Ta có: \(\frac{1}{2^2}< \frac{1}{1.2}=1-\frac{1}{2}\)

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

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

\(\frac{1}{2013^2}< \frac{1}{2012.2013}=\frac{1}{2012}-\frac{1}{2013}\)

=> \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2013^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2012}-\frac{1}{2013}=1-\frac{1}{2013}< 1\)

=> \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2013^2}< 1\)

Mà \(\frac{2014}{2013}>1\)

=> \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2013^2}< \frac{2014}{2013}\)

Ta có: \(M=1\cdot2014+2\cdot2013+\cdots+2014\cdot1\)

\(=2\left(1\times2014+2\times2013+\cdots+1007\times1008\right)\)

\(=2\left\lbrack1\times\left(2015-1\right)+2\times\left(2015-2\right)+\cdots+1007\times\left(2015-1007\right)\right\rbrack\)

\(=2\cdot\left\lbrack2015\times\left(1+2+\cdots+1007\right)-\left(1^2+2^2+\cdots+1007^2\right)\right\rbrack\)

\(=2\cdot\left\lbrack2015\times1007\times\frac{1008}{2}-\frac{1007\times\left(1007+1\right)\times\left(2\times1007+1\right)}{6}\right\rbrack\)

\(=2\cdot\left\lbrack2015\times1007\times504-1007\times168\times2015\right\rbrack=2\times2015\times1007\times168\left(3-1\right)=4\times168\times2015\times1007\)

\(N=1+\left(1+2\right)+\cdots+\left(1+2+\cdots+2014\right)\)

\(\) \(=\frac{1\times2}{2}+\frac{2\times3}{2}+\cdots+\frac{2014\times2015}{2}\)

\(=\frac12\times\left(1\times2+2\times3+\cdots+2014\times2015\right)\)

\(=\frac12\times\left\lbrack1\times\left(1+1\right)+2\times\left(2+1\right)+\cdots+2014\times\left(2014+1\right)\right\rbrack\)

\(=\frac12\times\left\lbrack\left(1\times1+2\times2+\cdots+2014\times2014\right)+\left(1+2+\cdots+2014\right)\right\rbrack\)

\(=\frac12\times\left\lbrack\frac{2014\times\left(2014+1\right)\times\left(2\times2014+1\right)}{6}+\frac{2014\times2015}{2}\right\rbrack\)

\(=\frac12\times\left\lbrack1007\times2015\times1343+1007\times2015\right\rbrack=\frac12\times1007\times2015\times\left(1343+1\right)\)

=1007x2015x672

=4x168x2015x1007

Do đó: M=N

\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{2014^2}-1\right)\)

\(-A=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{2014^2}\right)\)

\(-A=\frac{3}{2\cdot2}\cdot\frac{8}{3\cdot3}\cdot\frac{15}{4\cdot4}\cdot...\cdot\frac{4056195}{2014\cdot2014}\)

\(-A=\frac{\left(1\cdot3\right)\left(2\cdot4\right)\left(3\cdot5\right)...\left(2013\cdot2015\right)}{\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)...\left(2014\cdot2014\right)}\)

\(-A=\frac{\left(1\cdot2\cdot3\cdot...\cdot2013\right)\left(3\cdot4\cdot5\cdot...\cdot2015\right)}{\left(2\cdot3\cdot4\cdot...\cdot2014\right)\left(2\cdot3\cdot4\cdot...\cdot2014\right)}\)

\(-A=\frac{1\cdot2015}{2014\cdot2}=\frac{2015}{4028}\)

\(A=\frac{-2015}{4028}\)

Bảo Online Math làm cho

$A=\frac{1}{2^2-1}+\frac{1}{3^2-1}+...+\frac{1}{2014^2-1}=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2013.2014}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2013}-\frac{1}{2014}=1-\frac{1}{2014}=\frac{2013}{2014}>-\frac{1}{2}$