Find a closed-form solution to the following summation

Question

I am solving a summation that appears in a paper, it claims that $$\sum_{n=1}^{\infty} \binom{2n}{n+k}z^n=\bigg(\frac{4z}{(1+\sqrt{1-4z})^2}\bigg)^k$$ I found this identity here in equation (66) $$\sum_{n=0}^{\infty} \binom{2n+k}{n}z^n=1+\sum_{n=1}^{\infty} \binom{2n+k}{n}z^n=1+\sum_{n=1}^{\infty} \binom{2n+k}{n+k}z^n =\bigg(\frac{2}{1+\sqrt{1-4z}}\bigg)^k\frac{1}{\sqrt{1-4z}}$$ I didn't found the paper that has the proof of the identity above, but I notice that $$\binom{2n}{n+k}+\sum_{j=0}^{k-1} \binom{2n+j}{n+k-1}=\binom{2n+k}{n+k}$$ Therefore we obtain $$\sum_{n=1}^{\infty} \binom{2n}{n+k}z^n=\bigg(\frac{2}{1+\sqrt{1-4z}}\bigg)^k\frac{1}{\sqrt{1-4z}}-1-\sum_{n=1}^{\infty} \sum_{j=0}^{k-1} \binom{2n+j}{n+k-1}z^n$$ Yet I have no idea for the last summation above, any help would be appreciated.

Answer 1

We use an approach based upon the Lagrange inversion theorem and prove for $k$ being a non-negative integer \begin{align*} \color{blue}{\sum_{n=0}^\infty\binom{2n+k}{n}z^n=\left(\frac{2}{1+\sqrt{1-4z}}\right)^k\frac{1}{\sqrt{1-4z}}\qquad\qquad |z|<\frac{1}{4}} \end{align*}

We follow thereby the paper Lagrange Inversion: when and how by R. Sprugnoli etal. We consider the generating function \begin{align*} F(z)=\sum_{n=0}^{\infty}a_nz^n=\sum_{n=0}^\infty\binom{2n+k}{n}z^n \end{align*} and apply formula (G6) from the paper. The formula (G6) tells us that if there are functions $A(u)$ and $\phi(u)$, so that the coefficient $a_n$ admits a representation \begin{align*} a_n=[u^n]A(u)\phi(u)^n \end{align*} where $[u^n]$ denotes the coefficient of $u^n$ of a series, then the following is valid: \begin{align*} F(z)&=\sum_{n=0}^{\infty}[u^n]A(u)\phi(u)^nz^n\\ &=\left.\frac{A(u)}{1-z\phi^{\prime}(u)}\right|_{u=z\phi(u)}\tag{1} \end{align*}

We obtain \begin{align*} \color{blue}{F(z)}&\color{blue}{=\sum_{n=0}^\infty\binom{2n+k}{n}z^n}\\ &=\sum_{n=0}^\infty[u^n](1+u)^{2n+k}z^n\tag{2}\\ &=\left.\frac{(1+u)^k}{1-z\,\frac{d}{du}(1+u)^2}\right|_{u=z(1+u)^2}\tag{3}\\ &=\frac{(1+u)^k}{1-\frac{u}{(1+u)^2}\cdot 2(1+u)}\tag{4}\\ &=\frac{(1+u)^{k+1}}{1-u}\tag{5}\\ &=\frac{\left(\frac{2}{1+\sqrt{1-4z}}\right)^{k+1}}{2-\frac{2}{1+\sqrt{1-4z}}}\tag{6}\\ &\,\,\color{blue}{=\left(\frac{2}{1+\sqrt{1-4z}}\right)^k\frac{1}{\sqrt{1-4z}}}\tag{7} \end{align*} and the claim follows.

Comment:

• In (2) we use the coefficient of operator $[z^n]$ to denote the coefficient of a series. We observe we can apply the Lagrange Inversion theorem with $A(u)=(1+u)^k$ and $\phi(u)=(1+u)^2$.

• In (3) we use formula (1) with $A$ and $\phi$ as stated in the comment above.

• In (4) we substitute $z=\frac{u}{1+u}$ and do the derivation.

