Đặt \(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{n}{3^n}\)

\(\Rightarrow3A=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{n}{3^{n-1}}\)

\(\Rightarrow3A-A=\left(1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{n}{3^{n-1}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{n}{3^n}\right)\)

\(\Rightarrow2A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{n-1}}-\frac{n}{3^n}\)

Đặt \(S=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{n-1}}\)

\(\Rightarrow3S=3+1+\frac{1}{3}+...+\frac{1}{3^{n-2}}\)

\(\Rightarrow3S-S=\left(3+1+...+\frac{1}{3^{n-2}}\right)-\left(1+\frac{1}{3}+...+\frac{1}{3^{n-1}}\right)\)

\(\Rightarrow2S=3-\frac{1}{3^{n-1}}< 3\)

\(\Rightarrow2S< 3\)

\(\Rightarrow S< \frac{3}{2}\)

\(\Rightarrow2A< \frac{3}{2}\)

\(\Rightarrow A< \frac{3}{4}\left(đpcm\right)\)

Thanks bn :))

Đặt biểu thức trên là A

\(A^2=m+2\sqrt{m-1}+m-2\sqrt{m-1}-2\sqrt{\left(m+2\sqrt{m-1}\right)\left(m-2\sqrt{m-1}\right)}\)

\(A^2=2m-2\sqrt{m^2-4\left(m-1\right)}=2m-2\sqrt{m^2-4m+4}\)

\(A^2=2m-2\sqrt{\left(m-2\right)^2}=2m-2\left(m-2\right)=2m-2m+4=4\)

\(\Rightarrow A=\pm2\)

\(M=\left(\frac{1}{x^2+y^2}+\frac{1}{2xy}\right)+\frac{1}{2xy}\ge\frac{4}{x^2+y^2+2xy}+\frac{2}{\left(x+y\right)^2}\)

\(=\frac{6}{\left(x+y\right)^2}=6\)

Đẳng thức xảy ra khi \(x=y=\frac{1}{2}\)