Question

Câu hỏi : Chứng minh : 1/3 + 2/3^2 + 3/3^3 + 4/3^4 +...+ 2009/3^2009 < 3/4

Answer 1

\(P=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{2008}{3^{2008}}+\frac{2009}{3^{2009}}\)

\(\Rightarrow3P=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{2009}{3^{2008}}\)

\(\Rightarrow2P=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2008}}-\frac{2009}{3^{2009}}=A-\frac{2009}{3^{2009}}\)

\(A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}\)

\(\Rightarrow3A=3+1+\frac{1}{3}+...+\frac{1}{3^{2007}}\)

\(\Rightarrow2A=3-\frac{1}{3^{2008}}< 3\Rightarrow A< \frac{3}{2}\)

\(\Rightarrow2P=A-\frac{2009}{2^{2009}}< A< \frac{3}{2}\Rightarrow P< \frac{3}{4}\)