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It's not difficult to show that
$$(1-z^2)^{-1/2}=\sum_{n=0}^\infty \binom{2n}{n}2^{-2n}z^{2n}$$
On the other hand, we have $(1-z^2)^{-1}=\sum z^{2n}$. Squaring the first power series and comparing terms gives us
$$\sum_{k=0}^n \binom{2k}{k}\binom{2(n-k)}{n-k}2^{-2n}=1$$
that is,
$$\sum_{k=0}^n \binom{2k}{k}\binom{2(n-k)}{n-k}=2^{2n}$$
My question: is there a more direct, combinatorial proof of this identity? I've been racking my brains trying to come up with one but I'm not having much success.
It is possible to give a direct combinatorial proof, but it is quite difficult to find it.
One possibility is to use paths between points with integer coordinates and steps $(1,1)$ and $(1,-1)$.
1) $\binom{2n}{n}$ counts all paths from $(0,0)$ to $(2n,0)$.
2) $2^{2n}$ counts all paths starting from $(0,0)$ with $2n$ steps.
3) $\binom{2n}{n}$ counts all paths with $2n$ steps that never touch the $x$-axis again after the start. (This one is not obvious, but can be proved with a bijection.)
Now you can conclude that all paths are a concatenation of a path that returns a certain number of times to the $x$-axis and a path that never does.
Note that the main difficulty here was that the two binomial coefficients are interpreted differently.
Edited to add reference: In Richard P. Stanley: Enumerative Combinatorics Volume 1, Chapter 1, Solution to exercice 2c the following reference is given:
The problem of giving a combinatorial proof was raised by P. Veress and solved by G. Hajos in the 1930s. A recent proof appears in D.J. Kleitman, Studies in Applied Math. 54 (1975), 289 - 292. See also M. Sved, Math. Intelligencer, vol.6, no. 4 (1984), 44-45.
But I have not looked to check which article gives the proof I have outlined above.
Here is another proof, one that I slightly prefer. I'll start with the hardest part.
Lemma. The number of all words of length $n$ in the alphabet $\{A,B\}$ such that no prefix (left factor) of it contains more letters $B$ than $A$, is $\binom n{\lceil n/2\rceil}$.
Instead of these words one may also take, interpreting $A$ as an up-step and $B$ as a down-step, paths as in the answer by Phira that never go below the horizontal axis; or one can formulate as ballot sequences as in Bertand's ballot problem, with the difference that we allow $B$ to catch up with $A$ without overtaking, and that the (non-negative) size of the eventual lead of $A$ is not fixed.
Proof. The following step can be applied to any word for which some prefix does contain more letters $B$ than$~A$: find the smallest prefix for which the excess of the number of its letters $B$ over its letters $A$ is maximal among all prefixes, and change its last letter (which is a $B$) into $A$. There is an inverse step that can be applied to any word with more letters $A$ than letters $B$ (or more generally to a word for which some suffix (right factor) has this property): find the smallest suffix for which the majority of its letters $A$ over its letters $B$ is maximal among all suffixes, and change its first letter (which is an $A$) into $B$. The easiest way to see that these are inverse operations is that the presence of subwords $AB$ has no effect on these operations (in particular the operations will never change such subwords), and that what remains when recursively removing such subwords is a word of the form $BB\ldots BAA\ldots A$, where the last $B$ respectively first $A$ will be changed. Now given a word of length $n$ with $\lceil n/2\rceil$ letters $A$ and $\lfloor n/2\rfloor$ letters $B$, one can iterate the first operation until no prefix contains more letters $B$ than $A$, and conversely given a word of length $n$ satisfying that condition, if there are $d\geq0$ more letters $A$ than $B$ in all, one can iterate the reverse operation $\lfloor d/2\rfloor$ times to obtain a word of length $n$ with $\lceil n/2\rceil$ letters $A$ and $\lfloor n/2\rfloor$ letters $B$. This bijection proves the lemma. QED
Now to prove the identity of the question, consider the words of length $2n+1$ in which the letters $A$ are in the majority; their number is $2^{2n+1}/2=2^{2n}$. Consider the longest prefix (possibly empty) in which there are as many letters $A$ as $B$; it has an even length $2k$, and given that length there are $\binom{2k}k$ possibilities for this prefix. The next letter is necessarily an $A$, and after that there is a suffix of length $2n-2k$ in which no prefix (of that suffix) contains more letters $B$ than $A$. By the lemma there are $\binom{2n-2k}{n-k}$ of them, whence the result.
There is also a probabilistic proof of this identity.
Start with an urn containing one red marble and one blue marble. Make a series of $n$ draws from the urn; for each draw, remove a random ball in the urn, then put it back, along with two extra balls of the same color. We then ask, what is the probability that exactly $k$ of the draws were red?
The probability that the first $k$ draws are red and the last $n-k$ are blue is $$ \frac12\cdot\frac{3}4\cdot\frac5{6}\cdots\cdot\frac{2k-1}{2k}\cdot\frac1{2k+2}\cdot\frac{3}{2k+4}\cdots\frac{2(n-k)-1}{2n}=\frac{(2k-1)!!(2(n-k)-1)!!}{(2n)!!} $$ where $n!!=\prod_{k=0}^{\lfloor n/2\rfloor-1}(n-2k)=n(n-2)(n-4)\cdots$.
It is not hard to see that every sequence of $k$ red and $n-k$ blue draws has this same probability; rearranging the order of draws just changes the order of the factors in the numerator. Therefore, the probability of $k$ red draws is, using the identities $(2n)!!=2^nn!$ and $(2k-1)!!=\frac{(2k)!}{(2k)!!}=\frac{(2k)!}{2^kk!}$, $$ \binom{n}k\frac{(2k-1)!!(2(n-k)-1)!!}{(2n)!!}=\frac{\binom{2k}k\binom{2(n-k)}{n-k}}{2^{2n}} $$ Since these probabilities must sum to $1$, the desired identity follows!
Here is another proof by Egecioglu. It was published as a technical report, not a journal paper, so it's not easy to find.
