对一些常见求和估阶，如：
$$\sum_{k=1}^nk^\alpha,\sum_{k=1}^n\frac{1}{k},\sum_{k=1}^n\ln k,\sum_{k=1}^n\ln^\beta k,\sum_{k=1}^n\frac{1}{\ln k},\sum^n_{k=1} k!,\sum_{k=1}^n k^n,\sum_{k=1}^n\frac{1}{k!}\sum_{j=1}^nj^{\frac{1}{j^k}}$$$$\sum_{k=1}^nk^\alpha=\frac{n^{\alpha+1}}{n+1}+\frac{1}{2}n^{\alpha}+o(n^{\alpha})$$$$\sum_{k=1}^n\frac{1}{k}=\ln n+\gamma+\frac{1}{2n}-\frac{1}{12n^2}+\frac{1}{120n^4}+O(\frac{1}{n^6})$$$$\sum_{k=1}^n\ln k=n\ln n-n+\frac{\ln n}{2}+o(\ln n)$$$$\sum_{k=1}^n\frac{1}{\ln k}=\frac{n}{\ln n}+o(\frac{n}{\ln n})$$以上都可由E-M公式得到
$$\sum^n_{k=1} k!=n!+o(n!)$$$$作比证明即可$$$$\sum_{k=1}^n k^n=\frac{e}{e-1}n^n+o(n^n)$$$$作比证明，详见倒序求和$$$$\sum_{k=1}^n\frac{1}{k!}=e-1$$$$由e^x的泰勒展开在1处取值得到 $$$$\sum_{j=1}^nj^{\frac{1}{j^k}}=n+o(n)\qquad\quad k=1,2,3\cdots$$$$由夹逼准则和stolz得到$$