Prove that $\sum_{m=0}^n \frac{(2m-1)!!}{(2m)!!}\cdot\frac{(2n-2m-1)!!}{(2n-2m)!!}=1$.

Question

How to prove the following expression: $$\sum_{m=0}^n \frac{(2m-1)!!}{(2m)!!}\cdot\frac{(2n-2m-1)!!}{(2n-2m)!!}=1.$$

I've tried mathematical induction but it's really hard to tie the next case with the supposed one, if you go just inside and try to evaluate the expression you will get some sum which is also hard to evaluate manually.

Is there some formula that can simplify this? Or just some clever approach that can do that?

Maybe we can play around with simple cases and build later ones based on them(as math induction does) and so therefore prove this, I have an idea that we can probably build some really easy equation that can represent all sequences that we need and from this suppose that it's proved for t and then prove this for $t+1$.

Take generating function both sides you should be able to get $\frac{1}{\sqrt{1-x}}\frac{1}{\sqrt{1-x}}=\frac1{1-x}$ which is indeed true — , Mar 13 '25 at 04:37
This question is similar to: How to prove this combinatorial equality?. If you believe it’s different, please [edit] the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. Found using an Approach0 search. — John Omielan, Mar 13 '25 at 05:10
it will be true if we get $$\frac{1}{ \sqrt{1 + x} } \times \frac{1}{ \sqrt{1 + x} } = \frac{1}{1+x}$$ , thanks a lot for a hint — xMellox, Mar 13 '25 at 05:15

Prove that $\sum_{m=0}^n \frac{(2m-1)!!}{(2m)!!}\cdot\frac{(2n-2m-1)!!}{(2n-2m)!!}=1$.

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