Question

Chứng minh rằng với mọi n E N, n>1 ta có:

\(\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+...+\frac{1}{n^2}<\frac{1}{4}\)

 

THANKS..............

Answer 1

ta có 1/2³<1/1*2*3      1/3³<1/2*3*4      1/4³<1/3*4*5 .... 1/n³<1/(n-1)*n*(n+1)

Vậy=1/2³+1/3³+...+1/n³<1/1*2*3+1/2*3*4+.....1/(n-1)*n*(n+1)

Ta có      1/1*2*3      +        1/2*3*4       +...+      1/(n-1)*n*(n+1)

 =1/2*(1/1*2-1/2*3   +      1/2*3-1/3*4    +...+  1/(n-1)*n-1/n*(n+1)

=1/2*(1/2-     1/6      +       1/6   -1/12+..........+1/(n-1)*n-1/n*(n+1)

=1/2*(1/2-1/n*(n+1))

=1/4-1/2n*(n+1)<1/4

Vì 1/2^3+1/3^3+..+1/n^3<1/4-1/2n*(n+1)<1/4

nên =>1/2^3+1/3^3+...+1/n^3<1/4

Answer 2

\(< \frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{\left(n-1\right).n}\)

\(< 2\cdot\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{\left(n-1\right).n}\right)\)

\(< \frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{4\cdot5}-\frac{1}{5\cdot6}+...+\frac{2}{\left(n-1\right)\cdot n}\)

\(< \frac{1}{2}\cdot\left(\frac{1}{2}-\frac{2}{\left(n-1\right)\cdot n}\right)\)

\(< \frac{1}{4}-\frac{1}{\left(n-1\right)\cdot n}\)

                                          ĐPCM