How do i prove that the following series diverges?

[Dell]

1/ln((n)!) [n goes from 2-infinity]

1/ln((2)!) + 1/ln((3)!) +1 /ln((4)!) +...+ 1/ln((n)!)

the first thing i thought of was de lambert

lim An+1/An
n->inf

=ln(x!)/ln((x+1)!)
=ln(x!)/ln(x!*(x+1))
=ln(x!)/[ln(x!)+ln(x+1)]

=1-ln(x+1)/[ln(x!)+ln(x+1)]=1
which doesn't tell me anything about the convergence /divegence of the series.

have i done something wrong somewhere here, or is there another way to solve this?

 

[CompuChip]

I don't know if this is the most efficient way, but if you note that
ln(n!) = ln(n) + ln(n - 1) + ... <= n ln(n)
you can show that
$$S = \sum_{n = 2}^\infty \frac{1}{\ln(n!)} \ge \sum_{n = 2}^\infty \frac{1}{n \ln n}$$