Question

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

Answer 1

ta có : \(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2}\)

\(\dfrac{1}{3^2}=\dfrac{1}{3.3}< \dfrac{1}{2.3}\)

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

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

đặt tổng đó là A

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

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

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

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

vậy A lớn hơn 1