下降阶乘幂

相关定义与定理

1. $x^{\underset{―}{m}}=x(x-1)\cdots (x-m+1)(m\in {\mathbb{N}}_{+}).$
2. $x^{\bar{m}}=x(x+1)\cdots (x+m-1)(m\in {\mathbb{N}}_{+}).$
3. $x^{\underset{―}{0}}=x^{\bar{0}}=1.$
4. $n!=n^{\underset{―}{n}}=1^{\bar{n}}.$
5. $\Delta \text{qty}(x^{\underset{―}{m}})=m{x}^{\underset{―}{m-1}}.$
6. ${\displaystyle g(x)=\Delta f(x)⟺\sum g(x)\delta x=f(x)+C.}$
7. ${\displaystyle {\sum }_a^{b}g(x)\delta x=\text{eval}{f(x)}_a^b=f(b)-f(a).}$
8. ${\displaystyle {\sum }_a^{b}g(x)\delta x=\sum _{k=a}^{b-1}g(k)=\sum _{a⩽k<b}g(k)(b⩾a).}$
9. ${\displaystyle {\sum }_a^{b}g(x)\delta x+{\sum }_b^{c}g(x)\delta x={\sum }_a^{c}g(x)\delta x.}$
10. ${\displaystyle \sum _{0⩽k<n}{k}^{\underset{―}{m}}=\text{eval}{{\displaystyle \frac{k^{\underset{―}{m+1}}}{m+1}}}_0^n={\displaystyle \frac{n^{\underset{―}{m+1}}}{m+1}}(m,n\in {\mathbb{N}}_{+}).}$
11. $x^{\underset{―}{-m}}={\displaystyle \frac{1}{(x+1)(x+2)\cdots (x+m)}}(m>0).$
12. $x^{\underset{―}{m+n}}=x^{\underset{―}{m}}(x-m{)}^{\underset{―}{n}}(m,n\in \mathbb{Z}).$
13. ${\displaystyle {\sum }_a^b{x}^{\underset{―}{m}}\delta x=\{\begin{array}{ll}\text{eval}{{\displaystyle \frac{x^{\underset{―}{m+1}}}{m+1}}}_a^b,& m\ne -1\\ \text{eval}{H_x}_a^b,& m=-1\end{array}}$ ，
    其中 ${\displaystyle H_x=\sum _{k=1}^x{\displaystyle \frac{1}{k}}}$ 称为调和级数.
14. $\mathrm{E}f(x)=f(x+1).$
15. ${\displaystyle \sum u\Delta v=uv-\sum \mathrm{E}v\Delta u.}$

例题

判断级数 ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}{\displaystyle \frac{1+{\displaystyle \frac{1}{2}}+\cdots +{\displaystyle \frac{1}{n}}}{(n+1)(n+2)}}}$ 的敛散性，若收敛，则求其和.

解答

令 ${\displaystyle H_n=\sum _{k=1}^n{\displaystyle \frac{1}{k}},S_n=\sum _{k=1}^n{\displaystyle \frac{H_k}{(k+1)(k+2)}}=\sum _{\begin{array}{c}1⩽k<n+1\\ k\in {\mathbb{N}}_{+}\end{array}}{\displaystyle \frac{H_k}{(k+1)(k+2)}}={\sum }_1^{n+1}{\displaystyle \frac{H_x}{(x+1)(x+2)}}\delta x}$ ，
那么记

$$u(x)=H_x,\Delta v(x)={\displaystyle \frac{1}{(x+1)(x+2)}}=-{\displaystyle \frac{1}{x+2}}-\text{qty}(-{\displaystyle \frac{1}{x+1}})$$

因此

$$\Delta u(x)=x^{\underset{―}{-1}},v(x)=-{\displaystyle \frac{1}{x+1}}=-x^{\underset{―}{-1}},\mathrm{E}v(x)=-(x+1{)}^{\underset{―}{-1}}$$

于是

$$\begin{array}{rl}{S}_n& ={\sum }_1^{n+1}{\displaystyle \frac{H_x}{(x+1)(x+2)}}\delta x=\text{eval}{-x^{\underset{―}{-1}}{H}_x}_1^{n+1}+{\sum }_1^{n+1}(x+1{)}^{\underset{―}{-1}}{x}^{\underset{―}{-1}}\delta x\\ & =-(n+1{)}^{\underset{―}{-1}}{H}_{n+1}+1^{\underset{―}{-1}}{H}_1+{\sum }_1^{n+1}{x}^{\underset{―}{-2}}\delta x={\displaystyle \frac{1}{2}}-{\displaystyle \frac{H_{n+1}}{n+2}}-\text{eval}{x^{\underset{―}{-1}}}_1^{n+1}\\ & =1-{\displaystyle \frac{1}{n+2}}-{\displaystyle \frac{H_{n+1}}{n+2}}=1-{\displaystyle \frac{1}{n+1}}-{\displaystyle \frac{H_n}{n+2}}\end{array}$$

又因为

$$\underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{H_n}{n+2}}=\underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{\ln n+\gamma +\alpha (n)}{n+2}}=\underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{\ln n}{n+2}}=0(\alpha (n)\to 0)$$

所以 ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{S}_n=1}$ ，因此级数收敛，且其和为 1.