CMR : 1+$\frac{1}{1.2}$+ $\frac{1}{1.2.3}$ +…+$\frac{1}{1.2.3…n}$ <2

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)