Question

[min]

1/2² + 1/3² + 1/4²+..........+ 1/2015² + 1/2016² bé hơn 1

Answer 1

[Akai Haruma]

Lời giải:

$\frac{1}{2^2}< \frac{1}{1.2}$

$\frac{1}{3^2}< \frac{1}{2.3}$

....

$\frac{1}{2016^2}< \frac{1}{2015.2016}$

Do đó:

$\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2016^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2015.2016}$

Đặt vế phải là $A$

$A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{2015}-\frac{1}{2016}$

$=1-\frac{1}{2016}< 1$

Suy ra $\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2016^2}< A < 1$

Answer 2

[Nguyễn Huy Tú]

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

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

Vậy ta có đpcm