Can you be elegant

Calculus Level 3

$\frac{1}{2}-\displaystyle\lim_{ω\rightarrow \infty}\frac{(ω!)^{\frac{1}{ω}}}{ω}$ is

can't be determined positive negative 0

1 solution

Shivam Jadhav
Feb 19, 2016

By using $AM-GM$ $\frac{1+2+3+....+ω}{ω}\geq(ω!)^{\frac{1}{ω}}$ $\frac{(ω)(ω+1)}{2ω}\geq(ω!)^{\frac{1}{ω}}$ $\frac{ω+1}{2ω}\geq\frac{(ω!)^{\frac{1}{ω}}}{ω}$

$\displaystyle\lim_{ω\rightarrow \infty}\frac{1+ω}{2ω}\geq\displaystyle\lim_{ω\rightarrow \infty}\frac{(ω!)^{\frac{1}{ω}}}{ω}$

$\displaystyle\lim_{ω\rightarrow \infty}\frac{1}{2}\geq\displaystyle\lim_{ω\rightarrow \infty}\frac{(ω!)^{\frac{1}{ω}}}{ω}$

$\displaystyle\lim_{ω\rightarrow \infty}\frac{1}{2}-\displaystyle\lim_{ω\rightarrow \infty}\frac{(ω!)^{\frac{1}{ω}}}{ω}\geq0$

This solution is incomplete. First, it does not show that $\lim_{\omega\to\infty}\frac{(\omega!)^{1/\omega}}{\omega}$ exists. Also, if the limit does exist , it might be $\frac{1}{2}$ . Thus the answer could be "positive", "0", or "can't be determined".

Otto Bretscher - 5 years, 3 months ago

I have used Stirling formule and I found $\large \lim_{\omega\to \infty} \frac{(\omega!)^{\frac{1}{\omega}}}{\omega} = \frac{1}{e}$

Guillermo Templado - 5 years, 3 months ago

Yes that works.... or use a Riemann sum for $\ln(x)$

Otto Bretscher - 5 years, 3 months ago

@Otto Bretscher – Great, thank you

Guillermo Templado - 5 years, 3 months ago

@Guillermo Templado – Anytime, Compañero :)

Otto Bretscher - 5 years, 3 months ago

Can you be elegant

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