Câu hỏi 67
Chứng minh bất đẳng thức
$$
\frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots + \frac{1}{2022!} \gt \frac{1^2}{2!} + \frac{2^2}{3!} + \frac{3^2}{4!} + \dots + \frac{2022^2}{2023!}.
$$
$$
\frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots + \frac{1}{2022!} \gt \frac{1^2}{2!} + \frac{2^2}{3!} + \frac{3^2}{4!} + \dots + \frac{2022^2}{2023!}.
$$