Question

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

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

Answer 1

Sửa đề: Cho \(B=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2005}}\). CMR: \(B< \frac{1}{2}\)

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

\(\Rightarrow3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}\). Lại có:

\(3B-B=2B=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2005}}\right)\)

\(2B=1-\frac{1}{3^{2005}}< 1\Rightarrow B=\frac{1-\frac{1}{3^{2005}}}{2}< \frac{1}{2}^{\left(đpcm\right)}\)