How to prove the following equality: $$\sum_{m=0}^n\frac{(2m-1)!!}{(2m)!!}\cdot\frac{(2n-2m-1)!!}{(2n-2m)!!}=1$$ I have transformed the equality into $$\frac{{n \choose 2n}}{2^{2n}}\sum_{m=0}^n\frac{{m \choose n}^2}{{2m \choose 2n}}=1$$ But I still don't know how to prove the equality.
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The first sum can be seen as the coefficient of $x^n$ in the product between $\frac{1}{\sqrt{1-x}}$ and itself. – Jack D'Aurizio Aug 29 '19 at 18:52
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By applying the general binomial theorem, we can show that $$(1-x)^{-1/2} = \sum_{n=0}^\infty \binom{-1/2}{n} (-x)^n = \sum_{n=0}^\infty \frac{(2n-1)!!}{(2n)!!}x^n$$
By squaring both sides, we have $$\sum_{n=0}^\infty 1 \cdot x^n = (1-x)^{-1} = \sum_{n=0}^\infty \left(\sum_{m=0}^n\frac{(2m-1)!!}{(2m)!!}\frac{(2(n-m)-1)!!}{(2(n-m))!!}\right)x^n$$ from which the result follows by comparing coefficients.
Brian Moehring
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Since $(2m)!!=2^m m!$ and $(2m-1)!!=(2m)!/(2^m m!)$, we have that $$\sum_{m=0}^n\frac{(2m-1)!!}{(2m)!!}\cdot\frac{(2n-2m-1)!!}{(2n-2m)!!}= \sum_{m=0}^n \frac{(2m)!}{2^{2m} m!m!}\cdot \frac{(2(n-m))!}{2^{2(n-m)} (n-m)!(n-m)!}\\= \frac{1}{2^{2n}}\sum_{m=0}^n\binom{2m}{m}\binom{2(n-m)}{n-m}$$ which is equal to $1$ by the Identity for convolution of central binomial coefficients: $\sum\limits_{k=0}^n \binom{2k}{k}\binom{2(n-k)}{n-k}=2^{2n}$
Robert Z
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