# floor value of a function involving factorial sum

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

• @eccheng can you lease show me how you get it . Thanks – DXT Mar 2 at 11:34
• Sorry, I was wrong. Just look at the answer below. – eccheng Mar 2 at 12:02

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 $$\frac1{n-1} and this inequality then follows for all $$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.$$