Generating function for binomial coefficients $\binom{2n+k}{n}$ with fixed $k$

Question

Prove that $$ \frac{1}{\sqrt{1-4t}} \left(\frac{1-\sqrt{1-4t}}{2t}\right)^k = \sum\limits_{n=0}^{\infty}\binom{2n+k}{n}t^n, \quad \forall k\in\mathbb{N}. $$ I tried already by induction over $k$ but i have problems showing the statement holds for $k=0$ or $k=1$.

Answer 1

If we define $$ f_n(t)=\sum_{k=0}^\infty\binom{2k+n}{k}t^k\tag{1} $$ then we have $$ \begin{align} f_n'(t) &=\sum_{k=0}^\infty\binom{2k+n}{k}kt^{k-1}\\ &=\sum_{k=0}^\infty\binom{2k+n-1}{k-1}(2k+n)t^{k-1}\\ &=\sum_{k=0}^\infty\binom{2k+n+1}{k}(2k+n+2)t^k\\[6pt] &=(n+2)f_{n+1}(t)+2tf_{n+1}'(t)\tag{2} \end{align} $$

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If we define $$ \begin{align} g_n(t) &=\frac1{\sqrt{1-4t}}\left(\frac{1-\sqrt{1-4t}}{2t}\right)^n\\ &=\frac1{\sqrt{1-4t}}\left(\frac2{1+\sqrt{1-4t}}\right)^n\tag{3} \end{align} $$ then we have $$ g_n'(t) =\left(\frac{1}{\sqrt{1-4t}^3}+\frac{n+1}{1-4t}\right)\left(\frac2{1+\sqrt{1-4t}}\right)^{n+1}\tag{4} $$ and therefore $$ \begin{align} &(n+2)g_{n+1}(t)+2tg_{n+1}'(t)\\[9pt] &=\frac{n+2}{\sqrt{1-4t}}\left(\frac2{1+\sqrt{1-4t}}\right)^{n+1}\\ &+2t\left(\frac{1}{\sqrt{1-4t}^3}+\frac{n+2}{1-4t}\right)\left(\frac2{1+\sqrt{1-4t}}\right)^{n+2}\\ &=\frac{n+2}{\sqrt{1-4t}}\left(\frac2{1+\sqrt{1-4t}}\right)^{n+1}\\ &+(1-\sqrt{1-4t})\left(\frac{1}{\sqrt{1-4t}^3}+\frac{n+2}{1-4t}\right)\left(\frac2{1+\sqrt{1-4t}}\right)^{n+1}\\ &=\left(\frac{1}{\sqrt{1-4t}^3}+\frac{n+1}{1-4t}\right)\left(\frac2{1+\sqrt{1-4t}}\right)^{n+1}\\[9pt] &=g_n'(t)\tag{5} \end{align} $$

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Equations $(2)$ and $(5)$ ensure that $$ \begin{align} f_n'(t)&=(n+2)f_{n+1}(t)+2tf_{n+1}'(t)=2t^{-\frac n2}\left[t^{\frac{n+2}{2}}f_{n+1}(t)\right]'\\ g_n'(t)&=(n+2)g_{n+1}(t)+2tg_{n+1}'(t)=2t^{-\frac n2}\left[t^{\frac{n+2}{2}}g_{n+1}(t)\right]' \end{align}\tag{6} $$ Furthermore, it follows from $(1)$ and $(3)$ that $$ f_n(0)=g_n(0)=1\tag{7} $$

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The generalized binomial theorem yields $$ \begin{align} (1-4t)^{-1/2} &=1+\frac124\frac{t}{1!}+\frac12\frac324^2\frac{t^2}{2!}+\frac12\frac32\frac524^3\frac{t^3}{3!}+\dots\\ &=\sum_{k=0}^\infty\frac{(2k-1)!!}{k!}2^kt^k\\ &=\sum_{k=0}^\infty\frac{(2k)!}{2^kk!k!}2^kt^k\\ &=\sum_{k=0}^\infty\binom{2k}{k}t^k\tag{8} \end{align} $$ Equation $(8)$ ensures that $f_0(t)=g_0(t)$.

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Equations $(6)$, $(7)$, and $(8)$ ensure that $$ f_n(t)=g_n(t)\tag{9} $$ for all $n\ge0$. That is, $$ \frac1{\sqrt{1-4t}}\left(\frac{1-\sqrt{1-4t}}{2t}\right)^n =\sum_{k=0}^\infty\binom{2k+n}{k}t^k\tag{10} $$

