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)√2πn→0 as n→∞. I am not sure where I went wrong.
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