cho S=1+1/2+1/3+1/4+….+1/2^2017 Chứng minh S > 1009

Lời giải:

Ta có:

$S=1+(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+…+\frac{1}{2^{2017}})>1+(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+…+\frac{1}{2^{2017}.(2^{2017}+1)})$
Lại có:
$\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+…+\frac{1}{2^{2017}.(2^{2017}+1)}=\frac{2^{2017}}{2^{2017}+1}$~$1$
=>$\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+…+\frac{1}{2^{2017}.(2^{2017}+1)}=\frac{2^{2017}}{2^{2017}+1}$~$1-\frac{1}{2}$~$\frac{1}{2}$
$=>S>1+\frac{1}{2}>\frac{3}{2}$
Mà:$\frac{3}{2}>\frac{10}{9}$
=>$S=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+…+\frac{1}{2^{2017}}>\frac{10}{9}(đpcm)$