Ufyukyu

0 1 $begingroup$ I have proved by induction that the Statement above is true Here is my proof: Inductionbase: $frac{2!}{2^2}=frac{2}{4}$ Inductionstep: $frac{(n+1)!}{(n+1)^{n+1}}=frac{n!}{(n+1)^n}frac{n+1}{n+1}overset{(n+1>n)}{leq}frac{n!}{n^n}overset{text{IH}}{leq}frac{2}{n^2}$ But I still don't understand why my proof is valid, why it makes sense. For me it is just a Formula, can somebody explainme how someone came up with the idea that: $$frac{n!}{n^n}leqfrac{2}{n^2},forall ngeq 2$$ real-analysis proof-explanation share | cite | improve this question asked Jan 30 at 17:14 New2Math New2Math ...