Question

[Lê Phương Nga]

So sánh : A = \(\frac{1}{2^2}\)+ \(\frac{1}{3^2}\)+ \(\frac{1}{4^2}\)+ ..............+ \(\frac{1}{2018^2}\)với    B = \(\frac{75}{100}\)

Ta có  \(\frac{1}{3^2}\)< \(\frac{1}{2.3}\)                   \(\frac{1}{4^2}\)< \(\frac{1}{3.4}\)               \(\frac{1}{2018^2}\)< \(\frac{1}{2017.2018}\)

Suy ra : A < \(\frac{1}{2^2}\)+ \(\frac{1}{2.3}\)+ \(\frac{1}{3.4}\)+............................+ \(\frac{1}{2017.2018}\)

Gọi biểu thức \(\frac{1}{2.3}\)+ \(\frac{1}{3.4}\)+ ............... +  \(\frac{1}{2017.2018}\)là C 

\(\Rightarrow\)A < \(\frac{1}{2^2}\) +  C = \(\frac{1}{4}\) +  \(\frac{1}{2}\)-  \(\frac{1}{3}\)+ \(\frac{1}{3}\)- \(\frac{1}{4}\)+ ...................+ \(\frac{1}{2017}\)-   \(\frac{1}{2018}\)=  \(\frac{1}{4}\)+  \(\frac{1}{2}\)-  \(\frac{1}{2018}\)

\(\Rightarrow\)A < ( \(\frac{1}{4}\)+  \(\frac{1}{2}\))    -   \(\frac{1}{2018}\) = \(\frac{3}{4}\) - \(\frac{1}{2018}\)< \(\frac{3}{4}\)=  \(\frac{75}{100}\)

\(\Rightarrow\)A < B =  \(\frac{75}{100}\)( đpcm)