Asymptotics of a function involving factorials I want to understand the asymptotics of a function which depends on a parameter $N$ which is a large integer.
The function is of the form (variable $x$ is of order 1)
$$ F_N(x)=\sum_{n=0}^\infty f_{N,n}x^n.$$
I suspect that the coefficient $f_{N,N}$ dominates the asymptotics, so that $F_N(x)\approx f_{N,N}x^N$, in some sense. But I would like to actually prove this, and estimate the order of the error.
The coefficients are given by
$$ f_{N,n}=\sum_{k=0}^{{\rm min}(n,N)-1}\frac{k!(n-1-k)!(N-1-k)!}{(N+n-1-k)!}.$$
What I have done: I can see that if $n$ is small, then $f_{N,n}\sim N^{-n+1}$. On the other hand, if $n=N$ then the contribution to the sum from $k=N-1-m$ is of order $N^{-2m}$, so the sum is dominated by $m=0$ and we get a value of order 1.
Using the Stirling approximation blindly, without considering whether each factorial is large or not, I get an approximation for the summand which is
$$ \frac{k^k(n-k-1)^{n-k-1/2}(1-\frac{k+1}{N})^{N-k-1/2}}{N^n(1+\frac{n-k-1}{N})^{N+n-k-1/2}}$$
which could be useful. But this approximation is not valid in particular for the case $k=n-1=N-1$ which I think is the most important one.
How do I carry out a rigorous asymptotic analysis here?
 A: Just a hint as a possible way to go, through the Beta Function.
$$
\eqalign{
  & f_{\,N,\,n}  = \sum\limits_{k = 0}^{\min (N,n) - 1}
 {{{k!\left( {n - 1 - k} \right)!\left( {N - 1 - k} \right)!} \over {\left( {N + n - 1 - k} \right)!}}}  =   \cr 
  &  = n\sum\limits_{k = 0}^{\min (N,n) - 1}
 {{{\left( {k + 1 - 1} \right)!\left( {n - k - 1} \right)!\left( {n - 1} \right)!\left( {N - k - 1} \right)!}
 \over {\left( {n + 1 - 1} \right)!\left( {N + n - k - 1} \right)!}}}  =   \cr 
  &  = n\sum\limits_{k = 0}^{\min (N,n) - 1}
 {{\rm B}\left( {k + 1,n - k} \right){\rm B}\left( {n,N - k} \right)}  \cr} 
$$
Now, the integral representation of Beta looks to get clumsy.
On the other hand, Beta is rapidly decaying at growing of either of its arguments, so
it might be possible to try by truncating the sum.
Another way is that $F_{\, N}(x)$ actually reads
$$
\eqalign{
  & F_{\,N} (x) = \sum\limits_{n = 0}^\infty  {\sum\limits_{k = 0}^{\min (N,n) - 1}
 {{{k!\left( {n - 1 - k} \right)!\left( {N - 1 - k} \right)!}
 \over {\left( {N + n - 1 - k} \right)!}}} \;x^{\,n} }  =   \cr 
  &  = \sum\limits_n {\sum\limits_k
 {{{\left[ {0 \le k < n} \right]\left[ {0 \le k < N} \right]k!\left( {n - 1 - k} \right)!
\left( {N - 1 - k} \right)!} \over {\left( {N + n - 1 - k} \right)!}}x^{\,n} } }  =   \cr 
  &  = \sum\limits_k {\left[ {0 \le k < N} \right]k!\left( {N - 1 - k} \right)!
\sum\limits_n {{{\left[ {k < n} \right]\left( {n - 1 - k} \right)!}
 \over {\left( {N + n - 1 - k} \right)!}}x^{\,n} } }  =   \cr 
  &  = \sum\limits_{0 \le k < N} {k!\left( {N - 1 - k} \right)!
\sum\limits_{k < n} {{{1^{\,\overline {\,n - 1 - k\,} } }
 \over {1^{\,\overline {\,N + n - 1 - k\,} } }}x^{\,n} } }  =   \cr 
  &  = \sum\limits_{k = 0}^{N - 1} {k!\left( {N - 1 - k} \right)!
\sum\limits_{k + 1 \le n} {{{1^{\,\overline {\,n - 1 - k\,} } }
 \over {1^{\,\overline {\,n - 1 - k\,} } \left( {n - k} \right)^{\,\overline {\,N\,} } }}x^{\,n} } }  =   \cr 
  &  = \sum\limits_{k = 0}^{N - 1} {k!\left( {N - 1 - k} \right)!
\sum\limits_{k + 1 \le n} {{{x^{\,n} } \over {\left( {n - k} \right)^{\,\overline {\,N\,} } }}} }  =   \cr 
  &  = \left( {N - 1} \right)!\left( {\sum\limits_{k = 0}^{N - 1} {{{x^{\,k} }
 \over {\left( \matrix{  N - 1 \cr   k \cr}  \right)}}} } \right)
\left( {\sum\limits_{1 \le j} {{{x^{\,j} } \over {j^{\,\overline {\,N\,} } }}} } \right) \cr} 
$$
where:

*

*$[P]$ denotes the Iverson bracket;

*$ x^{\,\overline {\,k\,} } $ represents the  Rising Factorial.

I am sure you can continue by yourself.
