[数分提高]2014-2015-2第3教学周第2次课

求极限 $$\bex \vlm{n}\frac{1^k+2^k+\cdots+n^k}{n^{k+1}},\quad \vlm{n}\sex{\frac{1^k+2^k+\cdots+n^k}{n^{k}}-\frac{n}{k+1}}. \eex$$

 

解答: $$\beex \bea \mbox{第一个极限}&=\vlm{n}\frac{(n+1)^k}{(n+1)^{k+1}-n^{k+1}}\quad\sex{\mbox{Stolz  公式}}\\ &=\vlm{n}\frac{(n+1)^k}{(k+1)\xi_n^k}\quad\sex{f(t)=t^{k+1}\ra f(n+1)-f(n)=f'(\xi_n)}\\ &=\frac{1}{k+1}. \eea \eeex$$ $$\beex \bea \mbox{第二个极限}&=\vlm{n}\frac{(k+1)(1^k+2^k+\cdots+n^k)-n^{k+1}}{(k+1)n^k}\\ &=\frac{1}{k+1}\vlm{n} \frac{(k+1)(n+1)^k-[(n+1)^{k+1}-n^{k+1}]}{(n+1)^k-n^k}\\ &=\frac{1}{k+1}\vlm{n} \frac{(k+1)\frac{1}{n}\sex{1+\frac{1}{n}}^k-\sez{\sex{1+\frac{1}{n}}^k-1}}{\frac{1}{n}\sex{1+\frac{1}{n}}^k-1}\\ &=\frac{1}{k+1}\vlm{n} \frac{(k+1)t(1+t)^k-[(1+t)^{k+1}-1]}{t(1+t)^k-1}\quad\sex{t=\frac{1}{n}}\\ &=\frac{1}{k+1}\lim_{t\to 0} \frac{(k+1)t(1+t)^k-[(1+t)^{k+1}-1]}{kt^2}\quad\sex{(1+t)^k-1\sim kt}\\ &=\frac{1}{k+1}\lim_{t\to 0}\frac{(k+1)(1+t)^k+(k+1)t\cdot k(1+t)^{k-1}-(k+1)(1+t)^k}{2kt}\\ &=\frac{1}{k+1}\lim_{t\to 0}\frac{(k+1)kt(1+t)^{k-1}}{2kt}\\ &=\frac{1}{2}. \eea \eeex$$