A=\(\frac{\frac{1}{2018}+\frac{2}{2017}+\frac{3}{2016}+....+\frac{2017}{2}+\frac{2018}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2019}}\). Chứng minh rằng A là số nguyên

Mong mọi người giúp

\(A=\frac{1}{2017}+\frac{2}{2017^2}+\frac{3}{2017^3}+...+\frac{2017}{2017^{2017}}+\frac{2018}{2017^{2018}}\). Chứng minh rằng : A < \(\frac{2017}{2016^2}\)

Chứng minh rằng

\(\frac{1}{12}< \frac{1}{2^3}+\frac{1}{3^3}+...+\frac{1}{n^3}+\frac{1}{2017^3}< \frac{508}{2018}\)

(với mọi n>1)

Chứng minh A là một số nguyên dương : 

A = \(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}+\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}\right)^2+\)\(\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}\right)^2+...+\left(\frac{1}{2017}\right)^2\)

\(\text{Chứng minh rằng:}2017< \sqrt{\frac{2}{1}}+\sqrt[3]{\frac{3}{2}}+\sqrt[4]{\frac{4}{3}}+...+\sqrt[2018]{\frac{2018}{2017}}< 2018\)

Chứng minh rằng \(\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+...+\sqrt{1+\frac{1}{2017^2}+\frac{1}{2018^2}}< 2018\)

Chứng tỏ rằng: \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2017}}< \frac{1}{2}\)

Chứng minh rằng: 1+ \(\frac{1}{1!}\)+\(\frac{1}{2!}\)+\(\frac{1}{3!}\)+...+\(\frac{1}{2017!}\)<3

cho S =\(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2017^2}\)

chứng minh rằng S lớn hơn 2