Ta có :

\(B=\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+...+\frac{1}{2016}\)

\(B=\left(\frac{2015}{2}+1\right)+\left(\frac{2014}{3}+1\right)+...+\left(\frac{1}{2016}+1\right)+1\)

\(B=\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2016}+\frac{2017}{2017}\)

\(B=2017.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}\right)\)

\(\Rightarrow\frac{B}{A}=\frac{2017.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}=2017\)

Vậy \(\frac{B}{A}\)là số nguyên

1/2²+1/3²+1/4²+......+1/2015²+1/2016²  <  1/1.2+1/2.3+1/3.4+.....+1/2014.2015+1/2015.2016

Mà:  1/1.2+1/2.3+1/3.4+.....+1/2014.2015+1/2015.2016

=1-1/2+1/2-1/3+1/3-1/4+.......+1/2014-1/2015+1/2015-1/2016

=1-1/2016

=2016/2016-1/2016

=2015/2016 <1

Nên 1/2²+1/3²+1/4²+......+1/2015²+1/2016²  <  1

Với mọi số nguyên dương n ta có:

\(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{n\left(n+1\right)}=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n+1}\right)=\sqrt{n}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)\)

Ta có: \(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}<\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n}}=\frac{2}{\sqrt{n}}\)

\(\Rightarrow\frac{1}{\left(n+1\right)\sqrt{n}}<\sqrt{n}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\frac{2}{\sqrt{n}}=\frac{2}{\sqrt{n}}-\frac{2}{\sqrt{n+1}}\). Do đó ta có:

\(A<\frac{2}{\sqrt{1}}-\frac{2}{\sqrt{2}}+\frac{2}{\sqrt{2}}-\frac{2}{\sqrt{3}}+\frac{2}{\sqrt{3}}-\frac{2}{\sqrt{4}}+...+\frac{2}{\sqrt{2015}}-\frac{2}{\sqrt{2016}}=2-\frac{2}{\sqrt{2016}}<2\)

Vậy A < 2.

bằng 15 hay sao ý