Question

Chứng minh rằng:

\(\frac{2015}{4034}\)< \(\frac{1}{^{2^2}}\)+\(\frac{1}{3^2}\)+.....+\(\frac{1}{2016^2}\)<\(\frac{2015}{2016}\)

Answer 1

Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2016^2}\) ta  có : 

\(A>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2016.2017}\)

\(A>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2016}-\frac{1}{2017}\)

\(A>\frac{1}{2}-\frac{1}{2017}\)

\(A>\frac{2015}{4034}\) \(\left(1\right)\)

Lại có : 

\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2015.2016}\)

\(A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}\)

\(A< 1-\frac{1}{2016}\)

\(A< \frac{2015}{2016}\) \(\left(2\right)\)

Từ (1) và (2) suy ra : \(\frac{2015}{4034}< A< \frac{2015}{2016}\) ( đpcm ) 

Vậy \(\frac{2015}{4034}< \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2016^2}< \frac{2015}{2016}\)

Chúc bạn học tốt ~ 

Answer 2

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