Question

Chứng minh rằng tổng \(P=\frac{1}{3^2}-\frac{1}{3^4}+...+\frac{1}{3^{2006}}-\frac{1}{3^{2008}}\)  nhỏ hơn 0, 1

Answer 1

\(P=\frac{1}{3^2}-\frac{1}{3^4}+....+\frac{1}{3^{2006}}-\frac{1}{3^{2008}}\)

\(\Rightarrow9P=1-\frac{1}{3^2}+....+\frac{1}{3^{2004}}-\frac{1}{3^{2006}}\)

\(\Rightarrow9P+P=\left(1-\frac{1}{3^2}+....+\frac{1}{3^{2004}}-\frac{1}{3^{2006}}\right)+\left(\frac{1}{3^2}-\frac{1}{3^4}+....+\frac{1}{3^{2006}}-\frac{1}{3^{2008}}\right)\)

\(\Rightarrow10P=1-\frac{1}{3^{2008}}\)

\(\Rightarrow P=\frac{1}{10}-\frac{1}{3^{2008}\cdot10}< \frac{1}{10}=0,1\)

Vậy \(P< 0,1\)