下降阶乘幂

相关定义与定理

1. \(x^{\underline{m} }=x(x-1)\cdots(x-m+1)~~(m\in\mathbb{N}_+).\)
2. \(x^{\overline{m}}=x(x+1)\cdots(x+m-1)~~(m\in\mathbb{N}_+).\)
3. \(x^{\underline{0}}=x^{\overline{0}}=1.\)
4. \(n!=n^{\underline{n}}=1^{\overline{n}}.\)
5. \(\varDelta \qty(x^{\underline{m}})=mx^{\underline{m-1}}.\)
6. \(\displaystyle g(x)=\varDelta f(x)\iff \sum g(x)\delta x=f(x)+C.\)
7. \(\displaystyle \sum\nolimits_{a}^{b} g(x)\delta x=\eval{f(x)}_{a}^{b}=f(b)-f(a).\)
8. \(\displaystyle \sum\nolimits_{a}^{b}g(x)\delta x=\sum_{k=a}^{b-1}g(k)=\sum_{a\leqslant k<b}g(k)~~(b\geqslant a).\)
9. \(\displaystyle \sum\nolimits_{a}^{b}g(x)\delta x+\sum\nolimits_b^c g(x)\delta x=\sum\nolimits_a^cg(x)\delta x.\)
10. \(\displaystyle \sum_{0\leqslant k<n}k^{\underline{m}}=\eval{\dfrac{k^{\underline{m+1}}}{m+1}}_{0}^{n}=\dfrac{n^{\underline{m+1}}}{m+1}~~(m,n\in\mathbb{N}_+).\)
11. \(x^{\underline{-m}}=\dfrac{1}{(x+1)(x+2)\cdots(x+m)}~~(m>0).\)
12. \(x^{\underline{m+n}}=x^{\underline{m}}(x-m)^{\underline{n}} ~~(m,n\in\mathbb{Z}).\)
13. \(\displaystyle\sum\nolimits_a^b x^{\underline{m}}\delta x=\begin{cases} \eval{\dfrac{x^{\underline{m+1}}}{m+1}}_{a}^{b}, & m\neq -1 \\ \eval{H_x}_{a}^{b}, & m=-1 \end{cases}\)， 其中 \(\displaystyle H_x=\sum_{k=1}^{x}\dfrac{1}{k}\) 称为调和级数.
14. \(\mathrm{E}f(x)=f(x+1).\)
15. \(\displaystyle\sum u\varDelta v=uv-\sum\mathrm{E}v\varDelta u.\)

例题

判断级数 \(\displaystyle\sum_{n=1}^{\infty}\dfrac{1+\dfrac{1}{2}+\cdots+\dfrac{1}{n}}{(n+1)(n+2)}\) 的敛散性，若收敛，则求其和.

解答

令 \(\displaystyle H_n=\sum_{k=1}^{n}\dfrac{1}{k}~,S_n=\sum_{k=1}^{n}\dfrac{H_k}{(k+1)(k+2)}=\sum_{\substack{1\leqslant k<n+1\\k\in \mathbb{N}_+}}\dfrac{H_k}{(k+1)(k+2)}=\sum\nolimits_1^{n+1}\dfrac{H_x}{(x+1)(x+2)}\delta x\)， 那么记 \[u(x)=H_x~,\varDelta v(x)=\dfrac{1}{(x+1)(x+2)}=-\dfrac{1}{x+2}-\qty(-\dfrac{1}{x+1})\] 因此 \[\varDelta u(x)=x^{\underline{-1}},~v(x)=-\dfrac{1}{x+1}=-x^{\underline{-1}},~\mathrm{E}v(x)=-(x+1)^{\underline{-1}}\] 于是 \[\begin{align*} S_n & =\sum\nolimits_1^{n+1}\dfrac{H_x}{(x+1)(x+2)}\delta x=\eval{-x^{\underline{-1}}H_x}_{1}^{n+1}+\sum\nolimits_{1}^{n+1}(x+1)^{\underline{-1}}x^{\underline{-1}}\delta x \\ & =-(n+1)^{\underline{-1}}H_{n+1}+1^{\underline{-1}}H_1+\sum\nolimits_{1}^{n+1}x^{\underline{-2}}\delta x=\dfrac{1}{2}-\dfrac{H_{n+1}}{n+2} -\eval{x^{\underline{-1}}}_{1}^{n+1} \\ & =1-\dfrac{1}{n+2}-\dfrac{H_{n+1}}{n+2}=1-\dfrac{1}{n+1}-\dfrac{H_n}{n+2} \end{align*}\] 又因为 \[\lim_{n\to\infty}\dfrac{H_n}{n+2}=\lim_{n\to\infty}\dfrac{\ln n+\gamma+\alpha(n)}{n+2}=\lim_{n\to\infty}\dfrac{\ln n}{n+2}=0~~(\alpha(n)\to0)\] 所以 \(\displaystyle\lim_{n\to\infty}S_n=1\)，因此级数收敛，且其和为 1.