplifies to $\frac1{1-x}$ for all $n$, so in the end we really are looking at one power series.

Answer 2

It might also help to make the two meanings of $n$ explicit by using different letter for the sum index. For arbitrary but fixed integer $n$, we have generating functions $$f(x)=\sum_{k=0}^\infty a_kx^k=(1-x)^n, g(x)=\sum_{k=0}^\infty b_kx^k=\frac1{(1-x)^{n+1}},$$ with $$f(x)g(x)=\sum_{m=0}^\infty\left(\sum_{k=0}^ma_{k}b_{m-k}\right)x^m=(1-x)^n\cdot\frac1{(1-x)^{n+1}}=\frac{1}{1-x}.$$ Now to be precise, we should compare coefficients of $x^n$ on both sides, then the intended identity follows. But the right side does not depend on $n$ anymore, which is partly what causes the confusion. However following slightly modified example shows why that it is important to look at that specific coefficient.

Instead of the original problem, consider $a_k=\binom{n}{k}$,$b_k=\binom{n}k$ (again for arbitrary but fixed integer $n$), then actually $f(x)=\sum_{k=0}^\infty a_kx^k=(1+x)^n$, $g(x)=\sum_{k=0}^\infty b_kx^k=(1+x)^n$, and so $$ f(x)g(x)=(1+x)^n(1+x)^n=(1+x)^{2n}. $$ Now comparing coefficients at $x^m$ on both sides, we see

$$ \sum_{k=0}^m \binom nk \binom{n}{m-k}=\binom{2n}{m}. $$ If our goal now was to prove identity $\sum_{k=0}^n \binom nk \binom{n}{n-k}=\binom{2n}{n}$, we would have to look at coefficient of $x^n$, other coefficients would not help indeed (even though they give more general identity, but you get the point).

Notice that in both examples, $n$ is fixed integer, it is not used as an index in any sum, which hopefully helps to see through the argument.

Answer 3

$$ \begin{align} \sum_{k=0}^n(-1)^k\binom{n}{k}\binom{2n-k}{n} &=\sum_{k=0}^n(-1)^k\binom{n}{k}\binom{2n-k}{n-k}\tag1\\ &=\sum_{k=0}^n(-1)^n\binom{n}{k}\binom{-n-1}{n-k}\tag2\\ &=(-1)^n\binom{-1}{n}\tag3\\[9pt] &=1\tag4 \end{align} $$ Explanation:
$(1)$: symmetry of Pascal's Triangle: $\binom{n}{k}=\binom{n}{n-k}$
$(2)$: negative binomial coefficients
$(3)$: Vandermonde Identity
$(4)$: $\binom{-1}{n}=(-1)^n\binom{n}{n}$ (negative binomial coefficients)

The equation in step $(2)$ is very close to what you are looking at: the convolution of the coefficients for $(1-x)^n$ and $(1-x)^{-n-1}$. Vandermonde's Identity is based on just this sort of product and gives the coefficients for $(1-x)^{-1}$.

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\left.\sum_{k = 0}^{n}\pars{-1}^{k}{n \choose k}{2n - k \choose n} \right\vert_{\ n\ \in\ \mathbb{N}_{\large\ \geq\ 0}}} \\[5mm] = &\ \sum_{k = 0}^{n}\pars{-1}^{k}{n \choose k}\bracks{z^{n}}\pars{1 + z}^{2n - k} \\[5mm] = &\ \bracks{z^{n}}\pars{1 + z}^{2n}\sum_{k = 0}^{n} {n \choose k} \pars{-\,{1 \over 1 + z}}^{k} \\[5mm] = &\ \bracks{z^{n}}\pars{1 + z}^{2n} \pars{1 - {1 \over 1 + z}}^{n} \\[5mm] = &\ \bracks{z^{n}}\pars{1 + z}^{n}z^{n} = \bbx{\large 1} \\ & \end{align}