[文化课] 数列不等式专题

裂项

$$\frac{1}{n+k}=\frac{1}{k} (\frac{1}{n} - \frac{1}{n+k})$$

$$\frac{2n+1}{n^2 (n+1)^2}=\frac{1}{n^2} - \frac{1}{(n+1)^2}$$

$$\frac{2^n}{(2^n+1)(2^{n+1}+1)}=\frac{1}{2^n+1} - \frac{1}{2^{n+1}+1}$$

$$\frac{1}{n(n+1)(n+2)}=\frac{1}{2}[\frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)}]$$

$$\frac{n+2}{n(n+1) \cdot 2^{n+1}}=\frac{1}{n \cdot 2^n} - \frac{1}{(n+1) \cdot 2^{n+1}}$$

$$\frac{1}{\sqrt{n}+\sqrt{n+k}}=\frac{1}{k}(\sqrt{n+k} - \sqrt{n})$$

$$n \cdot n!=(n+1)!-n!$$

$$\binom{n}{m-1}=\binom{n+1}{m} - \binom{n}{m}$$

$$tan a_n \cdot tan a_{n+1}=\frac{tan a_{n+1} - tan a_n}{tan(a_{n+1}-a_n)}-1$$

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$(an+b) \cdot c^n$既可以错位相减，也可以裂项（待定系数）

如 $(2n-1) \cdot 3^n=((n+1)-2) \cdot 3^{n+1} - (n-2) \cdot 3^n$

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积分

$$\sum_{i=1}^n i^{-\frac{3}{2}} < 1 + \int_1^n x^{-\frac{3}{2}} \  \mathrm{d}x = 3-\frac{2}{\sqrt{n}} < 3$$

$$\sum_{i=1}^n \ln{i} > \int_1^n \ln{x} \ \mathrm{d}x = n\ln{n} - n + 1 > 2(\sqrt{n}-1)^2$$

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sp

$$\binom{n}{k} \frac{1}{n^k} = \frac{\prod_{i=n-k+1}^{n} i}{k! \cdot n^k} < \frac{1}{k!} < \frac{1}{(k-1)k}$$

$$(1+\frac{1}{n})^n = \sum_{i=1}^n \binom{n}{i} \frac{1}{n^i} < 2 + \sum_{i=2}^n \frac{1}{(k-1)k} = 3 - \frac{1}{n} < 3$$

当然$$\lim_{n \to +\infty}(n+\frac{1}{n})^n = e$$