# Scaling of a sum with factorials

I am interested in how the sum $$$$f(N)\equiv\frac 1{(N-1)!}\sum_{n=0}^{2N-1}\binom{2n+N}{2n-1}$$$$ scales for large $$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.

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.
• Thanks for the derivation. Do you infer the scaling as a geometric series from the fact that the ratio of the largest terms is independent on N, hence they behave as $x^N$?
• It is an approximation because I ignored the constants. If $N \gg 3$ it will be accurate for a while. As the terms decrease so rapidly, only the first few will be dominant in the sum. Feb 10, 2020 at 18:41
• @Graz It can be shown in this way that the approximation by a geometric series in fact gives the correct leading term. Fixing some typos, we have $$(N - 1)! \hspace {1.5px} f(N) \sim \frac {16} 9 \left( \frac {5^5} {2^8} \right)^{\! N} \sqrt {\frac {10} {\pi N}}, \quad N \to \infty.$$ Feb 19, 2020 at 15:17
• @Maxim: I don't see the typos you claim. The $3$ s in the denominator come from the $(3N-1)^{3N-1}$ term, the $\frac 6{25}$ from the $-2$ and $-1$s on the exponents. Feb 19, 2020 at 15:42
• The last term in the sum is $\binom {5 N - 2} {4 N - 3}$. Feb 19, 2020 at 16:21