floor value of a function involving factorial sum

Question

Evaluation of floor value of

$(2019!)\cdot (2018!+2017!+\cdots +2!+1!)^{-1}$

Try: I am trying to solve it using gamma function.

$\displaystyle (n-1)!=\Gamma (n)=\int^{\infty}_{0}e^{-x}\cdot x^{n-1}dx$

and $\displaystyle \sum^{2018}_{k=1}k!=\sum^{2018}_{k=1}\Gamma(k+1)=\sum^{2018}_{k=1}\int^{\infty}_{0}e^{-x}\cdot x^{k}dx$

$\displaystyle =\int^{\infty}_{0}e^{-x}\sum^{2018}_{k=1}x^{k}dx=\int^{\infty}_{0}e^{-x}\bigg(\frac{x^{2019}-x}{x-1}\bigg)dx$

So our expression is $$(2019!)\cdot \frac{1}{\displaystyle \int^{\infty}_{0}e^{-x}\bigg(\frac{x^{2019}-x}{x-1}\bigg)dx}$$

Now i did not know how can i bound it

Could some help me to solve it thanks

Answer 1

For $n\geqslant 5$, we have $$\begin{align}1!+2!+\cdots +(n-1)!&\gt (n-3)!+(n-2)!+(n-1)! \\\\&=(n-3)!\bigg(1+n-2+(n-2)(n-1)\bigg) \\\\&=(n-3)!\bigg(n^2-2n+1\bigg) \\\\&\gt (n-3)!\bigg(n^2-2n\bigg) \\\\&=\frac{n!}{n-1}\end{align}$$ from which we get $$\frac{n!}{1!+2!+\cdots +(n-1)!}\lt n-1\tag1$$

Also, for $n\geqslant 5$, we have $$\begin{align}1!+2!+\cdots +(n-1)!&\lt (n-3)(n-3)!+(n-2)!+(n-1)! \\\\&=(n-3)!\bigg(n-3+n-2+(n-2)(n-1)\bigg) \\\\&=(n-3)!\bigg(n^2-n-3\bigg) \\\\&\lt (n-3)!\bigg(n^2-n\bigg) \\\\&=\frac{n!}{n-2}\end{align}$$ from which we get $$n-2\lt\frac{n!}{1!+2!+\cdots +(n-1)!}\tag2$$

From $(1)(2)$, we have, for $n\geqslant 5$, $$n-2\lt\frac{n!}{1!+2!+\cdots +(n-1)!}\lt n-1$$ which implies $$\left\lfloor\frac{n!}{1!+2!+\cdots +(n-1)!}\right\rfloor=n-2$$

Therefore, the answer is $$\left\lfloor\frac{2019!}{1!+2!+\cdots +2018!}\right\rfloor=\color{red}{2017}$$

Answer 2

Let $$a_n=\frac{\sum_{k=1}^{n-1} k!}{n!}.$$ Then $$a_{n+1} =\frac{n!a_n+n!}{(n+1)!}=\frac{a_n+1}{n+1}.$$ Note $a_1=0$, $a_2=\frac12$, $a_3=\frac12$, $a_4=\frac38$, $a_5=\frac{11}{40}$. So for $n=5$, we have If $$\frac1{n-1}<a_n<\frac1{n-2}$$ and this inequality then follows for all $n\ge 5$ by induction: $$ a_{n+1}=\frac{a_n+1}{n+1}>\frac{\frac1{n-1}+1}{n+1}=\frac n{n^2-1}>\frac n{n^2}=\frac1n$$ $$ a_{n+1}=\frac{a_n+1}{n+1}<\frac{\frac1{n-2}+1}{n+1}=\frac1{n-1}\cdot\left(1-\frac{n-3}{n^2-n-2}\right)<\frac1{n-1}$$

It follows that $$ 2017<\frac1{a_{2019}}<2018.$$