裂项
$$frac{1}{n+k}=frac{1}{k} (frac{1}{n} - frac{1}{n+k})$$
$$frac{2n+1}{n^2 (n+1)^2}=frac{1}{n^2} - frac{1}{(n+1)^2}$$
$$frac{2^n}{(2^n+1)(2^{n+1}+1)}=frac{1}{2^n+1} - frac{1}{2^{n+1}+1}$$
$$frac{1}{n(n+1)(n+2)}=frac{1}{2}[frac{1}{n(n+1)} - frac{1}{(n+1)(n+2)}]$$
$$frac{n+2}{n(n+1) cdot 2^{n+1}}=frac{1}{n cdot 2^n} - frac{1}{(n+1) cdot 2^{n+1}}$$
$$frac{1}{sqrt{n}+sqrt{n+k}}=frac{1}{k}(sqrt{n+k} - sqrt{n})$$
$$n cdot n!=(n+1)!-n!$$
$$inom{n}{m-1}=inom{n+1}{m} - inom{n}{m}$$
$$tan a_n cdot tan a_{n+1}=frac{tan a_{n+1} - tan a_n}{tan(a_{n+1}-a_n)}-1$$
$(an+b) cdot c^n$既可以错位相减，也可以裂项（待定系数）
如 $(2n-1) cdot 3^n=((n+1)-2) cdot 3^{n+1} - (n-2) cdot 3^n$
积分
$$sum_{i=1}^n i^{-frac{3}{2}} < 1 + int_1^n x^{-frac{3}{2}} mathrm{d}x = 3-frac{2}{sqrt{n}} < 3$$
$$sum_{i=1}^n ln{i} > int_1^n ln{x} mathrm{d}x = nln{n} - n + 1 > 2(sqrt{n}-1)^2$$
sp
$$inom{n}{k} frac{1}{n^k} = frac{prod_{i=n-k+1}^{n} i}{k! cdot n^k} < frac{1}{k!} < frac{1}{(k-1)k}$$
$$(1+frac{1}{n})^n = sum_{i=1}^n inom{n}{i} frac{1}{n^i} < 2 + sum_{i=2}^n frac{1}{(k-1)k} = 3 - frac{1}{n} < 3$$
当然$$lim_{n o +infty}(n+frac{1}{n})^n = e$$