chứng tỏ rằng A=1+1\2^2+1\3^2+….+1\2016^2<7\4

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1. Ta có $a^2>a^2-1=a^2-ab+ab-1=(a-1)(a+1)$

   Từ đó ta có $\dfrac{1}{a^2}<\dfrac{1}{(a-1)(a+1)}$
   $\begin{array}{l}
   A = 1 + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{3^2}}} + … + \dfrac{1}{{{{2016}^2}}}\\
    = 1 + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{3^2}}} + … + \dfrac{1}{{{{2016}^2}}} < B = 1 + \dfrac{1}{{1.3}} + \dfrac{1}{{2.4}} + \dfrac{1}{{3.5}} + … + \dfrac{1}{{2014.2016}}\\
   B = 1 + \dfrac{1}{2}.\left( {\dfrac{1}{1} – \dfrac{1}{3} + \dfrac{1}{2} – \dfrac{1}{4} + \dfrac{1}{3} – \dfrac{1}{5} + … + \dfrac{1}{{2014}} – \dfrac{1}{{2016}}} \right) = 1 + \dfrac{1}{2}\left( {1 + \dfrac{1}{2} – \dfrac{1}{{2015}} – \dfrac{1}{{2017}}} \right) < 1 + \dfrac{1}{2}.\dfrac{3}{2} = \dfrac{7}{4}
   \end{array}$

    

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