Question

Chứng minh rằng :

1/1.3 + 1/3.5 + 1/5.7 + ... + 1/(2n-1)(2n+1) < 1/2

Answer 1

\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+....+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+....+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\right)\)

=\(\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+....+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\)

=\(\dfrac{1}{2}\left(1-\dfrac{1}{2n+1}\right)\)

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

Answer 2

đặt A=\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+....+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\)

=> 2A=\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+......+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\)

<=> 2A=\(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{7}+.....+\dfrac{1}{2n-2}-\dfrac{1}{2n+1}\)

<=>2A=\(1-\dfrac{1}{2n+1}\)

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

<=> A=\(\dfrac{1}{2}-\dfrac{1}{2\left(2n+1\right)}\)

=>\(A< \dfrac{1}{2}\) (đpcm)