Question

Cho A 1 phần 1mũ 2 1 phần 2 mũ 2 ... 1 phần 2017 mũ 2 chứng minh A nhỏ hơn 1

Answer 1

Ta có : \(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2017^2}\)

\(=1+\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{2017.2017}\)

Vì \(1=1\)

\(\frac{1}{2.2}< \frac{1}{1.2}\)

\(\frac{1}{3.3}< \frac{1}{2.3}\)

\(...\)

\(\frac{1}{2017.2017}< \frac{1}{2016.2017}\)

\(\Rightarrow A< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2016.2017}\)

         \(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2016}-\frac{1}{2017}\)

         \(=2-\frac{1}{2017}< 2\)

\(\Rightarrowđpcm\)