Đáp án:
Đặt `A = 1/(1!) + 1/(2!) + 1/(3!) + ... + 1/(2021!)`
`⇔ A = 1 + 1/(1 × 2) + 1/(1 × 2 × 3) + ... + 1/(1 × 2 × 3 × ... ×2021)`
Vì \(\left\{ \begin{array}{l}1 = 1\\ \dfrac{1}{1×2}=\dfrac{1}{1×2}\\ \dfrac{1}{1×2×3} < \dfrac{1}{2×3}\\..........\\ \dfrac{1}{1×2×3×...×2021} < \dfrac{1}{2020 × 2021}\end{array} \right.\)
Nên `A < 1 + 1/(1 × 2) + 1/(2 × 3) + ... + 1/(2020 × 2021)`
`-> A < 1 + 1 - 1/2 + 1/2 - 1/3 + ... + 1/2020 - 1/2021`
`-> A < 1 + 1 + (- 1/2 + 1/2 - 1/3 + ... + 1/2020) - 1/2021`
`-> A < 2 - 1/2021`
Ta thấy : `2 - 1/2021 < 2`
`-> A < 2 - 1/2021 < 2`
`-> A < 2`
hay `A < P`
