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a, 1/5+1/6+1/7+1/8+1/9 < 1/5.5=1 (1)
1/10+1/11+1/12+1/13+1/14+1/15+1/16+1/17 < 1/10.7 < 1/10.10 < 1 (2)
Từ (1) và (2) , suy ra 1/5+1/6+1/7+...+1/17 < 1+1 =2
Suy ra , 1/5+1/6+1/7+...+1/17 < 2
b, Ta cần c/m 1/13+1/25+1/41+1/61+1/85+1/113 < 3/10 (Vì 1/2 - 1/5 = 3/10)
1/13+1/25+1/41+1/61+1/85+1/113 < 1/10+1/25+1/25+1/25+1/25+1/25
1/13+1/25+1/41+1/61+1/85+1/113 < 1/10 + 5/25 = 1/10+1/5 = 3/10
Suy ra , 1/5+1/13+1/25+1/41+1/61+1/85+1/113 < 1/2
B=2+1/1.2+2+1/2.3+........+2+1/9.10
B=2.9+1/1.2+1/2.3+........+1/9.10
B=18+9/10
Xét vế trái : \(T=\frac{1}{5}+\frac{1}{13}+\frac{1}{25}+...+\frac{1}{221}\)
Ta có : \(T< \frac{1}{5}+\frac{1}{12}+\frac{1}{24}+...+\frac{1}{220}\)
\(=\frac{1}{5}+\frac{1}{2}\left(\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\right)=\frac{1}{5}+\frac{1}{2}\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)\)
\(=\frac{1}{5}+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)\)
\(=\frac{1}{5}+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{11}\right)< \frac{1}{5}+\frac{1}{4}\Rightarrow T< \frac{9}{20}\)
đặt A=\(\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{25}+\dfrac{1}{41}+\dfrac{1}{61}+\dfrac{1}{85}+\dfrac{1}{113}\)
= \(\dfrac{1}{5}+(\dfrac{1}{13}+\dfrac{1}{25}+\dfrac{1}{41})+(\dfrac{1}{61}+\dfrac{1}{85}+\dfrac{1}{113})\)
=> A< \(\dfrac{1}{5}+(\dfrac{1}{12}+\dfrac{1}{12}+\dfrac{1}{12})+(\dfrac{1}{60}+\dfrac{1}{60}+\dfrac{1}{60})\)
A<\(\dfrac{1}{5}+\dfrac{1}{4}+\dfrac{1}{20}\)=\(\dfrac{1}{2}\)
vậy A<\(\dfrac{1}{2}\),<2=> A<2 (đpcm)