I have been trying to show that $$\Bigg[\sum_{n=0}^{\infty}\frac{{2n}\choose{n}}{4^n}x^n\Bigg]^2=\sum_{n=0}^{\infty}x^n \text{ where }0\leq x<1$$ using only combinatorics.

So far, here is what I have tried:

\begin{align} \Bigg[\sum_{n=0}^{\infty}\frac{{2n}\choose{n}}{4^n}x^n\Bigg]^2 &= \sum_{n=0}^{\infty}\sum_{m=0}^\infty\frac{{{2n}\choose{n}} {{2m}\choose{m}}}{4^{m+n}}x^{m+n}\\ &=\sum_{k=0}^{\infty}\sum_{m=0}^{k}\frac{{{2(k-m)}\choose{k-m}} {{2m}\choose{m}}}{4^{k}}x^{k}\\ \end{align}

So, it suffices to show the following: $$\sum_{m=0}^{k}{{2(k-m)}\choose{k-m}} {{2m}\choose{m}}=4^k$$ I can rewrite this like so: $$\sum_{m=0}^{k}\frac{(2k-2m)!(2m)!}{\big[(k-m)!\big]^2\big(m!\big)^2}=4^k$$

Unfortunately, I don't no how to procede from here. Any help would be greatly appreciated.


1 Answer 1


It’s easiest to start from the fact that the central binomial coefficients have the generating function

$$\frac1{\sqrt{1-4x}}=\sum_{n\ge 0}\binom{2n}nx^n\;,$$

so that

$$\sum_{n\ge 0}\frac{\binom{2n}n}{4^n}x^n=\sum_{n\ge 0}\binom{2n}n\left(\frac{x}4\right)^n=\frac1{\sqrt{1-4\left(\frac{x}4\right)}}=\frac1{\sqrt{1-x}}\;,$$

and now the result is immediate when you recall that

$$\sum_{n\ge 0}x^n=\frac1{1-x}\;.$$

  • $\begingroup$ Isn't the product in my question the same as the one in your answer? Also, how would you go about showing that $$\frac{1}{\sqrt{1-4x}}=\sum_{n\geq0}{{2n}\choose{n}}x^n$$ $\endgroup$
    – Hrhm
    Oct 12, 2016 at 3:05
  • $\begingroup$ @Hrhm: Yes, I was temporarily misled by your intermediate step. You can find a proof of the generating function here. You can also deduce it from the generating function for the Catalan numbers, which is derived here. $\endgroup$ Oct 12, 2016 at 3:18

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