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Bài 1 :
Ta có;\(\frac{1}{21}+\frac{1}{22}+\frac{1}{23}+...+\frac{1}{30}>\frac{1}{30}.10=\frac{1}{3}\)
\(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}>\frac{1}{60}.30>\frac{1}{30}.24=\frac{2}{5}\)
Do đó :
\(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{60}>\frac{1}{3}+\frac{2}{5}=\frac{11}{15}\left(1\right)\)
Mặt khác :
\(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{40}< \frac{1}{20}.20=1\)
\(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{60}< \frac{1}{40}.20=\frac{1}{2}\)
Do đó :
\(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{60}< 1+\frac{1}{2}=\frac{3}{2}\left(2\right)\)
Từ (1 ) và (2) ta suy ra điều phải chứng minh
Bài 2 :
Đặt \(S=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{63}\)
MỘT MẶT ,TA CÓ THỂ VIẾT
\(S=\left(1+\frac{1}{2}\right)+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)\)\(+\left(\frac{1}{9}+\frac{1}{10}+...+\frac{1}{16}\right)+\left(\frac{1}{17}+\frac{1}{18}+...+\frac{1}{32}\right)\)\(+\left(\frac{1}{33}+\frac{1}{34}+...+\frac{1}{63}+\frac{1}{64}\right)-\frac{1}{64}\)
\(>\frac{1}{2}.2+\frac{1}{4}.2+\frac{1}{8}.4+\frac{1}{16}.8+\frac{1}{32}.16+\frac{1}{64}.32-\frac{1}{64}\)\(=\frac{7}{2}-\frac{1}{64}=\frac{223}{64}>\frac{192}{64}=3\left(1\right)\)
Mặt khác ,ta lại có\(S=1+\left(\frac{1}{2}+\frac{1}{3}\right)+\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}\right)\)\(+\left(\frac{1}{8}+\frac{1}{9}+...+\frac{1}{15}\right)+\left(\frac{1}{16}+\frac{1}{17}+...+\frac{1}{31}\right)\)\(+\left(\frac{1}{32}+\frac{1}{33}+...+\frac{1}{63}\right)< \)\(1+\frac{1}{2}.2+\frac{1}{4}.4+\frac{1}{8}.8+\frac{1}{16}.16+\frac{1}{32}.32=6\left(2\right)\)
Từ (1) và (2 ) ta kết luận \(3< S< 6\)
Chúc bạn học tốt ( -_- )
A=1+(1/2 + 1/3 + 1/4)+(1/5 + 1/6 + 1/7 + 1/8)+(1/9+...+1/16)+(1/17+...+1/32)+(1/33+...+1/64)
A>1+(1/2 + 1/4 + 1/4)+(1/8+ 1/8+ 1/8+ 1/8)+(1/16+1/16+...+1/16)+(1/64+...+1/64)
A>1 + 1 + 1/2 + 1/2 + 1/2+ 1/2
A>4
1/2+1/3+1/4+...+1/63>1/31+1/31+...+1/31(62 số hạng 1/31)
hay 1/2+1/3+1/4+...+1/63>62 x 1/31
nên 1/2+1/3+1/4+...+1/63>2(dpcm)
\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+......+\frac{3}{43.46}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{43}-\frac{1}{46}\)
\(=1-\frac{1}{46}< 1\)
Vậy \(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+......+\frac{3}{43.46}< 1\)
\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}\)
Ta có : \(\frac{1}{2^2}=\frac{1}{2\cdot2}< \frac{1}{1\cdot2}\)
\(\frac{1}{3^2}=\frac{1}{3\cdot3}< \frac{1}{2\cdot3}\)
...
\(\frac{1}{8^2}=\frac{1}{8\cdot8}< \frac{1}{7\cdot8}\)
Cộng vế theo vế
\(\Rightarrow B=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{7\cdot8}\)
\(\Rightarrow B< \frac{1}{1}-\frac{1}{8}=\frac{7}{8}\)
Lại có \(\frac{7}{8}< 1\)
Theo tính chất bắc cầu => \(B< \frac{7}{8}< 1\)
\(\Rightarrow B< 1\left(đpcm\right)\)
Ta có \(D=\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{10^2}< \frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{9.10}.\)
Mà \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}=\frac{9}{10}< 1\)
\(\Rightarrow D< 1\)
Vậy \(D< 1\)
Ta có: 1/22 < 1/1.2
1/32 < 1/2.3
1/42 < 1/3.4
......
1/102 < 1/9.10
=> D < 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/9.10
=> D < 1 -1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/9 -1/10
=> D < 1 - 1/10
=> D < 9/10
=. D < 9/10 < 1
=> D < 1 ( đpcm )