Calculation of $\frac{n!}{1!+2!+3!+\cdots+(n-1)!}$

Question

Calculation of $\displaystyle \left[\frac{n!}{1!+2!+3!+\cdots+(n-1)!}\right] = $

Where $n\geq 4$ and $n\in \mathbb{N}$ and $\left[x\right] =$ Greatest Integer of $x$

My Try :: For Upper Bond::

$n! = n.(n-1)! = \{(n-1)+1\}.(n-1)! = (n-1).(n-1)!+(n-1)!$

$ = (n-1).(n-1)!+(n-1).(n-2)!=(n-1).(n-1)!+\{(n-2)+1\}.(n-2)!$

Now I did not understand How can i proceed further.

plz help me , Thanks

Answer 1

The answer is $n-2$.

Let the denominator be $D_n$.

Hint: Show that $(n-2) D_n < n! < (n-1) D_n$.

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For the upper bound,

Since $n \geq 4$, show that $ (n-1)\left[(n-1)! + (n-2)! + (n-3)!\right] > n! $.

For the lower bound, proceed by induction. Given that $(k-2)D_k < k!$, then

$(k-1) D_{k+1} = (k-1)D_k + (k-1) k! < \frac{k-1}{k} k! + (k-1) k! < (k+1) k!$