Question

Cho A = 1+ \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{2015^2}+\frac{1}{2016^2}\) .

Chứng minh rằng: A < 1\(\frac{3}{4}\)

 

 

Answer 1

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Answer 2

Ta có

\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{2016^2}\)

\(\Rightarrow A< 1+\frac{1}{4}+\frac{1}{2.3}+......+\frac{1}{2015.2016}\)

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

\(\Rightarrow A< 1\frac{3}{4}-\frac{1}{2016}< 1\frac{3}{4}\)

=> đpcm