Đặt A=\(\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{2011^2}\)
Ta có:\(\frac{1}{4^2}< \frac{1}{3\cdot4}=\frac{1}{3}-\frac{1}{4}\)
\(\frac{1}{5^2}< \frac{1}{4\cdot5}=\frac{1}{4}-\frac{1}{5}\)
.............................
\(\frac{1}{2011^2}< \frac{1}{2010\cdot2011}=\frac{1}{2010}-\frac{1}{2011}\)
\(\Rightarrow A< \frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\cdot\cdot\cdot+\frac{1}{2010}-\frac{1}{2011}\)
\(=\frac{1}{3}-\frac{1}{2011}< \frac{1}{3}\)
Vậy A<\(\frac{1}{3}\)hay \(\frac{1}{4^2}+\frac{1}{5^2}+\cdot\cdot\cdot+\frac{1}{2011^2}< \frac{1}{3}\)
\(\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{2011^2}< \frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{2010\cdot2011}\)
Gọi \(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{2010\cdot2011}\)là \(S\)
Ta có:
\(S=\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{2010\cdot2011}\)
\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2010}-\frac{1}{2011}\)
\(=\frac{1}{3}-\frac{1}{2011}< \frac{1}{3}\)
Vì \(\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{2011^2}< S\)mà \(S< \frac{1}{3}\)\(\Rightarrow\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{2011^2}< \frac{1}{3}\)
\(Let\)\(A=\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{2011^2}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
\(\frac{1}{5^2}< \frac{1}{4.5}\)
\(\frac{1}{6^2}< \frac{1}{5.6}\)
\(.................\)
\(\frac{1}{2011^2}< \frac{1}{2010.2011}\)
_____________________________
\(A< \frac{1}{3}-\frac{1}{2011}< \frac{1}{3}\)
\(\Rightarrow A< \frac{1}{3}\left(đpcm\right)\)
