When looking at the central binomial coefficients (A000984), I found that if you took the ratio of consecutive numbers in the sequence it approached 1/4.
the sequence being; 1, 2, 6, 20, 70, etc.
the ratios being; 1/2, 2/6, 6/20, 20/70, etc.
I was wondering how to prove this conjecture by finding the limit of this sequence of ratios and what that equation would be.
The $n^{th}$ term of this sequence is given by $a_n=\binom{2n-2}{n-1}=\dfrac{(2n-2)!}{(n-1)!(n-1)!}$
Now, $a_{n+1}=\dfrac{(2n)!}{n!n!}=\dfrac{2n(2n-1)(2n-2)!}{n^2(n-1)!(n-1)!}$ $\Rightarrow \dfrac{a_n}{a_{n+1}}=\dfrac{n}{4n-2}$
So, the limit of the ratios is $\dfrac{1}{4}$.