Question

Chứng minh rằng

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

Answer 1

\(Vì\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};.....;\dfrac{1}{n^2}< \dfrac{1}{(n-1).n}\)

\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+....+\dfrac{1}{n^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+....+\dfrac{1}{\left(n-1\right).1}< 1\)\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+....+\dfrac{1}{n^2}< 1\left(đpcm\right)\)

vậy \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+....+\dfrac{1}{n^2}< 1\)

Answer 2

ĐặtA= \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+....+\dfrac{1}{n^2}\)

Do \(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)

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

.............

\(\dfrac{1}{n^2}< \dfrac{1}{\left(n-1\right)n}\)

Cộng vế với vế ta suy ra : A<\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.......+\dfrac{1}{\left(n-1\right)n}=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-....-\dfrac{1}{\left(n-1\right)}+\dfrac{1}{n-1}-\dfrac{1}{n}\)

=\(1-\dfrac{1}{n}\)

Mà 1-\(\dfrac{1}{n}\)<1

=> A<1 (đpcm)

Answer 3

Ta có: \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};\dfrac{1}{4^2}< \dfrac{1}{3.4};...;\dfrac{1}{n^2}< \dfrac{1}{\left(n-1\right)n}\)

=>\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right)n}\)

mà \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right)n}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)

\(=1-\dfrac{1}{n}< 1\)

=>\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}< 1\) (đpcm)

Answer 4

Đặt A=\(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^2}\)+\(\dfrac{1}{4^2}\)+....+ \(\dfrac{1}{n^2}\)

Ta có: \(\dfrac{1}{2^2}\)<\(\dfrac{1}{1.2}\)=1-\(\dfrac{1}{2}\)

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

...............

\(\dfrac{1}{n^2}\)<\(\dfrac{1}{\left(n-1\right).n}\)=\(\dfrac{1}{n-1}\)-\(\dfrac{1}{n}\)

->\(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^2}\)+\(\dfrac{1}{4^2}\)+...+\(\dfrac{1}{n^2}\)<1-\(\dfrac{1}{2}\)+\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)+.....+\(\dfrac{1}{n-1}\)-\(\dfrac{1}{n}\)

=> A<1-\(\dfrac{1}{n}\)<1

vậy A<1 ( điều phải chứng minh)