Sum of factorial

[Guest]

Prove that [tex]\frac{n!}{1(n-1)}+\frac{n!}{2(n-2)}+\cdots+\frac{n!}{\lfloor\frac{n}{2}\rfloor(n-\lfloor\frac{n}{2}\rfloor)}\cdot\frac{1}{1+r}=(n-1)!\left(1+\frac{1}{2}+\cdots+\frac{1}{n-1}\right)[/tex] for all [tex]n\ge 3[/tex] where r is the residue of n-1 when divided by 2.

[Guest]

Make life simpler by dividing both sides by [tex](n-1)![/tex].
Notice that [tex]\frac{n}{a(n-a)} = \frac{1}{a}+\frac{1}{n-a}[/tex] and the result should drop out.

Hope this helps,

R. Baber.