Let $a >b$ positive numbers. Find the minimum value of $\tfrac{a!+1}{b!+1}$

Question

Let $a >b$ be positive numbers. Find the minimum value of $$\dfrac{a!+1}{b!+1}$$

My try to solve the question:

I take $b=a-1$ to solve question so $$\dfrac{a!+1}{(a-1)!+1}=\dfrac{a}{(a-1)!+1}+\dfrac{1}{(a-1)!+1}$$ That's all. How can I get a solution? Any hint? Thank you.

Answer 1

HINT let's recall $f(n) = \dfrac{(n+1)! + 1}{n! + 1}$.

Note that $f(n + 1) > f(n)$ for any $n$. Also note that $\dfrac{(n+k + 1)! + 1}{n! + 1} \ge f(n)$ for any $k \ge 0$.

Answer 2

How $a > b$ then $a! > b!$. Soon, for $n \in \mathbb{N}$ $$ \dfrac{(a+n)! + 1 }{b! + 1}> \dfrac{a! + 1}{b! + 1} > 1. $$