Limit of function involving gamma and digamma function $\lim\limits_{n \to X} \frac{1-g-\psi\left(2-n\right)}{\left(1-n\right)!}$

Question

I am trying to calculate the following limit:

$$\lim_{n \to X} \frac{1-g-\psi\left(2-n\right)}{\left(1-n\right)!}$$ where $X$ is a positive integer. However, I am stuck. I tried writing the digamma function using harmonic numbers but I could not make them work as the input is negative. I also tried using asymptotic formulae (i.e. $\ln(x)$ for $\psi(x)$) but again this did not work as the argument is negative. How could I proceed?

Answer 1

Brief answer.

I know that: $$\psi(s+1)=-\gamma+\sum_{m=1}^\infty\left[\frac{1}{m}-\frac{1}{m+s}\right]$$So, the expression inside the limit (the limitand? ;)) is, for integer $X>1$: $$\frac{1+\sum_{n=1}^\infty\left[\frac{1}{m+1-n}-\frac{1}{m}\right]}{\Gamma(2-n)}\\=(2-n)(3-n)\cdots(X-n)\frac{1-\frac{1}{X-1}+\frac{1}{X-n}+\sum_{m=1\\m\neq X-1}^\infty\left[\frac{1}{m}-\frac{1}{m+1-n}\right]}{\Gamma(1+X-n)}$$If $X=1$, the limit is clearly $0$ as $n\to X$. Else, let's simplify this as: $$(2-n)\cdots(X-n-1)\cdot\frac{1+X-n-\frac{X-n}{X-1}+(X-n)\sum\cdots}{\Gamma(1+X-n)}\overset{n\to X}{\longrightarrow}(-1)^X\cdot(X-2)!$$

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