Question

so sánh A=\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2011}}+\frac{1}{3^{2012}}\) với \(\frac{1}{2}\)

Answer 1

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 \(A< \frac{1}{2}\)