Trích:
Nguyên văn bởi Lonely  Tính $\lim_{n\to +\infty}(\sqrt[n]{1}+\sqrt[n]{2}+\cdots+\sqrt[n]{2007}-2006)^n $. |
$=\lim_{n\to +\infty}(1+\sqrt[n]{2}+\cdots+\sqrt[n]{2007}-2006)^{\frac{1}{\sqrt[n]{2}+\cdots+\sqrt[n]{2007}-2006}\times n.(\sqrt[n]{2}+\cdots+\sqrt[n]{2007}-2006)} $
$=e^{\lim_{n\to +\infty}n.(\sqrt[n]{2}+\cdots+\sqrt[n]{2007}-2006)} $ (Vì $\lim_{x\rightarrow 0}(1+x)^{\frac{1}{x}}=e $)
$=e^{{\lim_{n\to +\infty}(\frac{2^{\frac{1}{n}}-1}{\frac{1}{n}})+(\frac{3^{\frac{1}{n}}-1}{\frac{1}{n}})+...(\frac{2007^{\frac{1}{n}}-1}{\frac{1}{n}})} $
$=e^{\ln2+\ln3+...+\ln2007} $ (Vì $\lim_{x\rightarrow 0}\frac{a^x-1}{x}=\lim_{x\rightarrow 0}\frac{e^{x\ln a}-1}{x\ln a}\times \ln a=\ln a $)
$=2.3...2007=2007! $
[RIGHT][I][B]Nguồn: MathScope.ORG[/B][/I][/RIGHT]