• In (5) we simplify the expression.

• In (6) we recall $u=z(1+u)^2$ which is a quadratic equation in $u=u(z)$. We calculate \begin{align*} u(z)&=\frac{1}{2z}\left(1-2z-\sqrt{1-4z}\right)\\ &=\frac{1}{2z}\left(1-\sqrt{1-4z}\right)-1\\ &=\frac{1}{2z}\,\frac{1-(1-4z)}{1+\sqrt{1-4z}}-1\\ &=\frac{2}{1+\sqrt{1-4z}}-1 \end{align*} and take the solution with the minus sign, since this one can be expanded as generating function.

• In (7) we make again some simplifications to obtain the wanted form.

Note: The relationship with OPs first stated formula is given via \begin{align*} \color{blue}{\sum_{n=0}^\infty\binom{2n}{n+k}z^n} &=\sum_{n=k}^\infty \binom{2n}{n+k}z^n\tag{8}\\ &=\sum_{n=k}^\infty \binom{2n}{n-k}z^n\tag{9}\\ &=\sum_{n=0}^\infty \binom{2n+2k}{n}z^{n+k}\tag{10}\\ &=z^k\left(\frac{2}{1+\sqrt{1-4z}}\right)^{2k}\frac{1}{\sqrt{1-4z}}\tag{11}\\ &\,\,\color{blue}{=\left(\frac{4z}{\left(1+\sqrt{1-4z}\right)^2}\right)^k\frac{1}{\sqrt{1-4z}}} \end{align*}

Comment:

• In (8) we start the right-hand side with index $n=k$, since $\binom{2n}{n+k}=0$ if $n<k$.

• In (9) we use the binomial identity $\binom{p}{q}=\binom{p}{p-q}$.

• In (10) we shift the index to start with $n=0$ again.

• In (11) we use the identity (7) from above with $k\to 2k$ and make finally some simplifications.

Answer 2

The identity

$$\sum_{n\ge 0} {2n+k\choose n} z^n = \left(\frac{2}{1+\sqrt{1-4z}} \right)^k \frac{1}{\sqrt{1-4z}} = Q_k(z)$$

can also be proved with the Cauchy Coefficient Formula, radius of convergence is the distance to the nearest singularity which is at $1/4.$

We obtain

$$[z^n] Q_k(z) = \frac{1}{2\pi i} \int_{|z|=\varepsilon} \frac{1}{z^{n+1}} \left(\frac{2}{1+\sqrt{1-4z}} \right)^k \frac{1}{\sqrt{1-4z}} \; dz.$$

Now we put $w=\sqrt{1-4z}$ with the branch cut on $[1/4,\infty)$ (principal branch of the logarithm) so that we have analyticity in a neighborhood of the origin and $dw = -2 \frac{1}{\sqrt{1-4z}} \; dz$. With $\sqrt{1-4z} = 1-2z-\cdots$ the image of $|z|=\varepsilon$ makes one turn around $w=1$ plus lower order fluctuations so that it may be deformed to a small circle and we get with $z=(1-w^2)/4$

$$- \frac{1}{2} \frac{1}{2\pi i} \int_{|w-1|=\gamma} \frac{4^{n+1}}{(1-w^2)^{n+1}} \left(\frac{2}{1+w} \right)^k \; dw \\ = - \frac{2^{2n+k+1}}{2\pi i} \int_{|w-1|=\gamma} \frac{1}{(1-w)^{n+1}} \frac{1}{(1+w)^{n+1+k}} \; dw \\ = \frac{(-1)^n\times 2^{n}}{2\pi i} \int_{|w-1|=\gamma} \frac{1}{(w-1)^{n+1}} \frac{1}{(1+(w-1)/2)^{n+1+k}} \; dw$$

This yields by the Cauchy Residue Theorem

$$(-1)^n 2^n \times (-1)^n \frac{1}{2^n} {n+n+k\choose n} = {2n+k\choose n}$$

as claimed.

Remark. This identity and the proof appeared at the following MSE link from which it may be obtained by a straightforward square root manipulation.