Answer 2

Due to a recent comment on my other answer, I took a second look at this question and tried to apply a double generating function. $$ \begin{align} &\sum_{n=0}^\infty\sum_{k=-n}^\infty\binom{2n+k}{n}x^ny^k\\ &=\sum_{n=0}^\infty\sum_{k=n}^\infty\binom{k}{n}\frac{x^n}{y^{2n}}y^k\\ &=\sum_{n=0}^\infty\frac{x^n}{y^{2n}}\frac{y^n}{(1-y)^{n+1}}\\ &=\frac1{1-y}\frac1{1-\frac{x}{y(1-y)}}\\ &=\frac{y}{y(1-y)-x}\\ &=\frac1{\sqrt{1-4x}}\left(\frac{1+\sqrt{1-4x}}{1+\sqrt{1-4x}-2y}-\frac{1-\sqrt{1-4x}}{1-\sqrt{1-4x}-2y}\right)\\ &=\frac1{\sqrt{1-4x}}\left(\frac{1+\sqrt{1-4x}}{1+\sqrt{1-4x}-2y}+\color{#C00000}{\frac{2x/y}{1+\sqrt{1-4x}-2x/y}}\right)\tag{1} \end{align} $$ The term in red contains those terms with negative powers of $y$. Eliminating those terms yields $$ \begin{align} \sum_{n=0}^\infty\sum_{k=0}^\infty\binom{2n+k}{n}x^ny^k &=\frac1{\sqrt{1-4x}}\frac{1+\sqrt{1-4x}}{1+\sqrt{1-4x}-2y}\\ &=\frac1{\sqrt{1-4x}}\sum_{k=0}^\infty\left(\frac{2y}{1+\sqrt{1-4x}}\right)^k\\ &=\frac1{\sqrt{1-4x}}\sum_{k=0}^\infty\left(\frac{1-\sqrt{1-4x}}{2x}\right)^ky^k\tag{2} \end{align} $$ Equating identical powers of $y$ in $(2)$ shows that $$ \sum_{n=0}^\infty\binom{2n+k}{n}x^n=\frac1{\sqrt{1-4x}}\left(\frac{1-\sqrt{1-4x}}{2x}\right)^k\tag{3} $$

Answer 3

We proceed by induction on $k$.

For $k = 0$, we have $\displaystyle \sum\limits_{m=0}^{\infty} \binom{2m}{m}t^m = \frac{1}{\sqrt{1-4t}}$

and for $k = 1$, we have

\begin{align*} \sum\limits_{m=0}^{\infty} \binom{2m+1}{m+1}t^m &= \sum\limits_{m=0}^{\infty} \left(2 - \frac{1}{m+1}\right)\binom{2m}{m}t^m \\&= \frac{2}{\sqrt{1-4t}} - C(t)\\&= \frac{2}{\sqrt{1-4t}} -\frac{2}{1+\sqrt{1-4t}} \\&= \frac{2}{\sqrt{1-4t}(1+\sqrt{1-4t})}\end{align*}

Where, $\displaystyle C(t) = \sum\limits_{m=0}^{\infty} \frac{1}{m+1}\binom{2m}{m}z^m = \frac{2}{1+\sqrt{1-4t}}$ is the generating function of Catalan Numbers.

Assuming the result holds for $k$ we prove for $k+1$:

\begin{align*}\sum\limits_{m=0}^{\infty} \binom{2m+k+1}{m+k+1}t^m &= \sum\limits_{m=0}^{\infty} \left(2 - \frac{k+1}{m+k+1}\right)\binom{2m+k}{m+k}t^m\\ &= \frac{2^{k+1}(1+\sqrt{1-4t})^{-k}}{\sqrt{1-4t}} - \frac{k+1}{t^{k+1}}\int_0^t \sum\limits_{m=0}^{\infty} \binom{2m+k}{m+k}t^{m+k}\,\mathrm{d}t\\ &= \frac{2^{k+1}(1+\sqrt{1-4t})^{-k}}{\sqrt{1-4t}} - \frac{2^k(k+1)}{t^{k+1}}\int_0^t \frac{t^k(1+\sqrt{1-4t})^{-k}}{\sqrt{1-4t}}\,\mathrm{d}t\\ &= \frac{2^{k+1}(1+\sqrt{1-4t})^{-k}}{\sqrt{1-4t}} - \frac{k+1}{2^kt^{k+1}}\int_0^t \frac{(1-\sqrt{1-4t})^{k}}{\sqrt{1-4t}}\,\mathrm{d}t\\ &= \frac{2^{k+1}(1+\sqrt{1-4t})^{-p}}{\sqrt{1-4t}} - \frac{k+1}{2^{k+1}t^{k+1}} \frac{(1-\sqrt{1-4t})^{k+1}}{k+1}\\ &= \frac{2^{k+1}(1+\sqrt{1-4t})^{-k}}{\sqrt{1-4t}} - 2^{k+1}(1+\sqrt{1-4t})^{-(k+1)}\\ &= \frac{2^{k+1}(1+\sqrt{1-4t})^{-(k+1)}}{\sqrt{1-4t}}\\ \end{align*}

Answer 4

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{{1 \over \root{1 - 4t}}\,\pars{1 - \root{1-4t} \over 2t}^{k} =\sum_{n = 0}^{\infty}{2n + k \choose n}t^{n}\,,\quad \forall k\ \in\ \mathbb{N}}$

