Que es mayor $(1000)^{1000}$ o $(1001)^{999}$

Question

Sé que esto tiene que ser hecho usando el binomio de expansión. He intentado $(1001)^{999}=(1000+1)^{999}$ pero no llegar a ninguna parte.

Answer 1

He aquí una solución sin calculadora. Deje $x = 1000$. Entonces

$$\begin{align}&x^x \gt (x+1)^{x-1}\\ \iff &x\cdot x^{x-1} \gt (x+1)^{x-1}\\ \iff &x \gt (1+\frac{1}{x})^{x-1}\\ \iff &x(1+\frac{1}{x}) \gt (1+\frac{1}{x})^x\\ \iff &x+1 \gt (1+\frac{1}{x})^x\end{align}$$

pero el lado derecho es menos de $e$ y el LHS es enorme.

Answer 2

$$\begin{gathered} \frac{{\left( {1000 + 1} \right)^{\,999} }} {{1000^{\,1000} }} = \frac{{\left( {1000 + 1} \right)^{\,1000} }} {{\left( {1000 + 1} \right)1000^{\,1000} }} = \hfill \\ = \frac{{\left( {1 + 1/1000} \right)^{\,1000} }} {{\left( {1000 + 1} \right)}} < \frac{e} {{1001}} < 1 \hfill \\ \end{reunieron} $$

Answer 3

En primer lugar, observar:

$1000^{1000}>1001^{999}\iff1000^{1000/999}>1001$

Entonces, a la conclusión de:

$1000^{1000/999}=1000^{1.\overline{001}}>1000^{1.001}>1001$

Answer 4

$$1000^{1000}\overset{?}<1001^{999}\iff\ln{\left((1000)^{1000}\right)}\overset{?}<\ln{\left((1001)^{999}\right)}\iff \frac{\ln(1000)}{999}\overset{?}<\frac{\ln(1001)}{1000}$$ This motivates to consider the function $$f(x)=\frac{\ln(x+1)}{x}$$ with $f'(x)=\dfrac{\frac1{x+1}x\ln(x+1)}{x^2}$. The numerator is obviously negative for large numbers of $x$, since $\frac{x}{x+1}<1$ but $\ln(x+1)\gg 1$, hence the function is decreasing for large $x$. Hence $$f(1000)>f(1001)\implies \frac{\ln(1000)}{999}>\frac{\ln(1001)}{1000}\implies 1000^{1000}>1001^{999}$$

Answer 5

SUGERENCIA: Supongamos:$$n^n\geq (n+1)^{n-1}$$ $$\Longleftrightarrow \frac{\ln{n}}{n-1}\geq \frac{\ln(n+1)}{n}$$ Por lo tanto,vamos:$$f{(x)}=\frac{\ln{x}}{x-1},x>1$$ Tenemos:$$f^{'}(x)=\frac{1-\frac{1}{x}-\ln{x}}{(x-1)^2}=\frac{g(x)}{(x-1)^2}$$ Entonces tenemos:$f^{'}(x)<0$

Así:$$f(x)<f(x+1)$$ $$\Longrightarrow \frac{\ln{n}}{n-1}<\frac{\ln(n+1)}{n}$$