During a tedious calculation, I arrived at a power series of the form which I want to compute:

$$ \sum\limits_{n=m}^\infty r^{2n} \binom{2n}{n-m} $$

But from the search I so far did in other threads, I saw only ad-hoc methods whose conditions I don't think suit here. My current attempts have not been fruitful, but I hope someone would perhaps nudge me in the right direction on how to deal with this.

  • $\begingroup$ To do what with the series? $\endgroup$ Oct 13, 2019 at 19:48
  • $\begingroup$ Give an explicit function for such a series $\endgroup$ Oct 13, 2019 at 19:49
  • $\begingroup$ Compute the series of $(1-4x^2)^{-1/2}$ using the binomial theorem. $\endgroup$ Oct 13, 2019 at 19:52
  • $\begingroup$ When you say binomial theorem, do you mean regarding it as a binomial series? $\endgroup$ Oct 13, 2019 at 20:02
  • $\begingroup$ This one. $\endgroup$ Oct 13, 2019 at 20:14

2 Answers 2


We find in H. Wilf's Generatingfunctionology formula (2.5.15) providing a generating function of the shifted central binomial coefficients in the form \begin{align*} \frac{1}{\sqrt{1-4x}}\left(\frac{1-\sqrt{1-4x}}{2x}\right)^k=\sum_{n}\binom{2n+k}{n}x^n\tag{1} \end{align*}

We can adapt (1) easily and obtain \begin{align*} \color{blue}{\sum\limits_{n=m}^\infty}&\color{blue}{ \binom{2n}{n-m}r^{2n}}\\ &=\sum_{n=0}^\infty\binom{2n+2m}{n}r^{2n+2m}\\ &=r^{2m}\sum_{n=0}^\infty\binom{2n+2m}{n}\left(r^2\right)^n\\ &\,\,\color{blue}{=\frac{1}{\sqrt{1-4r^2}}\left(\frac{1-\sqrt{1-4r^2}}{2r}\right)^{2m}} \end{align*}

  • $\begingroup$ Thank you for your answer, and even more for what seems to be a very useful reference. $\endgroup$ Oct 14, 2019 at 19:54
  • 2
    $\begingroup$ @Keen-ameteur: You're welcome. Good to see the reference is interesting for you. It is often cited (see e.g. this post). $\endgroup$ Oct 14, 2019 at 20:02

This answer is based upon the Lagrange inversion theorem. We follow the paper Lagrange Inversion: when and how by R. Sprugnoli etal. It is convenient to apply formula G6 from the paper, which states that if there are functions $F(u)$ and $\phi(u)$, so that \begin{align*} C_n=[u^n]F(u)\phi(u)^n\tag{1} \end{align*} the following is valid \begin{align*} f(r)&=\sum_{n=0}^{\infty}C_nr^n\\ &=\sum_{n=0}^{\infty}[u^n]F(u)\phi(u)^nr^n\\ &=\left.\frac{F(u)}{1-r\phi^{\prime}(u)}\right|_{u=r\phi(u)}\tag{2} \end{align*} $[u^n]$ is the coefficient of operator denoting the coefficient of $u^n$ in a series.

We obtain \begin{align*} \color{blue}{\sum_{n=m}^\infty r^{2n}\binom{2n}{n-m}}&=\sum_{n=0}^\infty\binom{2n+2m}{n}r^{2n+2m}\tag{3}\\ &=r^{2m}\sum_{n=0}^\infty[u^n](1+u)^{2n+2m}r^{2n}\tag{4}\\ &=r^{2m}\left.\frac{(1+u)^{2m}}{1-r^2\cdot 2(1+u)}\right|_{u=r^2(1+u)^2}\tag{5}\\ &=r^{2m}\frac{(1+u)^{2m+1}}{1-u}\tag{6}\\ \end{align*}


  • In (3) we shift the index to start with $n=0$.

  • In (4) we apply the coefficient of operator $[u^n](1+u)^{2n+2m}=\binom{2n+2m}{n}$.

  • In (5) we apply (2) with $F(u)=(1+u)^{2m}, \phi(u)=(1+u)^2$ evaluated at $r^2$.

  • In (6) we substitute $r^2=\frac{u}{(1+u)^2}$ and do some simplifications.

The next step is to solve the quadratic equation $u=r^2(1+u)^2$ (see (5)) and take the solution $u=u(r)$ which can be expanded as generating function around $0$. We obtain \begin{align*} u_{0,1}(r)=\frac{1}{2r^2}\left(1\pm\sqrt{1-4r^2}\right)-1 \end{align*}

We take the solution $u_0=\frac{1}{2r^2}\left(1\color{blue}{-}\sqrt{1-4r^2}\right)-1$ and obtain \begin{align*} 1+u&=\frac{1}{2r^2}\left(1-\sqrt{1-4r^2}\right)\\ 1-u&=2-\frac{1}{2r^2}\left(1-\sqrt{1-4r^2}\right)\\ &=\frac{\sqrt{1-4r^2}}{2r^2}\left(1-\sqrt{1-4r^2}\right) \end{align*}

Substituting the results for $1+u$ and $1-u$ in (6) we obtain \begin{align*} \color{blue}{\sum_{n=m}r^{2n}\binom{2n}{n-m}}&=r^{2m}\frac{(1+u)^{2m+1}}{1-u}\\ &=\frac{r^{2m}}{\sqrt{1-4r^2}}\left(\frac{1-\sqrt{1-4r^2}}{2r^2}\right)^{2m}\\ &\,\,\color{blue}{=\frac{1}{\sqrt{1-4r^2}}\left(\frac{1-\sqrt{1-4r^2}}{2r}\right)^{2m}} \end{align*}


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