Scaling of a sum with factorials

Question

I am interested in how the sum \begin{equation} f(N)\equiv\frac 1{(N-1)!}\sum_{n=0}^{2N-1}\binom{2n+N}{2n-1} \end{equation} scales for large $N$. So far, I tried to expand the binomial into factorials and to use Stirling's approximation with the terms involving $N$; this, however, could not help in finding a simple scaling law.

Answer 1

Here is a rough approach. The two largest summands are the last, ${5N-2 \choose 2N-1}$ and ${5N-4 \choose 2N-2}$ Their ratio is $\frac{(5N-2)!(2N-2)!(3N-2)!}{(5N-4)!(3N-1)!(2N-1)!}=\frac{(5N-2)(5N-3)}{(3N-1)(2N-1)}\approx \frac {25}6$. That ratio between terms will not change quickly, so we can imagine it being a geometric series with that ratio. The sum will then be $1+\frac 6{19}$ times the largest term, which we can expand by Stirling $$f(N) \approx \frac {25}{19}\frac{(5N-2)!}{(3N-1)!(2N-1)!}\\ \approx \frac {25}{19}\frac{(5N-2)^{5N-2}}{(3N-1)^{3N-1}(2N-1)^{2N-1}}\sqrt{\frac{5N-2}{2\pi (3N-1)(2N-1)}}\\ \approx \frac {25}{19}\left(\frac {5^5}{3^32^2}\right)^N\frac6{5^2}\sqrt{\frac{5}{12\pi N}}$$ where the exponential terms cancel. The term in parentheses is about $28.9$ so this gives an idea.