I'm trying to prove the binomial identity $$\sum_{k=0}^n {n\choose k}^2 = {2n\choose n}.$$ This is a problem from "Fourier Series and Integrals" by Mckean.
So far, I have computed sums of the form $\sum_{n=1}^\infty n^{-s},$ where $s$ is a positive even number. The trick then was to find a Fourier Series such that The Fourier coefficients were of the form $a\cdot n^{-s/2}$ where $a\in\mathbb{C}$. Lastly, I applied Plancherels identity to finish the computation.
However in this case, I do not see how to follow this method. They give me a hint in the exercise:
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$\Big[$ Hint: $\sum_{k=0}^n {n \choose k} e_k=(1+e_1)^n$; also look at $\sum_{k=0}^{2n}{2n\choose k}e_{k-n}=2^{2n}\cos^{2n}\pi x$ $\Big ]$.
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Right now, I am trying to figure out why we consider the summations in the hint and what the connection is between the Hint and the problem I want to solve.
Should I think of $\sum_{k=0}^n {n \choose k} e_k$ as the Fourier series expansion of $f=(1+e_1)^n$? If this is the case, Plancherels identity tells me that $$\sum_{k=1}^{\infty} |\hat{f}(k)|^2=||f||_{L^2}^2.$$ Furthermore, $$\hat{f}(k)={n\choose k}.$$ And thus we have the left hand side in the Binomial identity (except that it starts from $k=0$ but in Plancherels identity we start from $k=1$?).
Lastly, I just 'hope' that $||f||_{L^2}^2=2^{2n}\cos^{2n}\pi x$? Then $$||f||_{L^2}^2=\int_{S^1} |(1+e_1)^n|^2 dn=\int_{S^1} |(1+e_1)|^{2n} dn=\cdots$$
Perhaps I should integrate the above over $x$ instead of $n$, since $e_1=e^{2\pi i x}$? I am pretty confused right now since there are quite a few variables to keep track on.
Is this a good approach or should I do something different? Any help would be really appreciated!
Thanks.