常见数列处理方法

通项

$$1^2+2^2+3^2+\cdots+n^2=\dfrac{n(n+1)(2n+1)}{6}$$

$$1^3+2^3+3^3+\cdots+n^3=\left[ \dfrac{n(n+1)}{2}\right]^2$$

$$\dfrac{1}{n(n+1)(n+2)}=\dfrac{1}{2}\left[\dfrac{1}{n(n+1)}-\dfrac{1}{(n+1)(n+2)}\right]$$

$$\dfrac{1}{n(n+1)(n+2)(n+3)}=\dfrac{1}{3}\left[\dfrac{1}{n(n+1)(n+2)}-\dfrac{1}{(n+1)(n+2)(n+3)}\right]$$

放缩

糖水：

$$\dfrac{b+m}{a+m}>\dfrac{b}{a} \ \ \ (a>b>0,m>0)$$

$\dfrac{1}{\sqrt{n}}$:

$$\dfrac{2}{\sqrt{n+1}+\sqrt{n}}<\dfrac{1}{\sqrt{n}}<\dfrac{2}{\sqrt{n}+\sqrt{n-1}} \ \ \ (n \ge 2)$$

$\dfrac{1}{n^2}$：

$$\dfrac{1}{n^2+n}<\dfrac{1}{n^2}<\dfrac{1}{n^2-\frac{1}{4}}<\dfrac{1}{n^2-1}<\dfrac{1}{n^2-n} \ \ \ (n \ge 2)$$

$\dfrac{1}{n^3}$：

$$\dfrac{1}{n^3}<\dfrac{1}{n(n^2-1)}=\dfrac{1}{2}\left[\dfrac{1}{(n-1)n}-\dfrac{1}{n(n+1)}\right] \ \ \ (n \ge 2)$$

$\dfrac{1}{n\sqrt{n}}$：

$$2\left(\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\right)<\dfrac{1}{n\sqrt{n}}<2\left(\dfrac{1}{\sqrt{n-1}}-\dfrac{1}{\sqrt{n}}\right) \ \ \ (n \ge 2)$$