[数分提高]2014-2015-2第1教学周第1次课

求极限 $$\bex \vlm{n}\dfrac{(n^2+1)(n^2+2)\cdots(n^2+n)}{(n^2-1)(n^2-2)\cdots(n^2-n)}. \eex$$

 

 

 

解答: 还记得对数不等式么: $$\bex \dfrac{x}{1+x}<\ln(1+x)<x,\quad x>0. \eex$$ 我们有 $$\beex \bea \ln\dfrac{n^2+i}{n^2-i}&=\ln\sex{1+\dfrac{2i}{n^2-i}} <\dfrac{2i}{n^2-i}\leq \dfrac{2i}{n^2-n},\\ \ln\dfrac{n^2+i}{n^2-i}&>\dfrac{\dfrac{2i}{n^2-i}}{1+\dfrac{2i}{n^2-i}} =\dfrac{2i}{n^2+i}\geq \dfrac{2i}{n^2+n}. \eea \eeex$$ 相加而有 $$\bex 1< \ln \prod_{k=1}^n \dfrac{n^2+i}{n^2-i}<\dfrac{n^2+n}{n^2-n}. \eex$$ 令 $n\to\infty$ 即得原极限 $=e$.

 

 

试证: $$\bex 0<e-\sex{1+\frac{1}{n}}^n<\frac{e}{n}. \eex$$

 

 

 

证明: 还觉得我们上课时讲的那个不等式么: $$\bex \sex{1+\frac{1}{n}}^n<e<\sex{1+\frac{1}{n}}^{n+1}. \eex$$ 于是 $$\bex 0<e-\sex{1+\frac{1}{n}}^n<\sex{1+\frac{1}{n}}^{n+1} -\sex{1+\frac{1}{n}}^n =\sex{1+\frac{1}{n}}^n\frac{1}{n} <\frac{e}{n}. \eex$$