Question

Cho \(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{2005}}\)

CMR:\(B< \frac{1}{2}\)

Answer 1

Ta có \(B=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2005}}\)

\(\Rightarrow\frac{1}{3}.B=\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2006}}\)

\(\Rightarrow B-\frac{1}{3}.B=\frac{1}{3}-\frac{1}{3^{2006}}\)

\(\frac{2}{3}.B=\frac{1}{3}-\frac{1}{3^{2006}}\)

\(B=\left(\frac{1}{3}-\frac{1}{3^{2006}}\right):\frac{2}{3}\)

\(B=\frac{1}{3}:\frac{2}{3}-\frac{1}{3^{2006}}:\frac{2}{3}=\frac{1}{2}-\frac{1}{2.3^{2005}}< \frac{1}{2}\)