G.I.F

very very good ques but no one here is so intelligent to give the ans

My hunch is that for all n>3 (integers) the following inequality is valid

(n-2)<\frac{n!}{1!+2!+...+(n-1)!}<(n-1)

We will use Principle of Mathematical Induction to prove that.

P(4) is true.

Let P(n) be true. We will prove P(n+1) is true.

We have,

(n-2)<\frac{n!}{1!+2!+...+(n-1)!}<(n-1)

\Leftrightarrow \frac{1}{n-2}>\frac{1!+2!+...+(n-1)!}{n!}>\frac{1}{n-1}

\Leftrightarrow \frac{n!(n-1)}{(n+1)!(n-2)}>\frac{1!+2!+3!+..n!}{(n+1)!}>\frac{n!n}{(n+1)!(n-1)}

\Leftrightarrow \frac{(n+1)(n-2)}{n-1}<\frac{(n+1)!}{1!+2!+3!+..n!}<n-\frac{1}{n}

Further we have n-\frac{1}{n}<n

and (n-1)<\frac{(n+1)(n-2)}{(n-1)} for all n>3

So we get (n-1)<\frac{(n+1)!}{1!+2!+3!+...+n!}<n

P(n+1) is true. So, statement is proved.

Therefore \left[\frac{n!}{1!+2!+3!+..(n-1)!} \right] = (n-2) for all n>3

So, we get the required answer as 2008

Yes Shubhodip Right answer.