Cómo probar que: $[12\sqrt[n]{n!}]{\leq}7n+5$, $n\in N$
Sé que $\lim_{n\to \infty } (1+ \frac{7}{7n+5} )^{ n+1}=e$ y $\lim_{n\to \infty } \sqrt[n+1]{n+1} =1$.
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Marco Cantarini
Puntos
10794
Estoy suponiendo que $\left[x\right] $ is the floor function. For $n=1,2 $ works. So assume that $n\geq3 $. Using the bound $$n!\leq\sqrt{2\pi n}\left(\frac{n}{e}\right)^{n}e^{1/\left(12n\right)} $$ we have $$\left[12\sqrt[n]{n!}\right]\leq12\sqrt[n]{n!}\leq12\left(2\pi n\right)^{1/\left(2n\right)}e^{1/\left(12n^{2}\right)-1}n $$ so we have to prove now that $$12\left(2\pi n\right)^{1/\left(2n\right)}e^{1/\left(12n^{2}\right)-1}n\leq7n+5 $$ and we can prove it by induction. For $n=3 $ obras, por lo que consideramos \begin{align*} 12\left(2\pi\left(n+1\right)\right)^{1/\left(2\left(n+1\right)\right)}e^{1/\left(12\left(n+1\right)^{2}\right)-1}\left(n+1\right)= & 12\left(2\pi\left(n+1\right)\right)^{1/\left(2\left(n+1\right)\right)}e^{1/\left(12\left(n+1\right)^{2}\right)-1}n\\ + & 12\left(2\pi\left(n+1\right)\right)^{1/\left(2\left(n+1\right)\right)}e^{1/\left(12\left(n+1\right)^{2}\right)-1}\tag{1} \end{align*} y desde $12\left(2\pi x\right)^{1/\left(2x\right)}e^{1/\left(12x^{2}\right)-1}$ es monótona decreciente de la función para $x\geq1 $ we observe that $$ 12\left(2\pi\left(n+1\right)\right)^{1/\left(2\left(n+1\right)\right)}e^{1/\left(12\left(n+1\right)^{2}\right)-1}n\leq12\left(2\pi n\right)^{1/\left(2n\right)}e^{1/\left(12n^{2}\right)-1}n $$ hence $$ (1)\leq7n+5+12\left(2\pi\left(n+1\right)\right)^{1/\left(2\left(n+1\right)\right)}e^{1/\left(12\left(n+1\right)^{2}\right)-1}\leq7\left(n+1\right)+5 $$ since $$12\left(2\pi\left(n+1\right)\right)^{1/\left(2\left(n+1\right)\right)}e^{1/\left(12\left(n+1\right)^{2}\right)-1}\leq12\left(8\pi\right)^{1/8}e^{1/192-1}<7. $$
