let $b_{n,k}$ be the numbers defined formally by $$X^n=\sum_{k=0}^n b_{n,k}\binom{X}{k}$$ where $\binom{X}{n}=\frac{1}{n!}\prod_{k=0}^{n-1}(X-k)$.
I am looking for an equivalent of $b_{n,k}$ when $k$ is fixed and $n\to+\infty$.
Thanks in advance for any answer.
You are looking for a scaled version of the Stirling numbers of the second kind:
Note that $\binom{X}{n} = \frac{(x)_n}{n!}$ where $(x_n)$ is the falling factorial (or Pochhammer symbol).
that is not a hard question !
$X^{n}=b_{1}X/1!+b_{2}X(X-1)/2!+b_{3}X(X-1)(X-2)/3!+......$$\Longrightarrow$$X^{n}-1=b_{1}X/1!-1+b_{2}X(X-1)/2!+b_{3}X(X-1)(X-2)/3!+......=$$b_{1}X-1+b_{2}X(X-1)/2!+b_{3}X(X-1)(X-2)/3!+......$$\Longrightarrow$$1+X+X^2+X^3+......$$=b_{1}+\frac{b_{1}-1}{x-1}+b_{2}X/2!+b_{3}X(X-2)/3!+......$$\Longrightarrow$$b_{1}=1$
$X+X^2+X^3+......=b_{2}X/2!+b_{3}X(X-2)/3!+......$$\Longrightarrow$$1+X+X^2+X^3+......=b_{2}/2!+b_{3}(X-2)/3!+......$$\Longrightarrow$$\sum_{2}^{n}(-1)^n$$\frac{b_{n}}{n}=1$
then, we repeat this procedure and you can get your solution at last !
is it helpful ? thank you !
