# Alternately Multiplying and Dividing Within Pascal's Triangle

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.

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

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:

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

• And $\frac{1}{4^n}\binom{2n}{n}$ approaches 0 as n approaches infinity? Jun 12, 2019 at 23:21
• 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.
– anon
Jun 12, 2019 at 23:27
• Apply Stirling’s formula Jun 13, 2019 at 1:29