Question

CMR

\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}<1\)

Answer 1

Ta thấy:\(\frac{1}{2^2}=\frac{1}{2.2}<\frac{1}{1.2}\)

\(\frac{1}{3^2}=\frac{1}{3.3}<\frac{1}{2.3}\)

\(\frac{1}{4^2}=\frac{1}{4.4}<\frac{1}{3.4}\)

...................

\(\frac{1}{100^2}=\frac{1}{100.100}<\frac{1}{99.100}\)

=>\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

=>\(A<1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

=>\(A<1-\frac{1}{100}<1-0=1\)

=>A<1

=>\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}<1\)