How to find the generating function for $a_n = \binom{2n}{n}$?

Question

How to find the generating function for $a_n = \binom{2n}{n}$?

Using Mathematica, I get $A(z) = \frac{1}{\sqrt{1-4z}}$ and I am able to verify it.

However, how to derive it, e.g., from basic generating functions?

Answer 1

Recall that the generating function for the Catalan numbers $C_n = \frac{1}{n+1}{2n \choose n}$ is given by $$F(z) = \frac{1-\sqrt{1-4z}}{2z}$$

Now we have

$$zF(z) = \sum_{n=0}^\infty C_nz^{n+1}$$

Taking the derivative yields

$$F(z) + zF'(z) = \sum_{n=0}^\infty (n+1)C_nz^n = \sum_{n=0}^\infty {2n \choose n} z^n$$

Simplifying gives

$$F(z) + zF'(z) = \frac{1}{\sqrt{1-4z}}$$

Answer 2

The answer from @Olivier lets you verify directly if you know the answer beforehand. In case you want to derive the functions, you can do as follows:

HINT

First we have

$${2n\choose n} =2\bigg(2-\frac{1}{n}\bigg) {2(n-1)\choose n-1}$$

With this, establish a differential equation for $A(z)$ and solve it.

Answer 3

$$\frac{1}{4^n}\binom{2n}{n}=\frac{2}{\pi}\int_{0}^{\pi/2}\cos^{2n}(\theta)\,d\theta \tag{1}$$ due to $2\cos\theta=e^{i\theta}+e^{-i\theta}$ and $\int_{0}^{2\pi}e^{in\theta}e^{-mi\theta}\,d\theta = 2\pi\delta(m,n)$.
Multiplying both sides of $(1)$ by $z^{2n}$ and summing over $n\geq 0$ we get $$ \sum_{n\geq 0}\binom{2n}{n}\frac{z^{2n}}{4^n} = \frac{2}{\pi}\int_{0}^{\pi/2}\frac{d\theta}{1-z^2\cos^2\theta}=\frac{2}{\pi}\int_{0}^{+\infty}\frac{du}{(1-z^2)+u^2}=\frac{1}{\sqrt{1-z^2}}\tag{2} $$ for any $z\in(-1,1)$. By letting $z=2\sqrt{x}$ we get $$ \sum_{n\geq 0}\binom{2n}{n}x^n = \frac{1}{\sqrt{1-4x}}\tag{3} $$ for any $x\in\left(-\frac{1}{4},\frac{1}{4}\right)$.

Answer 4

Hint. If you can observe that, for $n=0,1,2,\cdots,$ $$ \binom{2n}{n}=(-1)^n \binom{-1/2}{n}4^{n} $$then you get

$$ \sum_{n\ge0}\binom{2n}{n}x^n= \sum_{n\ge0} (-1)^n\binom{-1/2}{n}4^nx^n=(1-4x)^{-1/2} ,\qquad |x|<\frac 14, $$

by using the standard Newton's generalized binomial theorem.