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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)
1/5+1/13+1/25+1/41+1/61+1/85+1/113
=1/5+(1/13+1/25+1/41)+(1/85+1/61+1/113)<15+1/12+1/12+1/12+1/60+1/60+1/60
..............................................................<1/5+1/4+1/20
..............................................................<4/20+5/20+1/20
..............................................................<1/2
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