Chứng minh rằng:\(\frac{1}{5}+\frac{1}{15}+\frac{1}{25}+....+\frac{1}{1985}< \frac{9}{20}\)
mk làm thế này đúng ko mọi người
Đặt \(A=\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+......+\frac{1}{243}\)
\(A=\frac{1}{3}+\left(\frac{1}{5}+\frac{1}{7}+\frac{1}{9}\right)+\left(\frac{1}{11}+\frac{1}{13}+\frac{1}{15}+....+\frac{1}{27}\right)+\left(\frac{1}{29}+\frac{1}{31}+\frac{1}{33}+....+\frac{1}{81}\right)+\left(\frac{1}{83}+\frac{1}{85}+\frac{1}{87}+.....+\frac{1}{243}\right)\)
\(=>A>\frac{1}{3}+\frac{1}{9}.3+\frac{1}{27}.9+\frac{1}{81}.27+\frac{1}{243}.81=\frac{1}{3.5}=\frac{5}{3}\)
\(=>A>\frac{5}{3}>\frac{5}{4}=>A< \frac{5}{4}\)
\(=>\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+....+\frac{1}{397}< \frac{5}{4}\)
\(=>1+\frac{1}{3}+\frac{1}{7}+....+\frac{1}{397}< \frac{5}{4}\)
\(=>\frac{1}{5}.\left(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+....+\frac{1}{397}\right)< \frac{9}{4}.\frac{1}{5}\)
\(=>\frac{1}{5}+\frac{1}{15}+\frac{1}{25}+......+\frac{1}{1985}< \frac{9}{20}\)
