A không thể lớn hơn 1 được

Ta có:

\(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{49}>\frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}=\frac{40}{50}=\frac{4}{5}\)

\(\frac{1}{50}+\frac{1}{51}+...+\frac{1}{99}>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{50}{100}=\frac{1}{2}\)

Từ đây ta suy ra 

A > \(\frac{4}{5}+\frac{1}{2}+\frac{1}{100}=1,31>1\)  

Hình như có lần đã c/m A>6/5

Đặt vế trái của Bất đẳng thức la A

\(A< \frac{1}{8}+\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+\frac{1}{40}+\frac{1}{40}+\frac{1}{40}.\)

\(A< \frac{1}{8}+\frac{3}{10}+\frac{3}{40}=\frac{3}{10}< \frac{5}{10}=\frac{1}{2}\)

hhhhhhhhh

Ta thấy: \(\frac{1}{8}< \frac{1}{2}\)

             \(\frac{1}{11}< \frac{1}{2}\)

              \(\frac{1}{12}< \frac{1}{2}\) 

              \(\frac{1}{13}< \frac{1}{2}\)

              \(\frac{1}{41}< \frac{1}{2}\)

               \(\frac{1}{42}< \frac{1}{2}\)

               \(\frac{1}{43}< \frac{1}{2}\)

=> \(\frac{1}{8}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{41}+\frac{1}{42}+\frac{1}{43}< \frac{1}{2}\)

 Vì A > 1/91+1/91+...+1/91=1/91*91=1

 Vậy A>1

30 số hạng đầu lớn hơn 1 

\(\frac{1}{10}+\frac{1}{11}+..+\frac{1}{19}>\frac{1}{20}+\frac{1}{20}+..+\frac{1}{20}=\frac{1}{2}\)\(\frac{1}{2}\)

\(\frac{1}{20}+\frac{1}{21}+..+\frac{1}{29}>\frac{1}{30}+\frac{1}{30}+..+\frac{1}{30}=\frac{1}{3}\)

\(\frac{1}{30}+\frac{1}{31}+...+\frac{1}{39}>\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}=\frac{1}{4}\)

=> \(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{39}>\frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{13}{12}>1\)

\(A=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}\)

\(A=\frac{1}{10}+\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}\right)>\frac{1}{10}+\left(\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}\right)\)

\(A=\frac{1}{10}+\frac{99}{100}=1\)

=> A > 1

\(A=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}\)

\(A=\frac{1}{10}+\frac{1}{11}+...+\frac{1}{19}>\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}=\frac{10}{20}=\frac{1}{2}\)

\(A=\frac{1}{20}+\frac{1}{21}+...+\frac{1}{29}>\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}=\frac{10}{30}=\frac{1}{3}\)

\(A=\frac{1}{30}+\frac{1}{31}+...+\frac{1}{39}>\frac{1}{40}+\frac{1}{40}+... +\frac{1}{40}=\frac{10}{40}=\frac{1}{4}\)

\(\Rightarrow A>1\)

Ta thấy:1/10;1/11;1/12;1/13;...;1/99>1/100

=)1/10+1/11+1/12+1/13+...+1/100>1/100+1/100+1/100+1/100..+1/100=1/100.100=1

Vậy A>1

\(A=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}\)

\(=\frac{1}{10}+\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}\right)>\frac{1}{10}+\left(\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}\right)\)

\(=\frac{1}{10}+\frac{90}{100}>1\)

\(A>1\left(đpcm\right)\)

a>1(đpcm)

a

A=1+3+3²+...+3^30

3A=3(1+3+3²+...+3^30)

3A=3+3²+3^3+...+3^31

3A-A=3^31-1

=>A=3^31-1

Ta có:A=1/10+1/11+1/12+...+1/99+1/100

           =1/10+(1/11+1/12+...+1/100)

          >1/10+(1/100+1/100+1/100+...+1/100)

           =1/10+90/100=1/10+9/10=1

      Vậy A>1

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