Câu hỏi 110
Cho $a$, $b$, $c$ là số nguyên dương. Cho thấy rằng $$
\left( 1 + \frac{a-c}{b} \sum_{i=1}^{b} \frac{1}{c+i} \right)^{-b} \le \frac{a!\,(b+c)!}{(a+b)!\,c!} \le \left( 1 + \frac{c-a}{b} \sum_{i=1}^{b} \frac{1}{a+i} \right)^{b}.
$$
\left( 1 + \frac{a-c}{b} \sum_{i=1}^{b} \frac{1}{c+i} \right)^{-b} \le \frac{a!\,(b+c)!}{(a+b)!\,c!} \le \left( 1 + \frac{c-a}{b} \sum_{i=1}^{b} \frac{1}{a+i} \right)^{b}.
$$