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\(\frac{\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}+...+\frac{1}{3}+\frac{1}{2}}{2012+\frac{2012}{2}+\frac{2011}{3}+...+\frac{1}{2013}}\)
cậu viết sai đề sửa lại:
\(=\frac{\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}+...+\frac{1}{3}+\frac{1}{2}}{\left(1+\frac{2012}{2}\right)+\left(1+\frac{2011}{3}\right)+...+\left(1+\frac{1}{2013}\right)}\)
\(=\frac{\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}+...+\frac{1}{3}+\frac{1}{2}}{\frac{2014}{2}+\frac{2014}{3}+...+\frac{2014}{2013}}\)
\(=\frac{\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}+...+\frac{1}{3}+\frac{1}{2}}{2014.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}\right)}\)
\(=\frac{1}{2014}\)
\(\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}\right).x=\frac{2012}{1}+\frac{2011}{2}+\frac{2010}{3}+...+\frac{1}{2012}\)
Ta có: \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2010^2}<\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2009.2010}\)
\(<1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2009}-\frac{1}{2010}\)
\(<1-\frac{1}{2010}\)
\(<\frac{2009}{2010}<1\)
=>N<1