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Đặt \(A=1+\frac{1}{2}+...+\frac{1}{64}\)
Ta có: \(A=1+\frac{1}{2}+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+...+\frac{1}{8}\right)+\left(\frac{1}{9}+...+\frac{1}{16}\right)+\left(\frac{1}{17}+...+\frac{1}{32}\right)\)\(+\left(\frac{1}{33}+...+\frac{1}{64}\right)\)
Ta thấy : \(\frac{1}{3}+\frac{1}{4}>\frac{1}{4}+\frac{1}{4}=\frac{1}{2}\)
\(\frac{1}{5}+...+\frac{1}{8}>\frac{1}{8}+...+\frac{1}{8}=\frac{1}{8}.4=\frac{1}{2}\)
\(\frac{1}{9}+\frac{1}{16}>\frac{1}{16}+...+\frac{1}{16}=\frac{1}{16}.8=\frac{1}{2}\)
\(\frac{1}{17}+...+\frac{1}{32}>\frac{1}{32}+...+\frac{1}{32}=\frac{1}{32}.16=\frac{1}{2}\)
\(\frac{1}{33}+...+\frac{1}{64}>\frac{1}{64}+...+\frac{1}{64}=\frac{1}{64}.32=\frac{1}{2}\)
\(\Rightarrow A>1+\frac{1}{2}.6=4\)
Vậy \(A>4\)
\(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{63}+\frac{1}{64}>4.\)
Có :\(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{63}+\frac{1}{64}\)
\(=1+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+...+\frac{1}{8}\right)+\left(\frac{1}{9}+...+\frac{1}{16}\right)+\left(\frac{1}{17}+....+\frac{1}{32}\right)+\left(\frac{1}{33}+...+\frac{1}{64}\right)\)
Ta có \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{13}{12}\)
và \(\frac{1}{2}+\frac{1}{4}+\frac{1}{4}=1\)
\(\Rightarrow\frac{1}{2}+\frac{1}{3}+\frac{1}{4}>\frac{1}{2}+\frac{1}{4}+\frac{1}{4}\)
ta có \(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}=\frac{533}{840}\)
và \(\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{1}{2}\)
\(\Rightarrow\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}>\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}\)
tương tự như trên ta tính được
\(\frac{1}{9}+\frac{1}{10}+...+\frac{1}{16}>\frac{1}{16}\cdot8=\frac{1}{2}\)
\(\frac{1}{17}+\frac{1}{18}+...+\frac{1}{32}>\frac{1}{32}\cdot16=\frac{1}{2}\)
\(\frac{1}{33}+\frac{1}{34}+\frac{1}{35}+..+\frac{1}{64}>\frac{1}{64}\cdot32=\frac{1}{2}\)
\(\Rightarrow1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{63}+\frac{1}{64}>1+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+1\)
\(\Rightarrow1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{63}+\frac{1}{64}>4\)
đúng. vì các ps ko lớn hơn 1