Conjecture: $\sum\limits_{n\geq0}\left(\frac12\right)^n\prod\limits_{k=1}^{n}\frac{2n-2k+1}{2n-2k+2}=\sqrt2$ [duplicate]

I am trying to solve a problem I made form myself: proving that $$\sum_{n\geq0}\left(\frac12\right)^n\prod_{k=1}^{n}\frac{2n-2k+1}{2n-2k+2}=\sqrt2$$ The highly accurate powers of Desmos seem to confirm my hunch. But how do I prove it?

I was just messing around with products and generating functions, and then I noticed that the numerical value of the sum in question was suspiciously similar to $$\sqrt2$$, so I conjectured the result. Unfortunately I found this completely by accident and have absolutely no idea of how to prove it.

Feeble attempt:

Define $$S(x)=\sum_{n\geq0}x^n\prod_{k=1}^{n}\frac{2n-2k+1}{2n-2k+2}$$ Which Wolfram says is equal to $$S(x)=\sum_{n\geq0}x^n\frac{(1/2-n)_n}{(-n)_n}$$ With $$\displaystyle (x)_n=\frac{\Gamma(x+n)}{\Gamma(x)}$$. But that doesn't really make sense because $$\Gamma(0)$$ is undefined. So all in all I'm just confused.

Could I have some help?

Edit:

According to the comments, it suffices to prove that $$\sum_{n\geq1}\left(\frac12\right)^n\prod_{k=1}^{n}\frac{2n-2k+1}{2n-2k+2}=\sqrt2-1$$

• Satisfy yourself that $(0)_0=1$. – J.G. Jan 20 '19 at 22:30
• @J.G. I am now satisfied – clathratus Jan 20 '19 at 22:30
• In case of $n=0$, product in your sum is empty product, so you may regard it as 1. (Or just do the sum for $n\geq1$) – Seewoo Lee Jan 20 '19 at 22:31
• I don't think you need to include the $n=0$ case because the product is not defined for it. – Pixel Jan 20 '19 at 22:31
• @Antinous For $n=0$ the product is empty, and thus conventionally defined to equal $1$ – Learner Jan 20 '19 at 22:59

The products can be rewritten as $$\begin{eqnarray*} \sum_{n=0}^{\infty} \frac{(2n-1)!!}{(2n)!!} \frac{1}{2^n} = \sum_{n=0}^{\infty} \binom{2n}{n} \frac{1}{8^n} \end{eqnarray*}$$ Now use $$\begin{eqnarray*} \sum_{n=0}^{\infty} \binom{2n}{n} x^n = \frac{1}{\sqrt{1-4x}}. \end{eqnarray*}$$

Edit: $$\begin{eqnarray*} \prod_{k=1}^{n}\frac{2n-2k+1}{2n-2k+2}&=&\prod_{k=1}^{n}\frac{2k-1}{2k} =\frac{(2n-1)!!}{(2n)!!}\\ &=&\frac{(2n)!}{(2^n n!)^2}=\binom{2n}{n} \frac{1}{2^{2n}}. \end{eqnarray*}$$

• Woah seriously?? (+1) That was fast! By the way, could you include the proof for your magical rewriting of products? – clathratus Jan 20 '19 at 22:36
• Wait actually no: I'm not gonna accept your answer until you show in detail how $$\prod_{k=1}^{n}\frac{2n-2k+1}{2n-2k+2}=\frac{(2n-1)!!}{(2n)!!}=\frac1{2^{2n}}{2n\choose n}$$ – clathratus Jan 20 '19 at 22:47
• Sure ... give me a few mins ? – Donald Splutterwit Jan 20 '19 at 22:48
• Very nice. Thank you! – clathratus Jan 20 '19 at 23:42
• To add some kind of reference for the result on $\binom{2n}n$, I'll mention links some other posts on this site: Generating functions and central binomial coefficient and Show $\sum\limits_{n=0}^{\infty}{2n \choose n}x^n=(1-4x)^{-1/2}$. (And probably more posts about this can be found.) – Martin Sleziak Jan 21 '19 at 6:06

Another way is \eqalign{ & \prod\limits_{k = 1}^n {{{2n - 2k + 1} \over {2n - 2k + 2}}} = \prod\limits_{k = 1}^n {{{\left( {n - k} \right) + 1/2} \over {\left( {n - k} \right) + 1}}} = \prod\limits_{j = 0}^{n - 1} {{{j + 1/2} \over {j + 1}}} = \cr & = {{\left( {1/2} \right)^{\,\overline {\,n\,} } } \over {n!}} = {{\Gamma \left( {n + 1/2} \right)} \over {\Gamma \left( {1/2} \right)\Gamma \left( {n + 1} \right)}} = \left( \matrix{ n - 1/2 \cr n \cr} \right) = \left( { - 1} \right)^{\,n} \left( \matrix{ - 1/2 \cr n \cr} \right) \cr} and then apply the binomial expansion.

• Neat! thank you! – clathratus Jan 20 '19 at 23:57