area under ln graph

struggling with the following:

By examining the area under the graphs: y=ln x and y=ln (x-1) show that:

NlnN - N < ln(N!) < (N+1)ln(N+1) - N

as some help this is what i've thrown together so far:

Area under y=lnx graph = xlnx-x + 1 (considering only the positive portion)
Area under y=ln(x-1) = (x-1)ln(x-1) - (x-2)

so you can see a similarity there! just need someone to make the leap

Reply 1

Willa

ok in the past few mins i've made a little progress, but just still need to piece it together:

analysing those two integrals it is apparent that:

for lnx: xlnx-(x-1)
for ln(x-1): (x-1)ln(x-1) - (x-2)

so (x+1)ln(x+1) + x must be the area under the ln curve ln(x+1)....which is obviously more than the area under the ln x curve cos it's just shifted across.

now ln(n!) is approx. the integral of ln x from 1 to N....so area under lnx graph. But that is +1 more than nlnn-n...hence the show that should be finished

but is it accurate to say ln(n!) = nln N - n + 1 (cos for small N that's not that accurate)

Reply 2

BCHL85

So u already proved NlnN - N <ln(N!).
The remain can be proved in similar way
ln((N+1)!) = lnN! + ln(N+1)
int(1->N+1)lnx = ln(N!) + ln(N+1)
(N+1)ln(N+1) - N = ln(N!) +ln(N+1)
ln(N+1) > 0 due to N>= 1.
So ln(N!) < (N+1)ln(N+1) - N