Question

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Answer 1

TA có: \(\frac{1}{2^2}<\frac{1}{1\cdot2}=1-\frac12\)

\(\frac{1}{3^2}<\frac{1}{2\cdot3}=\frac12-\frac13\)

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\(\frac{1}{2012^2}<\frac{1}{2011\cdot2012}=\frac{1}{2011}-\frac{1}{2012}\)

Do đó: \(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{2012^2}<1-\frac12+\frac12-\frac13+\cdots+\frac{1}{2011}-\frac{1}{2012}\)

=>\(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{2012^2}<1-\frac{1}{2012}\)

=>\(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{2012^2}<1\)

=>\(0<\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{2012^2}<1\)

=>\(0+1<1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{2012^2}<1+1\)

=>1<S<2

=>S không là số tự nhiên