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$\frac{1}{k!}{\sum }_{i=2}^k(A_k^i-C_k^i)(\frac{3}{4}{)}^i=\frac{1}{k!}{\sum }_{i=2}^k(C_k^{i}i!-C_k^i)(\frac{3}{4}{)}^i=\frac{1}{k!}{\sum }_{i=2}^k{C}_k^{i}i!(\frac{3}{4}{)}^i-\frac{1}{k!}{\sum }_{i=2}^k{C}_k^i(\frac{3}{4}{)}^i$，对上两项分开求和
1.求$\frac{1}{k!}{\sum }_{i=2}^k{C}_k^{i}i!(\frac{3}{4}{)}^i$，因为 $C_k^i=\frac{k!}{i!(k-i)!}$，代入得 $\frac{1}{k!}{\sum }_{i=2}^k{C}_k^{i}i!(\frac{3}{4}{)}^i=(\frac{3}{4}{)}^k+{\sum }_{i=2}^{k-1}\frac{1}{(k-i)!}(\frac{3}{4}{)}^i=(\frac{3}{4}{)}^k+(\frac{3}{4}{)}^k{\sum }_{i=2}^{k-1}\frac{1}{(k-i)!}(\frac{4}{3}{)}^{(k-i)}$，又因为 ${\lim }_{k\to +\infty }{\sum }_{i=2}^{k-1}\frac{1}{(k-i)!}(\frac{4}{3}{)}^{(k-i)}=e^{4/3}$，于是原式 $={\lim }_{k\to +\infty }(\frac{3}{4}{)}^k+(\frac{3}{4}{)}^k{e}^{4/3}=0$
2.求$\frac{1}{k!}{\sum }_{i=2}^k{C}_k^i(\frac{3}{4}{)}^i$，因为 ${\sum }_{i=0}^k{C}_k^i(\frac{3}{4}{)}^i=(1+\frac{3}{4}{)}^k$，所以原式$={\lim }_{k\to +\infty }\frac{(1+\frac{3}{4}{)}^k-1-\frac{3}{4}k}{k!}=0$

综合式1，2，原式等于0