Ta có : \(\dfrac{1}{2^2}\)<\(\dfrac{1}{1.2}\); \(\dfrac{1}{3^2}\)<\(\dfrac{1}{2.3}\);.....;\(\dfrac{1}{2016^2}\)<\(\dfrac{1}{2015.2016}\)

⇒ A = \(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^2}\)+...+\(\dfrac{1}{2016^2}\)< \(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+...+\(\dfrac{1}{2015.2016}\)

⇒ A = \(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^2}\)+...+\(\dfrac{1}{2016^2}\) < 1 - \(\dfrac{1}{2016}\)= \(\dfrac{2015}{2016}\) (ĐCPCM)

ta thấy:

\(\dfrac{1}{2^2}=\dfrac{1}{4}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)

\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)

...

\(\dfrac{1}{2015^2}< \dfrac{1}{2014.2015}\)

=> A < \(\dfrac{1}{4}+\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2014.2015}\right)\)

=> A< \(\dfrac{1}{4}+\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2014}-\dfrac{1}{2015}\right)\)

<=> A< \(\dfrac{1}{4}+\left(\dfrac{1}{2}-\dfrac{1}{2015}\right)\) = \(\dfrac{3}{4}-\dfrac{1}{2015}\) < \(\dfrac{3}{4}\).

=> đpcm.

bn kiếm trên mạng đi nó có đấy

Ta có:

1/2^2 < 1/1.2

1/3^2 < 1/2.3

...

1/2015^2 < 1/2014.2015

Suy ra: 1/2^2 + 1/3^2 + 1/4^2+...+1/2015^2 < 1/1.2 +1/2.3+...+1/2014.2015

1/2^2 + 1/3^2 + 1/4^2+...+1/2015^2 < 1-1/2+1/2-1/3+...+1/2014-1/2015

1/2^2 + 1/3^2 + 1/4^2+...+1/2015^2 < 1-1/2015

1/2^2 + 1/3^2 + 1/4^2+...+1/2015^2 < 2014/2015

Mình nghĩ đây là cách làm, bạn thử dựa vào làm xem nhé!

Ta có: \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2015^2}=\frac{1}{2^2}+\left(\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2015^2}\right)\)

Nhận xét: \(\frac{1}{2^2}=\frac{1}{4}\)

\(\frac{1}{3^2}< \frac{1}{2.3}\)

\(\frac{1}{4^2}< \frac{1}{3.4}\)

......

\(\frac{1}{2015^2}< \frac{1}{2014.2015}\)

\(\Rightarrow A< \frac{1}{4}+\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}\right)\)

\(\Rightarrow A< \frac{1}{4}+\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}\right)\)

\(\Rightarrow A< \frac{1}{4}+\left(\frac{1}{2}-\frac{1}{2015}\right)=\frac{1}{4}+\frac{1}{2}-\frac{1}{2015}=\frac{3}{4}-\frac{1}{2015}< \frac{3}{4}\)

Vậy A < 3/4

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

\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}+\frac{1}{2015.2016}\)

\(A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}+\frac{1}{2015}-\frac{1}{2016}\)

\(A< 1-\frac{1}{2016}\)

\(A< \frac{2015}{2016}\left(đpcm\right)\)

\(A=\frac{1}{2.2}+\frac{1}{3.3}+.....+\frac{1}{2016.2016}< \frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{2015.2016}\)

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

\(=1-\frac{1}{2016}\)

\(=\frac{2015}{2016}\)

\(\Rightarrow A< \frac{2015}{2016}\)

Vì \(\frac{1}{2^2}< \frac{1}{1\cdot2};\frac{1}{3^2}< \frac{1}{2\cdot3};...;\frac{1}{2015^2}< \frac{1}{2014\cdot2015}\)

\(\Rightarrow A< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{2014\cdot2015}\)

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

\(=1-\frac{1}{2015}< 1\)

Vậy \(A< 1\left(đpcm\right)\)

Ta có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{2015^2}< \frac{1}{2014.2015}\)

\(\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}\)

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

\(=1-\frac{1}{2015}< 1^{\left(đpcm\right)}\)