I recently noticed that, if you take the $(2n)$th row of Pascal's triangle, and alternately multiply and divide the numbers that crop up, almost everything cancels. For example, the 4th row (1 4 6 4 1) becomes $1/4 \times 6/4 \times 1$, which simplifies to $3/8$. I performed the same procedure for some later rows, and got $5/16$, $35/128$, $63/256$, $(3 \times 7 \times 11)/(2^{10})$, and $(3 \times 11 \times 13)/(2^{11})$.

All of the denominators seem to be powers of 2, which I don't know how to prove. Also, the fractions seem to me to be (very) slowly approaching 0, which I also don't know how to prove (if it's even true). Any assistance with either question would be greatly appreciated- I'm not very well versed in methods of proof.

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    $\begingroup$ 4th row the way mathematicians count them. $\endgroup$
    – user645636
    Jun 12, 2019 at 23:14
  • $\begingroup$ Okay. I'll switch that, then. $\endgroup$ Jun 12, 2019 at 23:15

1 Answer 1


We can simplify each individual ratio:

$$ \frac{\displaystyle \binom{2n}{0}}{\displaystyle \binom{2n}{1}}\frac{\displaystyle \binom{2n}{2}}{\displaystyle \binom{2n}{3}}\cdots\frac{\displaystyle\binom{2n}{2n-2}}{\displaystyle \binom{2n}{2n-1}} = \frac{1}{2n}\frac{3}{2n-2}\cdots\frac{2n-1}{2} $$

Reorder the factors in the denominator, then square them while interlacing them up top:

$$ \frac{1\cdot2\cdot3\cdot4\cdots(2n-1)(2n)}{2^2\cdot 4^2\cdot6^2\cdots (2n)^2} =\frac{(2n)!}{\big[2^n n!\big]^2}=\frac{1}{4^n}\binom{2n}{n}. $$

  • $\begingroup$ And $\frac{1}{4^n}\binom{2n}{n}$ approaches 0 as n approaches infinity? $\endgroup$ Jun 12, 2019 at 23:21
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    $\begingroup$ Yes, see central binomial coefficient for estimates and elementary inequalities. In particular, $$\frac{1}{4^n}\binom{2n}{n}\sim \frac{1}{\sqrt{\pi n}}. $$ Not sure off the top of my head of an elementary proof of a sufficient inequality. $\endgroup$
    – runway44
    Jun 12, 2019 at 23:27
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    $\begingroup$ Apply Stirling’s formula $\endgroup$ Jun 13, 2019 at 1:29

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