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![](https://rs.olm.vn/images/avt/0.png?1311)
\(\text{Nhân S với 4 ta được :}\)
\(\text{4S = 4/(5x5) + 4/(9x9) + … + 1/(409x409)}\)
\(\text{Ta }co\)
4/(5x5) < 4/(3x7) = 1/3 – 1/7
4/(9x9) < 4/(7x11) = 1/7 – 1/11
4/(409x409) < 4/(407x411) = 1/407 – 1/411
Mà :
\(\text{4/(3x7) + 4/(7x11) + …. + 4/(407x411) = 1/3 – 1/411 = 136/411}\)
4S < 136/411
S < 34/411 < 34/408 = 1/12
Hay S < 1/12
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=\frac{3}{1^2\cdot2^2}+\frac{5}{2^2\cdot3^2}+.....+\frac{19}{9^2\cdot10^2}\)
\(\Rightarrow S=\frac{3}{1\cdot4}+\frac{5}{4\cdot9}+....+\frac{19}{81\cdot100}\)
\(\Rightarrow S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+....+\frac{1}{81}-\frac{1}{100}\)
\(\Rightarrow S=1-\frac{1}{100}=\frac{99}{100}< 1\left(ĐPCM\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(S=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^4+...+\left(\frac{1}{2}\right)^{2016}+\left(\frac{1}{2}\right)^{2017}\)
\(2S=1+\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{2015}+\left(\frac{1}{2}\right)^{2016}\)
\(2S-S=\left[1+\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{2015}+\left(\frac{1}{2}\right)^{2016}\right]-\left[\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^4+...+\left(\frac{1}{2}\right)^{2016}+\left(\frac{1}{2}\right)^{2017}\right]\)
\(S=1-\left(\frac{1}{2}\right)^{2017}< 1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2018}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2017}}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2017}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2018}}\right)\)
\(A=1-\frac{1}{2^{2018}}< 1\)
\(\Rightarrow A< 1\left(đpcm\right)\)
hok tốt .