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?

  • $\begingroup$ 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? $\endgroup$ Nov 11, 2022 at 21:55
  • $\begingroup$ Welcome. In general, you should show us your attempts even if they failed $\endgroup$
    – FShrike
    Nov 11, 2022 at 22:19

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

You can play with this function here.


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