求极限 $$ex vlm{n}frac{1^k+2^k+cdots+n^k}{n^{k+1}},quad vlm{n}sex{frac{1^k+2^k+cdots+n^k}{n^{k}}-frac{n}{k+1}}. eex$$
解答: $$eex ea mbox{第一个极限}&=vlm{n}frac{(n+1)^k}{(n+1)^{k+1}-n^{k+1}}quadsex{mbox{Stolz 公式}}\ &=vlm{n}frac{(n+1)^k}{(k+1)xi_n^k}quadsex{f(t)=t^{k+1} a f(n+1)-f(n)=f'(xi_n)}\ &=frac{1}{k+1}. eea eeex$$ $$eex ea mbox{第二个极限}&=vlm{n}frac{(k+1)(1^k+2^k+cdots+n^k)-n^{k+1}}{(k+1)n^k}\ &=frac{1}{k+1}vlm{n} frac{(k+1)(n+1)^k-[(n+1)^{k+1}-n^{k+1}]}{(n+1)^k-n^k}\ &=frac{1}{k+1}vlm{n} frac{(k+1)frac{1}{n}sex{1+frac{1}{n}}^k-sez{sex{1+frac{1}{n}}^k-1}}{frac{1}{n}sex{1+frac{1}{n}}^k-1}\ &=frac{1}{k+1}vlm{n} frac{(k+1)t(1+t)^k-[(1+t)^{k+1}-1]}{t(1+t)^k-1}quadsex{t=frac{1}{n}}\ &=frac{1}{k+1}lim_{t o 0} frac{(k+1)t(1+t)^k-[(1+t)^{k+1}-1]}{kt^2}quadsex{(1+t)^k-1sim kt}\ &=frac{1}{k+1}lim_{t o 0}frac{(k+1)(1+t)^k+(k+1)tcdot k(1+t)^{k-1}-(k+1)(1+t)^k}{2kt}\ &=frac{1}{k+1}lim_{t o 0}frac{(k+1)kt(1+t)^{k-1}}{2kt}\ &=frac{1}{2}. eea eeex$$