Question

[hoàng phạm]

Chứng minh rằng: 1 + 1/1+2 + 1/1+2+3 + 1/1+2+3+4 +...+ 1/1+2+...+n <2 với mọi n thuộc N, n>1

Answer 1

[Nobi Nobita]

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

Ta có: \(1+2=\frac{2.3}{2}\); \(1+2+3=\frac{3.4}{2}\);...........; \(1+2+3+......+n=\frac{n\left(n+1\right)}{2}\)

\(\Rightarrow A=1+\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+.....+\frac{1}{\frac{n\left(n+1\right)}{2}}\)

\(=1+\frac{2}{2.3}+\frac{2}{3.4}+......+\frac{2}{n\left(n+1\right)}\)

\(=1+2\left[\frac{1}{2.3}+\frac{1}{3.4}+.......+\frac{1}{n\left(n+1\right)}\right]\)

\(=1+2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+........+\frac{1}{n}-\frac{1}{n+1}\right)\)

\(=1+2\left(\frac{1}{2}-\frac{1}{n+1}\right)=1+1-\frac{2}{n+1}=2-\frac{2}{n+1}< 2\)( đpcm )