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Ta có:
\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2012}}\)
\(\Rightarrow3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2011}}\)
\(\Rightarrow3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2011}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2012}}\right)\)
\(\Rightarrow2A=1-\frac{1}{3^{2012}}\)
\(\Rightarrow A=\left(1-\frac{1}{3^{2012}}\right).\frac{1}{2}\)
\(\Rightarrow A=\frac{1}{2}-\frac{1}{3^{2012}}\)
Vì \(\frac{1}{2}-\frac{1}{3^{2012}}< \frac{1}{2}\) nên \(A< \frac{1}{2}\)
Vậy \(A< \frac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+.....+\frac{1}{3^{2012}}\)
\(\Rightarrow3A=1+\frac{1}{3}+\frac{1}{3^2}+......+\frac{1}{3^{2011}}\)
\(\Rightarrow3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+......+\frac{1}{3^{2011}}\right)\)\(-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+.....+\frac{1}{3^{2012}}\right)\)
\(\Rightarrow2A=1-\frac{1}{3^{2012}}\)
\(\Rightarrow A=\frac{1-\frac{1}{3^{2012}}}{2}\)
Vì \(1-\frac{1}{3^{2012}}< 1\Rightarrow A< \frac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
B=\(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2012}}\)
=>3B=\(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2011}}\)
=>3B-B=2B=1-\(\dfrac{1}{3^{2012}}\)
=>B=\(\dfrac{1}{2}-\dfrac{1}{2.3^{20112}}\)<1/2
vậy........
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+.....+\frac{1}{3^{2011}}+\frac{1}{3^{2012}}\)
\(\Rightarrow3Á=1+\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^{2010}}+\frac{1}{3^{2011}}\)
\(\Rightarrow3A-A=2A=1-\frac{1}{3^{2012}}\)
\(\Rightarrow A=\frac{1-\frac{1}{3^{2012}}}{2}< \frac{1}{2}\)
Vậy \(A< \frac{1}{2}\)