Prove inequality

Dec 2015
935
122
Earth
prove : \(\displaystyle n+1 > \dfrac{1}{2} \sqrt[n]{n!}(1+\dfrac{1}{n}) \; \:\) , \(\displaystyle n \in \mathbb{N}\).
 
Dec 2015
935
122
Earth
Is this a proof ? (by leaving the LHS constant we can use the range of RHS as a proof )
$ n+1 > \dfrac{1}{2} \sqrt[n]{n!}(1+\dfrac{1}{n} ) =\dfrac{1}{2n}\sqrt[n]{n!}(n+1) \; $ ; $ \; 1>\dfrac{1}{2n}\sqrt[n]{n!}=a_n $.

Since \(\displaystyle a_{n+1}-a_n < 0\) , the range of \(\displaystyle a_n\) is \(\displaystyle \lim_{n\rightarrow \infty} a_n =0 < a_n < a_1=1/2<1\).
 
Dec 2015
935
122
Earth
\(\displaystyle 1>\dfrac{1}{2n}\sqrt[n]{n!} \; \) ; \(\displaystyle \; (2n)^n > n!\).
 
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