Question

CMR

\(\dfrac{1}{1^2+2^2}+\dfrac{1}{2^2+3^2}+\dfrac{1}{3^2+4^2}+...+\dfrac{1}{2016^2+2017^2}< \dfrac{1}{2}\)

Answer 1

(a - b)² \(\ge0\Leftrightarrow a^2+b^2-2ab\ge0\Leftrightarrow a^2+b^2\ge2ab\)

=> \(\frac{1}{a^2+b^2}< \frac{1}{2ab}\left(a;b>0;a\ne b\right)\)

Áp dụng vào bài toán ta có:

\(\frac{1}{1^2+2^2}+\frac{1}{2^2+3^2}+\frac{1}{3^2+4^2}+...+\frac{1}{2016^2+2017^2}< \frac{1}{2}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}\right)\)

\(< \frac{1}{2}\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\right)\)

\(< \frac{1}{2}\left(1-\frac{1}{2017}\right)< \frac{1}{2}\left(đpcm\right)\)