With $\ds{0\ <\ t\ <\ {1 \over 4}}$: \begin{align}&\color{#c00000}{\sum_{n = 0}^{\infty}{2n + k \choose n}t^{n}} =\sum_{n = 0}^{\infty}\bracks{\oint_{\verts{z}\ =\ 1^{+}}{\pars{1 + z}^{2n + k} \over z^{n + 1}}% \,{\dd z \over 2\pi\ic}}t^{n} \\[5mm]&=\oint_{\verts{z}\ =\ 1^{+}}{\pars{1 + z}^{k} \over z}\sum_{n = 0}^{\infty} \bracks{\pars{1 + z}^{2}\,t \over z}^{n}\,{\dd z \over 2\pi\ic} =\oint_{\verts{z}\ =\ 1^{+}}{\pars{1 + z}^{k} \over z} {1 \over 1 - \pars{1 + z}^{2}\,t/z}\,{\dd z \over 2\pi\ic} \\[5mm]&=-\,{1 \over t}\oint_{\verts{z}\ =\ 1^{+}} {\pars{1 + z}^{k} \over z^{2} + \pars{2 - 1/t}z + 1} \,{\dd z \over 2\pi\ic} =-\,{1 \over t}\oint_{\verts{z}\ =\ 1^{+}} {\pars{1 + z}^{k} \over \pars{z - z_{-}}\pars{z - z_{+}}}\,{\dd z \over 2\pi\ic} \end{align}

where $\ds{z_{\pm}}$ are the roots of $\ds{z^{2} + \pars{2 - {1 \over t}}z + 1 = 0}$: $$ z_{\pm} = {1 \pm \root{1 - 4t} - 2t \over 2t} $$ such that $\ds{z_{+} > 1}$ and $\ds{\verts{z_{-}} < 1}$.

Then, \begin{align}&\color{#c00000}{\sum_{n = 0}^{\infty}{2n + k \choose n}t^{n}} =-\,{1 \over t}\,{\pars{1 + z_{-}}^{k} \over z_{-} - z_{+}} =-\,{1 \over t}\pars{-\,{t \over \root{1 - 4t}}} \pars{1 + {1 - \root{1 - 4t} - 2t\over 2t}}^{k} \end{align}

which simplifies to $$\color{#66f}{\large% {1 \over \root{1 - 4t}}\,\pars{1 - \root{1-4t} \over 2t}^{k} =\sum_{n = 0}^{\infty}{2n + k \choose n}t^{n}} $$

Answer 5

This can be shown using a variant of Lagrange Inversion. Introduce $$T(z) = w = \sqrt{1-4z}$$ so that $$z = \frac{1}{4} (1-w^2)$$ and $$dz = -\frac{1}{2} w \; dw.$$

Then we seek to compute $$[z^n] \frac{1}{T(z)} \left(\frac{1-T(z)}{2z}\right)^k = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} \frac{1}{T(z)} \left(\frac{1-T(z)}{2z}\right)^k dz.$$

Using the substitution this becomes $$- \frac{1}{2\pi i} \int_{|w-1| = \epsilon} \frac{4^{n+1}}{(1-w^2)^{n+1}} \frac{1}{w} \left(\frac{1-w}{1/2(1-w^2)}\right)^k \frac{1}{2} w \; dw \\ = - \frac{1}{2} \frac{1}{2\pi i} \int_{|w-1| = \epsilon} \frac{4^{n+1}}{(1-w^2)^{n+1}} \left(\frac{2}{1+w}\right)^k dw \\ = - \frac{1}{2\pi i} \int_{|w-1| = \epsilon} \frac{1}{(1-w)^{n+1}} \frac{2^{2n+k+1}}{(1+w)^{n+k+1}} dw \\ = (-1)^n \frac{1}{2\pi i} \int_{|w-1| = \epsilon} \frac{1}{(w-1)^{n+1}} \frac{2^{2n+k+1}}{(1+w)^{n+k+1}} dw.$$

Now expand the second fraction about $w=1$ to get $$\frac{1}{(1+w)^{n+k+1}} = \frac{1}{(2+(w-1))^{n+k+1}} = \frac{1}{2^{n+k+1}} \frac{1}{(1+1/2(w-1))^{n+k+1}} \\ = \frac{1}{2^{n+k+1}} \sum_{q\ge 0} {q+n+k\choose q} (-1)^q \frac{(w-1)^q}{2^q}.$$

We need $q=n$ for the residue in the integral, getting the final answer $$(-1)^n \times 2^{2n+k+1} \times \frac{1}{2^{n+k+1}} {2n+k\choose n} (-1)^n \frac{1}{2^n} = {2n+k\choose n}.$$

This MSE link has another calculation that is quite similar.

Answer 6

The methodology is that you have got to make the brackets as the form expandable as the sum of polynomials with binomial coefficients And guess the clues through differentiation. It works no matter what the k is