Đáp án:
Giải thích các bước giải:
Ta có:
$\dfrac{1}{1.2}=\dfrac{1}{1}-\dfrac{1}{2}$
$\dfrac{1}{1.2.3}=\dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3}$
$\dfrac{1}{1.2.3.4}<\dfrac{1}{3.4}=\dfrac{1}{3}-\dfrac{1}{4}$
....
$\dfrac{1}{1.2.3...(n-1)n}<\dfrac{1}{(n-1)n}=\dfrac{1}{n-1}-\dfrac{1}{n}$
Cộng vế với vế:
$⇒1+\dfrac{1}{1.2}+\dfrac{1}{1.2.3}+...+\dfrac{1}{1.2...n}<1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}$
$⇒1+\dfrac{1}{1.2}+\dfrac{1}{1.2.3}+...+\dfrac{1}{1.2...n}<2-\dfrac{1}{n}<2$
$⇒1+\dfrac{1}{1.2}+\dfrac{1}{1.2.3}+...+\dfrac{1}{1.2...n}<2$ (đpcm)
