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

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?

• Here is a tester. The limit for $n=47$ has this prime factorization including all prime numbers less than $n$. For smaller prime factors, they have a higher exponent. Maybe the pattern continues? Nov 11, 2022 at 21:55
• Welcome. In general, you should show us your attempts even if they failed Nov 11, 2022 at 22:19

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: $$(2-n)\cdots(X-n-1)\cdot\frac{1+X-n-\frac{X-n}{X-1}+(X-n)\sum\cdots}{\Gamma(1+X-n)}\overset{n\to X}{\longrightarrow}(-1)^X\cdot(X-2)!$$