# Prove inequality

#### idontknow

prove : $$\displaystyle n+1 > \dfrac{1}{2} \sqrt[n]{n!}(1+\dfrac{1}{n}) \; \:$$ , $$\displaystyle n \in \mathbb{N}$$.

#### idontknow

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 $$\displaystyle \lim_{n\rightarrow \infty} a_n =0 < a_n < a_1=1/2<1$$.

#### idontknow

$$\displaystyle 1>\dfrac{1}{2n}\sqrt[n]{n!} \;$$ ; $$\displaystyle \; (2n)^n > n!$$.

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