Wednesday, 18 July 2018

Limit of Stirling's approximation as n goes to infinity.

I would like to see some detailed solution for n!2πn(ne)n as n. I know that the answer is 1 but i am not sure why? Here is what is tried:



I rewrote the stirling's formula like this.
(e/n)(2e/n)(3e/n)(ne/n)2πn0 as n. I am not sure where I went wrong.

No comments:

Post a Comment

real analysis - How to find limhrightarrow0fracsin(ha)h

How to find lim without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...